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1 658 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics from the cold reservoir in exchnge for the lest mount of work. Therefore, for these devices operting in the cooling mode, we define the COP in terms of Q c : energy trnsferred t low temperture COP 1cooling mode2 5 work done on het pump 5 0 Q c 0 W (22.3) Exmple 22.2 Freezing Wter A good refrigertor should hve high COP, typiclly 5 or 6. In ddition to cooling pplictions, het pumps re ecoming incresingly populr for heting purposes. The energy-soring coils for het pump re locted outside uilding, in contct with the ir or uried in the ground. The other set of coils re in the uilding s interior. The circulting fluid flowing through the coils sors energy from the outside nd releses it to the interior of the uilding from the interior coils. In the heting mode, the COP of het pump is defined s the rtio of the energy trnsferred to the hot reservoir to the work required to trnsfer tht energy: energy trnsferred t high temperture COP 1heting mode Q h 0 work done on het pump W (22.4) If the outside temperture is 25 F (24 C) or higher, typicl vlue of the COP for het pump is out 4. Tht is, the mount of energy trnsferred to the uilding is out four times greter thn the work done y the motor in the het pump. As the outside temperture decreses, however, it ecomes more difficult for the het pump to extrct sufficient energy from the ir nd so the COP decreses. Therefore, the use of het pumps tht extrct energy from the ir, lthough stisfctory in moderte climtes, is not pproprite in res where winter tempertures re very low. It is possile to use het pumps in colder res y urying the externl coils deep in the ground. In tht cse, the energy is extrcted from the ground, which tends to e wrmer thn the ir in the winter. Q uick Quiz 22.2 The energy entering n electric heter y electricl trnsmission cn e converted to internl energy with n efficiency of 100%. By wht fctor does the cost of heting your home chnge when you replce your electric heting system with n electric het pump tht hs COP of 4.00? Assume the motor running the het pump is 100% efficient. () 4.00 () 2.00 (c) (d) A certin refrigertor hs COP of When the refrigertor is running, its power input is 500 W. A smple of wter of mss 500 g nd temperture 20.0 C is plced in the freezer comprtment. How long does it tke to freeze the wter to ice t 0 C? Assume ll other prts of the refrigertor sty t the sme temperture nd there is no lekge of energy from the exterior, so the opertion of the refrigertor results only in energy eing extrcted from the wter. olution Conceptulize Energy leves the wter, reducing its temperture nd then freezing it into ice. The time intervl required for this entire process is relted to the rte t which energy is withdrwn from the wter, which, in turn, is relted to the power input of the refrigertor. Ctegorize We ctegorize this exmple s one tht comines our understnding of temperture chnges nd phse chnges from Chpter 20 nd our understnding of het pumps from this chpter. Anlyze Use the power rting of the refrigertor to find the time intervl Dt required for the freezing process to occur: P 5 W Dt Dt 5 W P

2 22.3 Reversile nd Irreversile Processes continued Use Eqution 22.3 to relte the work W done on the het pump to the energy Q c extrcted from the wter: Use Equtions 20.4 nd 20.7 to sustitute the mount of energy Q c tht must e extrcted from the wter of mss m: Recognize tht the mount of wter tht freezes is Dm 5 2m ecuse ll the wter freezes: Dt 5 0Q c 0 P 1COP2 Dt 5 Dt 5 0 mc DT 1 L f Dm 0 P 1COP2 0 m 1c DT 2 L f 2 0 P 1COP2 ustitute numericl vlues: Dt kg J/kg # 8C C J/kg W s Finlize In relity, the time intervl for the wter to freeze in refrigertor is much longer thn 83.3 s, which suggests tht the ssumptions of our model re not vlid. Only smll prt of the energy extrcted from the refrigertor interior in given time intervl comes from the wter. Energy must lso e extrcted from the continer in which the wter is plced, nd energy tht continuously leks into the interior from the exterior must e extrcted Reversile nd Irreversile Processes In the next section, we will discuss theoreticl het engine tht is the most efficient possile. To understnd its nture, we must first exmine the mening of reversile nd irreversile processes. In reversile process, the system undergoing the process cn e returned to its initil conditions long the sme pth on PV digrm, nd every point long this pth is n equilirium stte. A process tht does not stisfy these requirements is irreversile. All nturl processes re known to e irreversile. Let s exmine the ditic free expnsion of gs, which ws lredy discussed in ection 20.6, nd show tht it cnnot e reversile. Consider gs in thermlly insulted continer s shown in Figure A memrne seprtes the gs from vcuum. When the memrne is punctured, the gs expnds freely into the vcuum. As result of the puncture, the system hs chnged ecuse it occupies greter volume fter the expnsion. Becuse the gs does not exert force through displcement, it does no work on the surroundings s it expnds. In ddition, no energy is trnsferred to or from the gs y het ecuse the continer is insulted from its surroundings. Therefore, in this ditic process, the system hs chnged ut the surroundings hve not. For this process to e reversile, we must return the gs to its originl volume nd temperture without chnging the surroundings. Imgine trying to reverse the process y compressing the gs to its originl volume. To do so, we fit the continer with piston nd use n engine to force the piston inwrd. During this process, the surroundings chnge ecuse work is eing done y n outside gent on the system. In ddition, the system chnges ecuse the compression increses the temperture of the gs. The temperture of the gs cn e lowered y llowing it to come into contct with n externl energy reservoir. Although this step returns the gs to its originl conditions, the surroundings re gin ffected ecuse energy is eing dded to the surroundings from the gs. If this Pitfll Prevention 22.2 All Rel Processes Are Irreversile The reversile process is n ideliztion; ll rel processes on the Erth re irreversile. Vcuum Gs t T i Insulting wll Memrne Figure 22.7 Aditic free expnsion of gs.

3 660 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics The gs is compressed slowly s individul grins of snd drop onto the piston. Energy reservoir Figure 22.8 A method for compressing gs in n lmost reversile isotherml process. Pitfll Prevention 22.3 Don t hop for Crnot Engine The Crnot engine is n ideliztion; do not expect Crnot engine to e developed for commercil use. We explore the Crnot engine only for theoreticl considertions. energy could e used to drive the engine tht compressed the gs, the net energy trnsfer to the surroundings would e zero. In this wy, the system nd its surroundings could e returned to their initil conditions nd we could identify the process s reversile. The Kelvin Plnck sttement of the second lw, however, specifies tht the energy removed from the gs to return the temperture to its originl vlue cnnot e completely converted to mechnicl energy y the process of work done y the engine in compressing the gs. Therefore, we must conclude tht the process is irreversile. We could lso rgue tht the ditic free expnsion is irreversile y relying on the portion of the definition of reversile process tht refers to equilirium sttes. For exmple, during the sudden expnsion, significnt vritions in pressure occur throughout the gs. Therefore, there is no well-defined vlue of the pressure for the entire system t ny time etween the initil nd finl sttes. In fct, the process cnnot even e represented s pth on PV digrm. The PV digrm for n ditic free expnsion would show the initil nd finl conditions s points, ut these points would not e connected y pth. Therefore, ecuse the intermedite conditions etween the initil nd finl sttes re not equilirium sttes, the process is irreversile. Although ll rel processes re irreversile, some re lmost reversile. If rel process occurs very slowly such tht the system is lwys very nerly in n equilirium stte, the process cn e pproximted s eing reversile. uppose gs is compressed isothermlly in piston cylinder rrngement in which the gs is in therml contct with n energy reservoir nd we continuously trnsfer just enough energy from the gs to the reservoir to keep the temperture constnt. For exmple, imgine tht the gs is compressed very slowly y dropping grins of snd onto frictionless piston s shown in Figure As ech grin lnds on the piston nd compresses the gs smll mount, the system devites from n equilirium stte, ut it is so close to one tht it chieves new equilirium stte in reltively short time intervl. Ech grin dded represents chnge to new equilirium stte, ut the differences etween sttes re so smll tht the entire process cn e pproximted s occurring through continuous equilirium sttes. The process cn e reversed y slowly removing grins from the piston. A generl chrcteristic of reversile process is tht no nonconservtive effects (such s turulence or friction) tht trnsform mechnicl energy to internl energy cn e present. uch effects cn e impossile to eliminte completely. Hence, it is not surprising tht rel processes in nture re irreversile The Crnot Engine In 1824, French engineer nmed di Crnot descried theoreticl engine, now clled Crnot engine, tht is of gret importnce from oth prcticl nd theoreticl viewpoints. He showed tht het engine operting in n idel, reversile cycle clled Crnot cycle etween two energy reservoirs is the most efficient engine possile. uch n idel engine estlishes n upper limit on the efficiencies of ll other engines. Tht is, the net work done y working sustnce tken through the Crnot cycle is the gretest mount of work possile for given mount of energy supplied to the sustnce t the higher temperture. Crnot s theorem cn e stted s follows: No rel het engine operting etween two energy reservoirs cn e more efficient thn Crnot engine operting etween the sme two reservoirs. In this section, we will show tht the efficiency of Crnot engine depends only on the tempertures of the reservoirs. In turn, tht efficiency represents the

4 22.4 The Crnot Engine 661 Hot reservoir t T h Q h Q hc Het engine Q c W W C Crnot het pump Cold reservoir t T c Q c C mximum possile efficiency for rel engines. Let us confirm tht the Crnot engine is the most efficient. We imgine hypotheticl engine with n efficiency greter thn tht of the Crnot engine. Consider Figure 22.9, which shows the hypotheticl engine with e 7 e C on the left connected etween hot nd cold reservoirs. In ddition, let us ttch Crnot engine etween the sme reservoirs. Becuse the Crnot cycle is reversile, the Crnot engine cn e run in reverse s Crnot het pump s shown on the right in Figure We mtch the output work of the engine to the input work of the het pump, W 5 W C, so there is no exchnge of energy y work etween the surroundings nd the engine het pump comintion. Becuse of the proposed reltion etween the efficiencies, we must hve 0 W 0 e. e C. 0 W C 0 0 Q h 0 0 Q hc 0 The numertors of these two frctions cncel ecuse the works hve een mtched. This expression requires tht From Eqution 22.1, the equlity of the works gives us Figure 22.9 A Crnot engine operted s het pump nd nother engine with proposed higher efficiency operte etween two energy reservoirs. The work output nd input re mtched. 0 Q hc 0. 0 Q h 0 (22.5) 0 W W C 0 0 Q h Q c Q hc Q c C 0 which cn e rewritten to put the energies exchnged with the cold reservoir on the left nd those with the hot reservoir on the right: 0 Q hc Q h Q c C Q c 0 (22.6) Note tht the left side of Eqution 22.6 is positive, so the right side must e positive lso. We see tht the net energy exchnge with the hot reservoir is equl to the net energy exchnge with the cold reservoir. As result, for the comintion of the het engine nd the het pump, energy is trnsferring from the cold reservoir to the hot reservoir y het with no input of energy y work. This result is in violtion of the Clusius sttement of the second lw. Therefore, our originl ssumption tht e 7 e C must e incorrect, nd we must conclude tht the Crnot engine represents the highest possile efficiency for n engine. The key feture of the Crnot engine tht mkes it the most efficient is its reversiility; it cn e run in reverse s het pump. All rel engines re less efficient thn the Crnot engine ecuse they do not operte through reversile cycle. The efficiency of rel engine is further reduced y such prcticl difficulties s friction nd energy losses y conduction. To descrie the Crnot cycle tking plce etween tempertures T c nd T h, let s ssume the working sustnce is n idel gs contined in cylinder fitted with movle piston t one end. The cylinder s wlls nd the piston re thermlly nonconducting. Four stges of the Crnot cycle re shown in Figure 22.10(pge 662), INTERFOTO/Almy di Crnot French engineer ( ) Crnot ws the first to show the quntittive reltionship etween work nd het. In 1824, he pulished his only work, Reflections on the Motive Power of Het, which reviewed the industril, politicl, nd economic importnce of the stem engine. In it, he defined work s weight lifted through height.

5 662 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics Figure The Crnot cycle. The letters A, B, C, nd D refer to the sttes of the gs shown in Figure The rrows on the piston indicte the direction of its motion during ech process. A B The gs undergoes n isotherml expnsion. Q h P A The work done during the cycle equls the re enclosed y the pth on the PV digrm. Q h B W eng Figure PV digrm for the Crnot cycle. The net work done W eng equls the net energy trnsferred into the Crnot engine in one cycle, Q h 2 Q c. D Q c T h C T c V Therml insultion d D A The gs undergoes n ditic compression. Energy reservoir t T h Q 0 Cycle Q 0 C D The gs undergoes n isotherml compression. Q c Energy reservoir t T c c B C The gs undergoes n ditic expnsion. Therml insultion nd the PV digrm for the cycle is shown in Figure The Crnot cycle consists of two ditic processes nd two isotherml processes, ll reversile: 1. Process A B (Fig ) is n isotherml expnsion t temperture T h. The gs is plced in therml contct with n energy reservoir t temperture T h. During the expnsion, the gs sors energy Q h from the reservoir through the se of the cylinder nd does work W AB in rising the piston. 2. In process B C (Fig ), the se of the cylinder is replced y thermlly nonconducting wll nd the gs expnds diticlly; tht is, no energy enters or leves the system y het. During the expnsion, the temperture of the gs decreses from T h to T c nd the gs does work W BC in rising the piston. 3. In process C D (Fig c), the gs is plced in therml contct with n energy reservoir t temperture T c nd is compressed isothermlly t temperture T c. During this time, the gs expels energy Q c to the reservoir nd the work done y the piston on the gs is W CD. 4. In the finl process D A (Fig d), the se of the cylinder is replced y nonconducting wll nd the gs is compressed diticlly. The temperture of the gs increses to T h, nd the work done y the piston on the gs is W DA.

6 22.4 The Crnot Engine 663 The therml efficiency of the engine is given y Eqution 22.2: e Q c 0 0 Q h 0 In Exmple 22.3, we show tht for Crnot cycle, 0 Q c 0 0 Q h 0 Hence, the therml efficiency of Crnot engine is 5 T c T h (22.7) e C T c (22.8) T h This result indictes tht ll Crnot engines operting etween the sme two tempertures hve the sme efficiency. 5 Eqution 22.8 cn e pplied to ny working sustnce operting in Crnot cycle etween two energy reservoirs. According to this eqution, the efficiency is zero if T c 5 T h, s one would expect. The efficiency increses s T c is lowered nd T h is rised. The efficiency cn e unity (100%), however, only if T c 5 0 K. uch reservoirs re not ville; therefore, the mximum efficiency is lwys less thn 100%. In most prcticl cses, T c is ner room temperture, which is out 300 K. Therefore, one usully strives to increse the efficiency y rising T h. Theoreticlly, Crnot-cycle het engine run in reverse constitutes the most effective het pump possile, nd it determines the mximum COP for given comintion of hot nd cold reservoir tempertures. Using Equtions 22.1 nd 22.4, we see tht the mximum COP for het pump in its heting mode is COP C 1heting mode2 5 0 Q h 0 W 0 Q h Q h Q c Q c T 5 T h c T h 2 T c 0 Q h 0 The Crnot COP for het pump in the cooling mode is COP C 1cooling mode2 5 T h 2 T c As the difference etween the tempertures of the two reservoirs pproches zero in this expression, the theoreticl COP pproches infinity. In prctice, the low temperture of the cooling coils nd the high temperture t the compressor limit the COP to vlues elow 10. Q uick Quiz 22.3 Three engines operte etween reservoirs seprted in temperture y 300 K. The reservoir tempertures re s follows: Engine A: T h K, T c K; Engine B: T h K, T c K; Engine C: T h K, T c K. Rnk the engines in order of theoreticlly possile efficiency from highest to lowest. T c T h WWEfficiency of Crnot engine 5 For the processes in the Crnot cycle to e reversile, they must e crried out infinitesimlly slowly. Therefore, lthough the Crnot engine is the most efficient engine possile, it hs zero power output ecuse it tkes n infinite time intervl to complete one cycle! For rel engine, the short time intervl for ech cycle results in the working sustnce reching high temperture lower thn tht of the hot reservoir nd low temperture higher thn tht of the cold reservoir. An engine undergoing Crnot cycle etween this nrrower temperture rnge ws nlyzed y F. L. Curzon nd B. Ahlorn ( Efficiency of Crnot engine t mximum power output, Am. J. Phys. 43(1), 22, 1975), who found tht the efficiency t mximum power output depends only on the reservoir tempertures T c nd T h nd is given y e C-A (T c /T h ) 1/2. The Curzon Ahlorn efficiency e C-A provides closer pproximtion to the efficiencies of rel engines thn does the Crnot efficiency.

7 664 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics Exmple 22.3 Efficiency of the Crnot Engine how tht the rtio of energy trnsfers y het in Crnot engine is equl to the rtio of reservoir tempertures, s given y Eqution olution Conceptulize Mke use of Figures nd to help you visulize the processes in the Crnot cycle. Ctegorize Becuse of our understnding of the Crnot cycle, we cn ctegorize the processes in the cycle s isotherml nd ditic. Anlyze For the isotherml expnsion (process A B in Fig ), find the energy trnsfer y het from the hot reservoir using Eqution nd the first lw of thermodynmics: In similr mnner, find the energy trnsfer to the cold reservoir during the isotherml compression C D : Divide the second expression y the first: (1) Apply Eqution to the ditic processes B C nd D A: Divide the first eqution y the second: ustitute Eqution (2) into Eqution (1): Finlize This lst eqution is Eqution 22.7, the one we set out to prove. Exmple 22.4 The tem Engine 0 Q h DE int 2 W AB W AB 0 5 nrt h ln V B 0 Q c DE int 2 W CD W CD 0 5 nrt c ln V C 0 Q c 0 5 T c ln 1V C /V D 2 0 Q h 0 T h ln 1V B /V A 2 T h V g21 B 5 T c V g21 C T h V g21 A 5 T c V g21 D V B V A (2) g21 V B 5 V C V D V A 5 V C V D g21 0 Q c 0 5 T c ln 1V C /V D 2 0 Q h 0 T h ln 1V B /V A 2 5 T c ln 1V C /V D 2 T h ln 1V C /V D 2 5 T c A stem engine hs oiler tht opertes t 500 K. The energy from the urning fuel chnges wter to stem, nd this stem then drives piston. The cold reservoir s temperture is tht of the outside ir, pproximtely 300 K. Wht is the mximum therml efficiency of this stem engine? T h V A V D olution Conceptulize In stem engine, the gs pushing on the piston in Figure is stem. A rel stem engine does not operte in Crnot cycle, ut, to find the mximum possile efficiency, imgine Crnot stem engine. Ctegorize We clculte n efficiency using Eqution 22.8, so we ctegorize this exmple s sustitution prolem. ustitute the reservoir tempertures into Eqution 22.8: e C T c K or 40.0% T h 500 K This result is the highest theoreticl efficiency of the engine. In prctice, the efficiency is considerly lower.

8 22.5 Gsoline nd Diesel Engines continued Wht If? uppose we wished to increse the theoreticl efficiency of this engine. This increse cn e chieved y rising T h y DT or y decresing T c y the sme DT. Which would e more effective? Answer A given DT would hve lrger frctionl effect on smller temperture, so you would expect lrger chnge in efficiency if you lter T c y DT. Let s test tht numericlly. Rising T h y 50 K, corresponding to T h K, would give mximum efficiency of 22.5 Gsoline nd Diesel Engines e C T c T h K 550 K Decresing T c y 50 K, corresponding to T c K, would give mximum efficiency of e C T c K T h 500 K Although chnging T c is mthemticlly more effective, often chnging T h is prcticlly more fesile. In gsoline engine, six processes occur in ech cycle; they re illustrted in Figure In this discussion, let s consider the interior of the cylinder ove the piston to e the system tht is tken through repeted cycles in the engine s opertion. For given cycle, the piston moves up nd down twice, which represents four-stroke cycle consisting of two upstrokes nd two downstrokes. The processes in the cycle cn e pproximted y the Otto cycle shown in the PV digrm in Figure (pge 666). In the following discussion, refer to Figure for the pictoril representtion of the strokes nd Figure for the significnce on the PV digrm of the letter designtions elow: Air nd fuel 1. During the intke stroke (Fig nd O A in Figure 22.13), the piston moves downwrd nd gseous mixture of ir nd fuel is drwn into the The intke vlve opens, nd the ir fuel mixture enters s the piston moves down. prk plug Piston The piston moves up nd compresses the mixture. The sprk plug fires nd ignites the mixture. The hot gs pushes the piston downwrd. The exhust vlve opens, nd the residul gs escpes. The piston moves up nd pushes the remining gs out. Exhust Intke Compression prk Power Relese Exhust c d e f Figure The four-stroke cycle of conventionl gsoline engine. The rrows on the piston indicte the direction of its motion during ech process.

9 666 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics P T A Q h B T C O C Aditic processes V 2 V 1 Figure PV digrm for the Otto cycle, which pproximtely represents the processes occurring in n internl comustion engine. D A Q c V cylinder t tmospheric pressure. Tht is the energy input prt of the cycle: energy enters the system (the interior of the cylinder) y mtter trnsfer s potentil energy stored in the fuel. In this process, the volume increses from V 2 to V 1. This pprent ckwrd numering is sed on the compression stroke (process 2 elow), in which the ir fuel mixture is compressed from V 1 to V During the compression stroke (Fig nd A B in Fig ), the piston moves upwrd, the ir fuel mixture is compressed diticlly from volume V 1 to volume V 2, nd the temperture increses from T A to T B. The work done on the gs is positive, nd its vlue is equl to the negtive of the re under the curve AB in Figure Comustion occurs when the sprk plug fires (Fig c nd B C in Fig ). Tht is not one of the strokes of the cycle ecuse it occurs in very short time intervl while the piston is t its highest position. The comustion represents rpid energy trnsformtion from potentil energy stored in chemicl onds in the fuel to internl energy ssocited with moleculr motion, which is relted to temperture. During this time intervl, the mixture s pressure nd temperture increse rpidly, with the temperture rising from T B to T C. The volume, however, remins pproximtely constnt ecuse of the short time intervl. As result, pproximtely no work is done on or y the gs. We cn model this process in the PV digrm (Fig ) s tht process in which the energy Q h enters the system. (In relity, however, this process is trnsformtion of energy lredy in the cylinder from process O A.) 4. In the power stroke (Fig d nd C D in Fig ), the gs expnds diticlly from V 2 to V 1. This expnsion cuses the temperture to drop from T C to T D. Work is done y the gs in pushing the piston downwrd, nd the vlue of this work is equl to the re under the curve CD. 5. Relese of the residul gses occurs when n exhust vlve is opened (Fig e nd D A in Fig ). The pressure suddenly drops for short time intervl. During this time intervl, the piston is lmost sttionry nd the volume is pproximtely constnt. Energy is expelled from the interior of the cylinder nd continues to e expelled during the next process. 6. In the finl process, the exhust stroke (Fig e nd A O in Fig ), the piston moves upwrd while the exhust vlve remins open. Residul gses re exhusted t tmospheric pressure, nd the volume decreses from V 1 to V 2. The cycle then repets. If the ir fuel mixture is ssumed to e n idel gs, the efficiency of the Otto cycle is 1 e Otto cycle2 (22.9) g21 1V 1 /V 2 2 where V 1 /V 2 is the compression rtio nd g is the rtio of the molr specific hets C P /C V for the ir fuel mixture. Eqution 22.9, which is derived in Exmple 22.5, shows tht the efficiency increses s the compression rtio increses. For typicl compression rtio of 8 nd with g 5 1.4, Eqution 22.9 predicts theoreticl efficiency of 56% for n engine operting in the idelized Otto cycle. This vlue is much greter thn tht chieved in rel engines (15% to 20%) ecuse of such effects s friction, energy trnsfer y conduction through the cylinder wlls, nd incomplete comustion of the ir fuel mixture. Diesel engines operte on cycle similr to the Otto cycle, ut they do not employ sprk plug. The compression rtio for diesel engine is much greter thn tht for gsoline engine. Air in the cylinder is compressed to very smll volume, nd, s consequence, the cylinder temperture t the end of the compression stroke is

10 22.6 Entropy 667 very high. At this point, fuel is injected into the cylinder. The temperture is high enough for the ir fuel mixture to ignite without the ssistnce of sprk plug. Diesel engines re more efficient thn gsoline engines ecuse of their greter compression rtios nd resulting higher comustion tempertures. Exmple 22.5 Efficiency of the Otto Cycle how tht the therml efficiency of n engine operting in n idelized Otto cycle (see Figs nd 22.13) is given y Eqution Tret the working sustnce s n idel gs. olution Conceptulize tudy Figures nd to mke sure you understnd the working of the Otto cycle. Ctegorize As seen in Figure 22.13, we ctegorize the processes in the Otto cycle s isovolumetric nd ditic. Anlyze Model the energy input nd output s occurring y het in processes B C nd D A. (In relity, most of the energy enters nd leves y mtter trnsfer s the ir fuel mixture enters nd leves the cylinder.) Use Eqution to find the energy trnsfers y het for these processes, which tke plce t constnt volume: B C Q h 5 nc V (T C 2 T B ) D A Q c 5 nc V (T D 2 T A ) ustitute these expressions into Eqution 22.2: (1) e Q c T D 2 T A 0 Q h 0 T C 2 T B Apply Eqution to the ditic processes A B nd C D: olve these equtions for the tempertures T A nd T D, noting tht V A 5 V D 5 V 1 nd V B 5 V C 5 V 2 : utrct Eqution (2) from Eqution (3) nd rerrnge: A B T A V A g21 5 T B V B g21 C D T C V C g21 5 T D V D g21 (2) T A 5 T B V B V A (3) T D 5 T C V C V D g21 g21 (4) T D 2 T A 5 V 2 T C 2 T B V 1 1 ustitute Eqution (4) into Eqution (1): e V 1 /V 2 2 g21 Finlize This finl expression is Eqution T B V 2 V 1 5 T C V 2 V Entropy The zeroth lw of thermodynmics involves the concept of temperture, nd the first lw involves the concept of internl energy. Temperture nd internl energy re oth stte vriles; tht is, the vlue of ech depends only on the thermodynmic stte of system, not on the process tht rought it to tht stte. Another stte vrile this one relted to the second lw of thermodynmics is entropy. Entropy ws originlly formulted s useful concept in thermodynmics. Its importnce grew, however, s the field of sttisticl mechnics developed ecuse the nlyticl techniques of sttisticl mechnics provide n lterntive mens of interpreting entropy nd more glol significnce to the concept. In sttisticl g21 g21 g21 Pitfll Prevention 22.4 Entropy Is Astrct Entropy is one of the most strct notions in physics, so follow the discussion in this nd the susequent sections very crefully. Do not confuse energy with entropy. Even though the nmes sound similr, they re very different concepts. On the other hnd, energy nd entropy re intimtely relted, s we shll see in this discussion.

11 668 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics mechnics, the ehvior of sustnce is descried in terms of the sttisticl ehvior of its toms nd molecules. We will develop our understnding of entropy y first considering some nonthermodynmic systems, such s pir of dice nd poker hnds. We will then expnd on these ides nd use them to understnd the concept of entropy s pplied to thermodynmic systems. We egin this process y distinguishing etween microsttes nd mcrosttes of system. A microstte is prticulr configurtion of the individul constituents of the system. A mcrostte is description of the system s conditions from mcroscopic point of view. For ny given mcrostte of the system, numer of microsttes re possile. For exmple, the mcrostte of 4 on pir of dice cn e formed from the possile microsttes 1 3, 2 2, nd 3 1. The mcrostte of 2 hs only one microstte, 1 1. It is ssumed ll microsttes re eqully prole. We cn compre these two mcrosttes in three wys: (1) Uncertinty: If we know tht mcrostte of 4 exists, there is some uncertinty s to the microstte tht exists, ecuse there re multiple microsttes tht will result in 4. In comprison, there is lower uncertinty (in fct, zero uncertinty) for mcrostte of 2 ecuse there is only one microstte. (2) Choice: There re more choices of microsttes for 4 thn for 2. (3) Proility: The mcrostte of 4 hs higher proility thn mcrostte of 2 ecuse there re more wys (microsttes) of chieving 4. The notions of uncertinty, choice, nd proility re centrl to the concept of entropy, s we discuss elow. Let s look t nother exmple relted to poker hnd. There is only one microstte ssocited with the mcrostte of royl flush of five spdes, lid out in order from ten to ce (Fig ). Figure shows nother poker hnd. The mcrostte here is worthless hnd. The prticulr hnd (the microstte) in Figure nd the hnd in Figure re eqully prole. There re, however, mny other hnds similr to tht in Figure 22.14; tht is, there re mny microsttes tht lso qulify s worthless hnds. If you, s poker plyer, re told your opponent holds mcrostte of royl flush in spdes, there is zero uncertinty s to wht five crds re in the hnd, only one choice of wht those crds re, nd low proility tht the hnd ctully occurred. In contrst, if you re told tht your opponent hs the mcrostte of worthless hnd, there is high uncertinty s to wht the five crds re, mny choices of wht they could e, nd high proility tht worthless hnd occurred. Another vrile in poker, of course, is the vlue of the hnd, relted to the proility: the higher the proility, the lower the vlue. The importnt point to tke wy from this discussion is tht uncertinty, choice, nd proility re relted in these situtions: if one is high, the others re high, nd vice vers. Another wy of descriing mcrosttes is y mens of missing informtion. For high-proility mcrosttes with mny microsttes, there is lrge mount Figure () A royl flush hs low proility of occurring. () A worthless poker hnd, one of mny. Cengge Lerning/George emple Cengge Lerning/George emple

12 22.6 Entropy 669 of missing informtion, mening we hve very little informtion out wht microstte ctully exists. For mcrostte of 2 on pir of dice, we hve no missing informtion; we know the microstte is 1 1. For mcrostte of worthless poker hnd, however, we hve lots of missing informtion, relted to the lrge numer of choices we could mke s to the ctul hnd tht is held. Q uick Quiz 22.4 () uppose you select four crds t rndom from stndrd deck of plying crds nd end up with mcrostte of four deuces. How mny microsttes re ssocited with this mcrostte? () uppose you pick up two crds nd end up with mcrostte of two ces. How mny microsttes re ssocited with this mcrostte? For thermodynmic systems, the vrile entropy is used to represent the level of uncertinty, choice, proility, or missing informtion in the system. Consider configurtion ( mcrostte) in which ll the oxygen molecules in your room re locted in the west hlf of the room nd the nitrogen molecules in the est hlf. Compre tht mcrostte to the more common configurtion of the ir molecules distriuted uniformly throughout the room. The ltter configurtion hs the higher uncertinty nd more missing informtion s to where the molecules re locted ecuse they could e nywhere, not just in one hlf of the room ccording to the type of molecule. The configurtion with uniform distriution lso represents more choices s to where to locte molecules. It lso hs much higher proility of occurring; hve you ever noticed your hlf of the room suddenly eing empty of oxygen? Therefore, the ltter configurtion represents higher entropy. For systems of dice nd poker hnds, the comprisons etween proilities for vrious mcrosttes involve reltively smll numers. For exmple, mcrostte of 4 on pir of dice is only three times s prole s mcrostte of 2. The rtio of proilities of worthless hnd nd royl flush is significntly lrger. When we re tlking out mcroscopic thermodynmic system contining on the order of Avogdro s numer of molecules, however, the rtios of proilities cn e stronomicl. Let s explore this concept y considering 100 molecules in continer. Hlf of the molecules re oxygen nd the other hlf re nitrogen. At ny given moment, the proility of one molecule eing in the left prt of the continer shown in Figure s result of rndom motion is 1 2. If there re two molecules s shown in Figure 22.15, the proility of oth eing in the left prt is , or 1 in 4. If there re three molecules (Fig c), the proility of them ll eing in the left portion t the sme moment is , or 1 in 8. For 100 independently moving molecules, the proility tht the 50 oxygen molecules will e found in the left prt t ny moment is Likewise, the proility tht the remining 50 nitrogen molecules will e found in the right prt t ny moment is Therefore, the proility of c Pitfll Prevention 22.5 Entropy Is for Thermodynmic ystems We re not pplying the word entropy to descrie systems of dice or crds. We re only discussing dice nd crds to set up the notions of microsttes, mcrosttes, uncertinty, choice, proility, nd missing informtion. Entropy cn only e used to descrie thermodynmic systems tht contin mny prticles, llowing the system to store energy s internl energy. Pitfll Prevention 22.6 Entropy nd Disorder ome textook tretments of entropy relte entropy to the disorder of system. While this pproch hs some merit, it is not entirely successful. For exmple, consider two smples of the sme solid mteril t the sme temperture. One smple hs volume V nd the other volume 2V. The lrger smple hs higher entropy thn the smller one simply ecuse there re more molecules in it. But there is no sense in which it is more disordered thn the smller smple. We will not use the disorder pproch in this text, ut wtch for it in other sources. Figure Possile distriutions of identicl molecules in continer. The colors used here exist only to llow us to distinguish mong the molecules. () One molecule in continer hs 1-in-2 chnce of eing on the left side. () Two molecules hve 1-in-4 chnce of eing on the left side t the sme time. (c) Three molecules hve 1-in-8 chnce of eing on the left side t the sme time.

13 670 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics finding this oxygen nitrogen seprtion s result of rndom motion is the product , which corresponds to out 1 in When this clcultion is extrpolted from 100 molecules to the numer in 1 mol of gs ( ), the seprted rrngement is found to e extremely improle! Conceptul Exmple 22.6 Let s Ply Mrles! uppose you hve g of 100 mrles of which 50 re red nd 50 re green. You re llowed to drw four mrles from the g ccording to the following rules. Drw one mrle, record its color, nd return it to the g. hke the g nd then drw nother mrle. Continue this process until you hve drwn nd returned four mrles. Wht re the possile mcrosttes for this set of events? Wht is the most likely mcrostte? Wht is the lest likely mcrostte? olution Becuse ech mrle is returned to the g efore the next one is drwn nd the g is then shken, the proility of drwing red mrle is lwys the sme s the proility of drwing green one. All the possile microsttes nd mcrosttes re shown in Tle As this tle indictes, there is only one wy Tle 22.1 Possile Results of Drwing Four Mrles from Bg Totl Numer Mcrostte Possile Microsttes of Microsttes All R RRRR 1 1G, 3R RRRG, RRGR, RGRR, GRRR 4 2G, 2R RRGG, RGRG, GRRG, 6 RGGR, GRGR, GGRR 3G, 1R GGGR, GGRG, GRGG, RGGG 4 All G GGGG 1 to drw mcrostte of four red mrles, so there is only one microstte for tht mcrostte. There re, however, four possile microsttes tht correspond to the mcrostte of one green mrle nd three red mrles, six microsttes tht correspond to two green mrles nd two red mrles, four microsttes tht correspond to three green mrles nd one red mrle, nd one microstte tht corresponds to four green mrles. The most likely mcrostte two red mrles nd two green mrles corresponds to the lrgest numer of choices of microsttes, nd, therefore, the most uncertinty s to wht the exct microstte is. The lest likely mcrosttes four red mrles or four green mrles correspond to only one choice of microstte nd, therefore, zero uncertinty. There is no missing informtion for the lest likely sttes: we know the colors of ll four mrles. We hve investigted the notions of uncertinty, numer of choices, proility, nd missing informtion for some non-thermodynmic systems nd hve rgued tht the concept of entropy cn e relted to these notions for thermodynmic systems. We hve not yet indicted how to evlute entropy numericlly for thermodynmic system. This evlution ws done through sttisticl mens y Boltzmnn in the 1870s nd ppers in its currently ccepted form s 5 k B ln W (22.10) where k B is Boltzmnn s constnt. Boltzmnn intended W, stnding for Whrscheinlichkeit, the Germn word for proility, to e proportionl to the proility tht given mcrostte exists. It is equivlent to let W e the numer of microsttes ssocited with the mcrostte, so we cn interpret W s representing the numer of wys of chieving the mcrostte. Therefore, mcrosttes with lrger numers of microsttes hve higher proility nd, equivlently, higher entropy. In the kinetic theory of gses, gs molecules re represented s prticles moving rndomly. uppose the gs is confined to volume V. For uniform distriution of gs in the volume, there re lrge numer of equivlent microsttes, nd the entropy of the gs cn e relted to the numer of microsttes corresponding to given mcrostte. Let us count the numer of microsttes y considering the

14 22.7 Chnges in Entropy for Thermodynmic ystems 671 vriety of moleculr loctions ville to the molecules. Let us ssume ech molecule occupies some microscopic volume V m. The totl numer of possile loctions of single molecule in mcroscopic volume V is the rtio w = V/V m, which is huge numer. We use lowercse w here to represent the numer of wys single molecule cn e plced in the volume or the numer of microsttes for single molecule, which is equivlent to the numer of ville loctions. We ssume the proilities of molecule occupying ny of these loctions re equl. As more molecules re dded to the system, the numer of possile wys the molecules cn e positioned in the volume multiplies, s we sw in Figure For exmple, if you consider two molecules, for every possile plcement of the first, ll possile plcements of the second re ville. Therefore, there re w wys of locting the first molecule, nd for ech wy, there re w wys of locting the second molecule. The totl numer of wys of locting the two molecules is W 5 w 3 w 5 w 2 5 (V/V m ) 2. (Uppercse W represents the numer of wys of putting multiple molecules into the volume nd is not to e confused with work.) Now consider plcing N molecules of gs in the volume V. Neglecting the very smll proility of hving two molecules occupy the sme loction, ech molecule my go into ny of the V/V m loctions, nd so the numer of wys of locting N molecules in the volume ecomes W 5 w N 5 (V/V m ) N. Therefore, the sptil prt of the entropy of the gs, from Eqution 22.10, is 5 k B ln W 5 k B ln V V m N 5 Nk B ln V V m 5 nr ln V V m (22.11) We will use this expression in the next section s we investigte chnges in entropy for processes occurring in thermodynmic systems. Notice tht we hve indicted Eqution s representing only the sptil portion of the entropy of the gs. There is lso temperture-dependent portion of the entropy tht the discussion ove does not ddress. For exmple, imgine n isovolumetric process in which the temperture of the gs increses. Eqution ove shows no chnge in the sptil portion of the entropy for this sitution. There is chnge in entropy, however, ssocited with the increse in temperture. We cn understnd this y ppeling gin to it of quntum physics. Recll from ection 21.3 tht the energies of the gs molecules re quntized. When the temperture of gs chnges, the distriution of energies of the gs molecules chnges ccording to the Boltzmnn distriution lw, discussed in ection Therefore, s the temperture of the gs increses, there is more uncertinty out the prticulr microstte tht exists s gs molecules distriute themselves into higher ville quntum sttes. We will see the entropy chnge ssocited with n isovolumetric process in Exmple Chnges in Entropy for Thermodynmic ystems Thermodynmic systems re constntly in flux, chnging continuously from one microstte to nother. If the system is in equilirium, given mcrostte exists, nd the system fluctutes from one microstte ssocited with tht mcrostte to nother. This chnge is unoservle ecuse we re only le to detect the mcrostte. Equilirium sttes hve tremendously higher proility thn nonequilirium sttes, so it is highly unlikely tht n equilirium stte will spontneously chnge to nonequilirium stte. For exmple, we do not oserve spontneous split into the oxygen nitrogen seprtion discussed in ection Wht if the system egins in low-proility mcrostte, however? Wht if the room egins with n oxygen nitrogen seprtion? In this cse, the system will progress from this low-proility mcrostte to the much-higher proility

15 672 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics Chnge in entropy for n infinitesiml process Chnge in entropy for finite process stte: the gses will disperse nd mix throughout the room. Becuse entropy is relted to proility, spontneous increse in entropy, such s in the ltter sitution, is nturl. If the oxygen nd nitrogen molecules were initilly spred evenly throughout the room, decrese in entropy would occur if the spontneous splitting of molecules occurred. One wy of conceptulizing chnge in entropy is to relte it to energy spreding. A nturl tendency is for energy to undergo sptil spreding in time, representing n increse in entropy. If sketll is dropped onto floor, it ounces severl times nd eventully comes to rest. The initil grvittionl potentil energy in the sketll Erth system hs een trnsformed to internl energy in the ll nd the floor. Tht energy is spreding outwrd y het into the ir nd into regions of the floor frther from the drop point. In ddition, some of the energy hs spred throughout the room y sound. It would e unnturl for energy in the room nd floor to reverse this motion nd concentrte into the sttionry ll so tht it spontneously egins to ounce gin. In the ditic free expnsion of ection 22.3, the spreding of energy ccompnies the spreding of the molecules s the gs rushes into the evcuted hlf of the continer. If wrm oject is plced in therml contct with cool oject, energy trnsfers from the wrm oject to the cool one y het, representing spred of energy until it is distriuted more evenly etween the two ojects. Now consider mthemticl representtion of this spreding of energy or, equivlently, the chnge in entropy. The originl formultion of entropy in thermodynmics involves the trnsfer of energy y het during reversile process. Consider ny infinitesiml process in which system chnges from one equilirium stte to nother. If dq r is the mount of energy trnsferred y het when the system follows reversile pth etween the sttes, the chnge in entropy d is equl to this mount of energy divided y the solute temperture of the system: d 5 dq r T (22.12) We hve ssumed the temperture is constnt ecuse the process is infinitesiml. Becuse entropy is stte vrile, the chnge in entropy during process depends only on the endpoints nd therefore is independent of the ctul pth followed. Consequently, the entropy chnge for n irreversile process cn e determined y clculting the entropy chnge for reversile process tht connects the sme initil nd finl sttes. The suscript r on the quntity dq r is reminder tht the trnsferred energy is to e mesured long reversile pth even though the system my ctully hve followed some irreversile pth. When energy is sored y the system, dq r is positive nd the entropy of the system increses. When energy is expelled y the system, dq r is negtive nd the entropy of the system decreses. Notice tht Eqution does not define entropy ut rther the chnge in entropy. Hence, the meningful quntity in descriing process is the chnge in entropy. To clculte the chnge in entropy for finite process, first recognize tht T is generlly not constnt during the process. Therefore, we must integrte Equ tion 22.12: f f dq r D 5 3 d 5 3 (22.13) i i T As with n infinitesiml process, the chnge in entropy D of system going from one stte to nother hs the sme vlue for ll pths connecting the two sttes. Tht is, the finite chnge in entropy D of system depends only on the properties of the initil nd finl equilirium sttes. Therefore, we re free to choose ny convenient reversile pth over which to evlute the entropy in plce of the ctul pth s long s the initil nd finl sttes re the sme for oth pths. This point is explored further on in this section.

16 22.7 Chnges in Entropy for Thermodynmic ystems 673 From Eqution 22.10, we see tht chnge in entropy is represented in the Boltzmnn formultion s D 5 k B ln W f W i (22.14) where W i nd W f represent the initl nd finl numers of microsttes, respectively, for the initil nd finl configurtions of the system. If W f > W i, the finl stte is more prole thn the the initil stte (there re more choices of microsttes), nd the entropy increses. Q uick Quiz 22.5 An idel gs is tken from n initil temperture T i to higher finl temperture T f long two different reversile pths. Pth A is t constnt pressure, nd pth B is t constnt volume. Wht is the reltion etween the entropy chnges of the gs for these pths? () D A. D B () D A 5 D B (c) D A, D B Q uick Quiz 22.6 True or Flse: The entropy chnge in n ditic process must e zero ecuse Q 5 0. Exmple 22.7 Chnge in Entropy: Melting A solid tht hs ltent het of fusion L f melts t temperture T m. Clculte the chnge in entropy of this sustnce when mss m of the sustnce melts. olution Conceptulize We cn choose ny convenient reversile pth to follow tht connects the initil nd finl sttes. It is not necessry to identify the process or the pth ecuse, whtever it is, the effect is the sme: energy enters the sustnce y het nd the sustnce melts. The mss m of the sustnce tht melts is equl to Dm, the chnge in mss of the higher-phse (liquid) sustnce. Ctegorize Becuse the melting tkes plce t fixed temperture, we ctegorize the process s isotherml. Anlyze Use Eqution 20.7 in Eqution 22.13, noting tht the temperture remins fixed: dq r D 5 3 T dq T r 5 Q r 5 L f Dm m T m Finlize Notice tht Dm is positive so tht D is positive, representing tht energy is dded to the sustnce. Entropy Chnge in Crnot Cycle Let s consider the chnges in entropy tht occur in Crnot het engine tht opertes etween the tempertures T c nd T h. In one cycle, the engine tkes in energy Q h from the hot reservoir nd expels energy Q c to the cold reservoir. These energy trnsfers occur only during the isotherml portions of the Crnot cycle; therefore, the constnt temperture cn e rought out in front of the integrl sign in Eqution The integrl then simply hs the vlue of the totl mount of energy trnsferred y het. Therefore, the totl chnge in entropy for one cycle is D 5 0 Q h 0 T h 2 0 Q c 0 T c (22.15) where the minus sign represents tht energy is leving the engine t temperture T c. In Exmple 22.3, we showed tht for Crnot engine, 0 Q c 0 0 Q h 0 5 T c T h T m 5 L f m T m

17 674 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics When the memrne is ruptured, the gs will expnd freely nd irreversily into the full volume. Vcuum Gs t T i in volume V i Insulting wll Memrne Figure Aditic free expnsion of gs. The continer is thermlly insulted from its surroundings; therefore, Q 5 0. Using this result in Eqution 22.15, we find tht the totl chnge in entropy for Crnot engine operting in cycle is zero: D 5 0 Now consider system tken through n ritrry (non-crnot) reversile cycle. Becuse entropy is stte vrile nd hence depends only on the properties of given equilirium stte we conclude tht D 5 0 for ny reversile cycle. In generl, we cn write this condition s C dq r T 5 0 1reversile cycle2 (22.16) where the symol r indictes tht the integrtion is over closed pth. Entropy Chnge in Free Expnsion Let s gin consider the ditic free expnsion of gs occupying n initil volume V i (Fig ). In this sitution, memrne seprting the gs from n evcuted region is roken nd the gs expnds to volume V f. This process is irreversile; the gs would not spontneously crowd into hlf the volume fter filling the entire volume. Wht is the chnge in entropy of the gs during this process? The process is neither reversile nor qusi-sttic. As shown in ection 20.6, the initil nd finl tempertures of the gs re the sme. To pply Eqution 22.13, we cnnot tke Q 5 0, the vlue for the irreversile process, ut must insted find Q r ; tht is, we must find n equivlent reversile pth tht shres the sme initil nd finl sttes. A simple choice is n isotherml, reversile expnsion in which the gs pushes slowly ginst piston while energy enters the gs y het from reservoir to hold the temperture constnt. Becuse T is constnt in this process, Eqution gives f dq r D 5 3 T 5 1 f T 3 dq r i For n isotherml process, the first lw of thermodynmics specifies tht e f i dq r is equl to the negtive of the work done on the gs during the expnsion from V i to V f, which is given y Eqution Using this result, we find tht the entropy chnge for the gs is i D 5 nr ln V f V i (22.17) Becuse V f. V i, we conclude tht D is positive. This positive result indictes tht the entropy of the gs increses s result of the irreversile, ditic expnsion. It is esy to see tht the energy hs spred fter the expnsion. Insted of eing concentrted in reltively smll spce, the molecules nd the energy ssocited with them re scttered over lrger region. Entropy Chnge in Therml Conduction Let us now consider system consisting of hot reservoir nd cold reservoir tht re in therml contct with ech other nd isolted from the rest of the Universe. A process occurs during which energy Q is trnsferred y het from the hot reservoir t temperture T h to the cold reservoir t temperture T c. The process s descried is irreversile (energy would not spontneously flow from cold to hot), so we must find n equivlent reversile process. The overll process is comintion of two processes: energy leving the hot reservoir nd energy entering the cold reservoir. We will clculte the entropy chnge for the reservoir in ech process nd dd to otin the overll entropy chnge.

18 22.7 Chnges in Entropy for Thermodynmic ystems 675 Consider first the process of energy entering the cold reservoir. Although the reservoir hs sored some energy, the temperture of the reservoir hs not chnged. The energy tht hs entered the reservoir is the sme s tht which would enter y mens of reversile, isotherml process. The sme is true for energy leving the hot reservoir. Becuse the cold reservoir sors energy Q, its entropy increses y Q /T c. At the sme time, the hot reservoir loses energy Q, so its entropy chnge is 2Q /T h. Therefore, the chnge in entropy of the system is Exmple 22.8 Aditic Free Expnsion: Revisited Let s verify tht the mcroscopic nd microscopic pproches to the clcultion of entropy led to the sme conclusion for the ditic free expnsion of n idel gs. uppose the idel gs in Figure expnds to four times its initil volume. As we hve seen for this process, the initil nd finl tempertures re the sme. (A) Using mcroscopic pproch, clculte the entropy chnge for the gs. olution Conceptulize Look ck t Figure 22.16, which is digrm of the system efore the ditic free expnsion. Imgine reking the memrne so tht the gs moves into the evcuted re. The expnsion is irreversile. Ctegorize We cn replce the irreversile process with reversile isotherml process etween the sme initil nd finl sttes. This pproch is mcroscopic, so we use thermodynmic vrile, in prticulr, the volume V. Anlyze Use Eqution to evlute the entropy chnge: D 5 nr ln V f 5 nr ln 4V i 5 nr ln 4 V i V i (B) Using sttisticl considertions, clculte the chnge in entropy for the gs nd show tht it grees with the nswer you otined in prt (A). olution D 5 Q T c 1 2Q T h 5 Q 1 T c 2 1 T h. 0 (22.18) This increse is consistent with our interprettion of entropy chnges s representing the spreding of energy. In the initil configurtion, the hot reservoir hs excess internl energy reltive to the cold reservoir. The process tht occurs spreds the energy into more equitle distriution etween the two reservoirs. Ctegorize This pproch is microscopic, so we use vriles relted to the individul molecules. Anlyze As in the discussion leding to Eqution 22.11, the numer of microsttes ville to single molecule in the initil volume V i is w i 5 V i /V m, where V i is the initil volume of the gs nd V m is the microscopic volume occupied y the molecule. Use this numer to find the numer of ville microsttes for N molecules: W i 5 w N i 5 V N i V m Find the numer of ville microsttes for N molecules in the finl volume V f 5 4V i : W f 5 V N f 5 4V N i V m V m continued

19 676 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics 22.8 continued Use Eqution to find the entropy chnge: D 5 k B ln W f W i 5 k B ln 4V N i 5 k V B ln 14 N 2 5 Nk B ln 4 5 nr ln 4 i Finlize The nswer is the sme s tht for prt (A), which delt with mcroscopic prmeters. Wht If? In prt (A), we used Eqution 22.17, which ws sed on reversile isotherml process connecting the initil nd finl sttes. Would you rrive t the sme result if you chose different reversile process? Answer You must rrive t the sme result ecuse entropy is stte vrile. For exmple, consider the two-step process in Figure 22.17: reversile ditic expnsion from V i to 4V i (A B) during which the temperture drops from T 1 to T 2 nd reversile isovolumetric process (B C) tht tkes the gs ck to the initil temperture T 1. During the reversile ditic process, D 5 0 ecuse Q r 5 0. For the reversile isovolumetric process (B C), use Eqution 22.13: Find the rtio of temperture T 1 to T 2 from Eqution for the ditic process: Figure (Exmple 22.8) A gs expnds to four times its initil volume nd ck to the initil temperture y mens of two-step process. f dq r D 5 3 T 5 3 T 1 5 4V i T 2 V i i g21 T 1 T 2 P T 1 T 2 A V i nc V dt 5 nc T V ln T 1 T g21 ustitute to find D: D 5 nc V ln 142 g21 5 nc V 1g 2 12 ln 4 We do indeed otin the exct sme result for the entropy chnge. Entropy sttement of the second lw of thermodynmics 22.8 Entropy nd the econd Lw 5 nc V C P C V 2 1 ln 4 5 n 1C P 2 C V 2 ln 4 5 nr ln 4 If we consider system nd its surroundings to include the entire Universe, the Universe is lwys moving towrd higher-proility mcrostte, corresponding to the continuous spreding of energy. An lterntive wy of stting this ehvior is s follows: The entropy of the Universe increses in ll rel processes. This sttement is yet nother wording of the second lw of thermodynmics tht cn e shown to e equivlent to the Kelvin-Plnck nd Clusius sttements. Let us show this equivlence first for the Clusius sttement. Looking t Figure 22.5, we see tht, if the het pump opertes in this mnner, energy is spontneously flowing from the cold reservoir to the hot reservoir without n input of energy y work. As result, the energy in the system is not spreding evenly etween the two reservoirs, ut is concentrting in the hot reservoir. Consequently, if the Clusius sttement of the second lw is not true, then the entropy sttement is lso not true, demonstrting their equivlence. C B 4V i V

20 22.8 Entropy nd the econd Lw 677 Hot reservoir t T h Q h Q c W eng Het engine W eng Cold reservoir t T c Het pump Q c For the equivlence of the Kelvin Plnck sttement, consider Figure 22.18, which shows the impossile engine of Figure 22.3 connected to het pump operting etween the sme reservoirs. The output work of the engine is used to drive the het pump. The net effect of this comintion is tht energy leves the cold reservoir nd is delivered to the hot reservoir without the input of work. (The work done y the engine on the het pump is internl to the system of oth devices.) This is foridden y the Clusius sttement of the second lw, which we hve shown to e equivlent to the entropy sttement. Therefore, the Kelvin Plnck sttement of the second lw is lso equivlent to the entropy sttement. When deling with system tht is not isolted from its surroundings, rememer tht the increse in entropy descried in the second lw is tht of the system nd its surroundings. When system nd its surroundings interct in n irreversile process, the increse in entropy of one is greter thn the decrese in entropy of the other. Hence, the chnge in entropy of the Universe must e greter thn zero for n irreversile process nd equl to zero for reversile process. We cn check this sttement of the second lw for the clcultions of entropy chnge tht we mde in ection Consider first the entropy chnge in free expnsion, descried y Eqution Becuse the free expnsion tkes plce in n insulted continer, no energy is trnsferred y het from the surroundings. Therefore, Eqution represents the entropy chnge of the entire Universe. Becuse V f. V i, the entropy chnge of the Universe is positive, consistent with the second lw. Now consider the entropy chnge in therml conduction, descried y Eqution Let ech reservoir e hlf the Universe. (The lrger the reservoir, the etter is the ssumption tht its temperture remins constnt!) Then the entropy chnge of the Universe is represented y Eqution Becuse T h. T c, this entropy chnge is positive, gin consistent with the second lw. The positive entropy chnge is lso consistent with the notion of energy spreding. The wrm portion of the Universe hs excess internl energy reltive to the cool portion. Therml conduction represents spreding of the energy more equitly throughout the Universe. Finlly, let us look t the entropy chnge in Crnot cycle, given y Eqution The entropy chnge of the engine itself is zero. The entropy chnge of the reservoirs is D 5 0 Q c 0 T c 2 0 Q h 0 T h Figure The impossile engine of Figure 22.3 trnsfers energy y work to het pump operting etween two energy reservoirs. This sitution is foridden y the Clusius sttement of the second lw of thermodynmics. In light of Eqution 22.7, this entropy chnge is lso zero. Therefore, the entropy chnge of the Universe is only tht ssocited with the work done y the engine. A portion of tht work will e used to chnge the mechnicl energy of system externl to the engine: speed up the shft of mchine, rise weight, nd so on. There is no chnge in internl energy of the externl system due to this portion

21 678 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics ummry Definitions The therml efficiency e of het engine is e ; W eng 0 Q h Q h Q c Q c 0 0 Q h 0 0 Q h 0 From microscopic viewpoint, the entropy of given mcrostte is defined s ; k B ln W (22.10) where k B is Boltzmnn s constnt nd W is the numer of microsttes of the system corresponding to the mcrostte. Concepts nd Principles of the work, or, equivlently, no energy spreding, so the entropy chnge is gin zero. The other portion of the work will e used to overcome vrious friction forces or other nonconservtive forces in the externl system. This process will cuse n increse in internl energy of tht system. Tht sme increse in internl energy could hve hppened vi reversile thermodynmic process in which energy Q r is trnsferred y het, so the entropy chnge ssocited with tht prt of the work is positive. As result, the overll entropy chnge of the Universe for the opertion of the Crnot engine is positive, gin consistent with the second lw. Ultimtely, ecuse rel processes re irreversile, the entropy of the Universe should increse stedily nd eventully rech mximum vlue. At this vlue, ssuming tht the second lw of thermodynmics, s formulted here on Erth, pplies to the entire expnding Universe, the Universe will e in stte of uniform temperture nd density. The totl energy of the Universe will hve spred more evenly throughout the Universe. All physicl, chemicl, nd iologicl processes will hve cesed t this time. This gloomy stte of ffirs is sometimes referred to s the het deth of the Universe. (22.2) A het engine is device tht tkes in energy y het nd, operting in cyclic process, expels frction of tht energy y mens of work. The net work done y het engine in crrying working sustnce through cyclic process (DE int 5 0) is W eng 5 Q h 2 Q c (22.1) where Q h is the energy tken in from hot reservoir nd Q c is the energy expelled to cold reservoir. The microstte of system is the description of its individul components. The mcrostte is description of the system from mcroscopic point of view. A given mcrostte cn hve mny microsttes. In reversile process, the system cn e returned to its initil conditions long the sme pth on PV digrm, nd every point long this pth is n equilirium stte. A process tht does not stisfy these requirements is irreversile. Two wys the second lw of thermodynmics cn e stted re s follows: It is impossile to construct het engine tht, operting in cycle, produces no effect other thn the input of energy y het from reservoir nd the performnce of n equl mount of work (the Kelvin Plnck sttement). It is impossile to construct cyclicl mchine whose sole effect is to trnsfer energy continuously y het from one oject to nother oject t higher temperture without the input of energy y work (the Clusius sttement).

22 Ojective Questions 679 Crnot s theorem sttes tht no rel het engine operting (irreversily) etween the tempertures T c nd T h cn e more efficient thn n engine operting reversily in Crnot cycle etween the sme two tempertures. The therml efficiency of het engine operting in the Crnot cycle is e C T c T h (22.8) The mcroscopic stte of system tht hs lrge numer of microsttes hs four qulities tht re ll relted: (1) uncertinty: ecuse of the lrge numer of microsttes, there is lrge uncertinty s to which one ctully exists; (2) choice: gin ecuse of the lrge numer of microsttes, there is lrge numer of choices from which to select s to which one exists; (3) proility: mcrostte with lrge numer of microsttes is more likely to exist thn one with smll numer of microsttes; (4) missing informtion: ecuse of the lrge numer of microsttes, there is high mount of missing informtion s to which one exists. For thermodynmic system, ll four of these cn e relted to the stte vrile of entropy. The chnge in entropy d of system during process etween two infinitesimlly seprted equilirium sttes is d 5 dq r T (22.12) where dq r is the energy trnsfer y het for the system for reversile process tht connects the initil nd finl sttes. Ojective Questions 1. The second lw of thermodynmics implies tht the coefficient of performnce of refrigertor must e wht? () less thn 1 () less thn or equl to 1 (c) greter thn or equl to 1 (d) finite (e) greter thn 0 2. Assume smple of n idel gs is t room temperture. Wht ction will necessrily mke the entropy of the smple increse? () Trnsfer energy into it y het. () Trnsfer energy into it irreversily y het. (c) Do work on it. (d) Increse either its temperture or its volume, without letting the other vrile decrese. (e) None of those choices is correct. 3. A refrigertor hs 18.0 kj of work done on it while 115 kj of energy is trnsferred from inside its interior. Wht is its coefficient of performnce? () 3.40 () 2.80 (c) 8.90 (d) 6.40 (e) Of the following, which is not sttement of the second lw of thermodynmics? () No het engine operting in cycle cn sor energy from reservoir nd use it entirely to do work. () No rel engine operting etween two energy reservoirs cn e more efficient The second lw of thermodynmics sttes tht when rel (irreversile) processes occur, there is sptil spreding of energy. This spreding of energy is relted to thermodynmic stte vrile clled entropy. Therefore, yet nother wy the second lw cn e stted is s follows: The entropy of the Universe increses in ll rel processes. The chnge in entropy of system during n ritrry finite process etween n initil stte nd finl stte is f dq r D 5 3 T i (22.13) The vlue of D for the system is the sme for ll pths connecting the initil nd finl sttes. The chnge in entropy for system undergoing ny reversile, cyclic process is zero. 1. denotes nswer ville in tudent olutions Mnul/tudy Guide thn Crnot engine operting etween the sme two reservoirs. (c) When system undergoes chnge in stte, the chnge in the internl energy of the system is the sum of the energy trnsferred to the system y het nd the work done on the system. (d) The entropy of the Universe increses in ll nturl processes. (e) Energy will not spontneously trnsfer y het from cold oject to hot oject. 5. Consider cyclic processes completely chrcterized y ech of the following net energy inputs nd outputs. In ech cse, the energy trnsfers listed re the only ones occurring. Clssify ech process s () possile, () impossile ccording to the first lw of thermodynmics, (c) impossile ccording to the second lw of thermodynmics, or (d) impossile ccording to oth the first nd second lws. (i) Input is 5 J of work, nd output is 4 J of work. (ii) Input is 5 J of work, nd output is 5 J of energy trnsferred y het. (iii) Input is 5 J of energy trnsferred y electricl trnsmission, nd output is 6 J of work. (iv) Input is 5 J of energy trnsferred y het, nd output is 5 J of energy trnsferred

23 680 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics y het. (v) Input is 5 J of energy trnsferred y het, nd output is 5 J of work. (vi) Input is 5 J of energy trnsferred y het, nd output is 3 J of work plus 2 J of energy trnsferred y het. 6. A compct ir-conditioning unit is plced on tle inside well-insulted prtment nd is plugged in nd turned on. Wht hppens to the verge temperture of the prtment? () It increses. () It decreses. (c) It remins constnt. (d) It increses until the unit wrms up nd then decreses. (e) The nswer depends on the initil temperture of the prtment. 7. A stem turine opertes t oiler temperture of 450 K nd n exhust temperture of 300 K. Wht is the mximum theoreticl efficiency of this system? () () (c) (d) (e) A thermodynmic process occurs in which the entropy of system chnges y 28 J/K. According to the second lw of thermodynmics, wht cn you conclude out the entropy chnge of the environment? () It must e 18 J/K or less. () It must e etween 18 J/K nd 0. (c) It must e equl to 18 J/K. (d) It must e 18 J/K or more. (e) It must e zero. 9. A smple of montomic idel gs is contined in cylinder with piston. Its stte is represented y the dot in the PV digrm shown in Figure OQ22.9. Arrows A through E represent isoric, isotherml, ditic, nd isovolumetric processes tht the smple cn undergo. In ech process except D, the volume chnges y fctor of 2. All five processes re reversile. Rnk the processes ccording to the chnge in entropy of the gs from the lrgest positive vlue to the Conceptul Questions 1. The energy exhust from certin col-fired electric generting sttion is crried y cooling wter into Lke Ontrio. The wter is wrm from the viewpoint of living things in the lke. ome of them congregte round the outlet port nd cn impede the wter flow. () Use the theory of het engines to explin why this ction cn reduce the electric power output of the sttion. () An engineer sys tht the electric output is reduced ecuse of higher ck pressure on the turine ldes. Comment on the ccurcy of this sttement. 2. Discuss three different common exmples of nturl processes tht involve n increse in entropy. Be sure to ccount for ll prts of ech system under considertion. 3. Does the second lw of thermodynmics contrdict or correct the first lw? Argue for your nswer. 4. The first lw of thermodynmics sys you cn t relly win, nd the second lw sys you cn t even rek even. Explin how this sttement pplies to prticulr device or process; lterntively, rgue ginst the sttement. 1. denotes nswer ville in tudent olutions Mnul/tudy Guide P B C A D lrgest-mgnitude negtive vlue. In your rnkings, disply ny cses of equlity. 10. An engine does 15.0 kj of work while exhusting 37.0 kj to cold reservoir. Wht is the efficiency of the engine? () () (c) (d) (e) The rrow OA in the PV P digrm shown in Figure OQ22.11 represents O reversile ditic expnsion of n idel gs. The B sme smple of gs, strting from the sme stte O, C A V now undergoes n ditic free expnsion to the Figure OQ22.11 sme finl volume. Wht point on the digrm could represent the finl stte of the gs? () the sme point A s for the reversile expnsion () point B (c) point C (d) ny of those choices (e) none of those choices E Figure OQ Energy is the mistress of the Universe, nd entropy is her shdow. Writing for n udience of generl reders, rgue for this sttement with t lest two exmples. Alterntively, rgue for the view tht entropy is like n executive who instntly determines wht will hppen, wheres energy is like ookkeeper telling us how little we cn fford. (Arnold ommerfeld suggested the ide for this question.) 6. () Give n exmple of n irreversile process tht occurs in nture. () Give n exmple of process in nture tht is nerly reversile. 7. The device shown in Figure CQ22.7, clled thermoelectric converter, uses series of semiconductor cells to trnsform internl energy to electric potentil energy, which we will study in Chpter 25. In the photogrph on the left, oth legs of the device re t the sme temperture nd no electric potentil energy is produced. When one leg is t higher temperture thn the other s shown in the photogrph on the right, however, electric potentil energy is produced s V

24 Prolems 681 the device extrcts energy from the hot reservoir nd drives smll electric motor. () Why is the difference in temperture necessry to produce electric potentil energy in this demonstrtion? () In wht sense does this intriguing experiment demonstrte the second lw of thermodynmics? Prolems Figure CQ22.7 The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide ection 22.1 Het Engines nd the econd Lw of Thermodynmics AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign 1. A prticulr het engine hs mechnicl power output of 5.00 kw nd n efficiency of 25.0%. The engine M expels J of exhust energy in ech cycle. Find () the energy tken in during ech cycle nd () the time intervl for ech cycle. 2. The work done y n engine equls one-fourth the energy it sors from reservoir. () Wht is its therml efficiency? () Wht frction of the energy sored is expelled to the cold reservoir? 3. A het engine tkes in 360 J of energy from hot reservoir nd performs 25.0 J of work in ech cycle. Find W () the efficiency of the engine nd () the energy expelled to the cold reservoir in ech cycle. 4. A gun is het engine. In prticulr, it is n internl comustion piston engine tht does not operte in cycle, ut comes prt during its ditic expnsion process. A certin gun consists of 1.80 kg of iron. It fires one 2.40-g ullet t 320 m/s with n energy efficiency of 1.10%. Assume the ody of the gun sors ll the energy exhust the other 98.9% nd increses uniformly in temperture for short time intervl efore Courtesy of PACO cientific Compny 8. A stem-driven turine is one mjor component of n electric power plnt. Why is it dvntgeous to hve the temperture of the stem s high s possile? 9. Discuss the chnge in entropy of gs tht expnds () t constnt temperture nd () diticlly. 10. uppose your roommte clens nd tidies up your messy room fter ig prty. Becuse she is creting more order, does this process represent violtion of the second lw of thermodynmics? 11. Is it possile to construct het engine tht cretes no therml pollution? Explin. 12. () If you shke jr full of jelly ens of different sizes, the lrger ens tend to pper ner the top nd the smller ones tend to fll to the ottom. Why? () Does this process violte the second lw of thermodynmics? 13. Wht re some fctors tht ffect the efficiency of utomoile engines? BIO Q/C it loses ny energy y het into the environment. Find its temperture increse. 5. An engine sors 1.70 kj from hot reservoir t 277 C nd expels 1.20 kj to cold reservoir t 27 C in ech cycle. () Wht is the engine s efficiency? () How much work is done y the engine in ech cycle? (c) Wht is the power output of the engine if ech cycle lsts s? 6. A multicylinder gsoline engine in n irplne, operting t rev/min, tkes in energy J nd exhusts J for ech revolution of the crnkshft. () How mny liters of fuel does it consume in 1.00 h of opertion if the het of comustion of the fuel is equl to J/L? () Wht is the mechnicl power output of the engine? Ignore friction nd express the nswer in horsepower. (c) Wht is the torque exerted y the crnkshft on the lod? (d) Wht power must the exhust nd cooling system trnsfer out of the engine? 7. uppose het engine is connected to two energy reservoirs, one pool of molten luminum (660 C) nd W the other lock of solid mercury (238.9 C). The engine runs y freezing 1.00 g of luminum nd melting 15.0 g of mercury during ech cycle. The het of

25 682 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics fusion of luminum is J/kg; the het of fusion of mercury is J/kg. Wht is the efficiency of this engine? ection 22.2 Het Pumps nd Refrigertors 8. A refrigertor hs coefficient of performnce equl W to The refrigertor tkes in 120 J of energy from cold reservoir in ech cycle. Find () the work required in ech cycle nd () the energy expelled to the hot reservoir. 9. During ech cycle, refrigertor ejects 625 kj of energy to high-temperture reservoir nd tkes in 550 kj of energy from low-temperture reservoir. Determine () the work done on the refrigernt in ech cycle nd () the coefficient of performnce of the refrigertor. 10. A het pump hs coefficient of performnce of 3.80 nd opertes with power consumption of W. () How much energy does it deliver into home during 8.00 h of continuous opertion? () How much energy does it extrct from the outside ir? 11. A refrigertor hs coefficient of performnce of The ice try comprtment is t C, nd the room temperture is 22.0 C. The refrigertor cn convert 30.0 g of wter t 22.0 C to 30.0 g of ice t C ech minute. Wht input power is required? Give your nswer in wtts. 12. A het pump hs coefficient of performnce equl to 4.20 nd requires power of 1.75 kw to operte. () How much energy does the het pump dd to home in one hour? () If the het pump is reversed so tht it cts s n ir conditioner in the summer, wht would e its coefficient of performnce? 13. A freezer hs coefficient of performnce of It is dvertised s using electricity t rte of 457 kwh/yr. () On verge, how much energy does it use in single dy? () On verge, how much energy does it remove from the refrigertor in single dy? (c) Wht mximum mss of wter t 20.0 C could the freezer freeze in single dy? Note: One kilowtt-hour (kwh) is n mount of energy equl to running 1-kW pplince for one hour. ection 22.3 Reversile nd Irreversile Processes ection 22.4 The Crnot Engine 14. A het engine opertes etween reservoir t 25.0 C nd one t 375 C. Wht is the mximum efficiency possile for this engine? 15. One of the most efficient het engines ever uilt is M col-fired stem turine in the Ohio River vlley, operting etween C nd 430 C. () Wht is its mximum theoreticl efficiency? () The ctul efficiency of the engine is 42.0%. How much mechnicl power does the engine deliver if it sors J of energy ech second from its hot reservoir? 16. Why is the following sitution impossile? An inventor comes to ptent office with the clim tht her het engine, which employs wter s working sustnce, hs thermodynmic efficiency of Although this efficiency is low compred with typicl utomoile engines, she explins tht her engine opertes etween n energy reservoir t room temperture nd wter ice mixture t tmospheric pressure nd therefore requires no fuel other thn tht to mke the ice. The ptent is pproved, nd working prototypes of the engine prove the inventor s efficiency clim. 17. A Crnot engine hs power output of 150 kw. The W engine opertes etween two reservoirs t 20.0 C nd 500 C. () How much energy enters the engine y het per hour? () How much energy is exhusted y het per hour? 18. A Crnot engine hs power output P. The engine opertes etween two reservoirs t temperture T c nd T h. () How much energy enters the engine y het in time intervl Dt? () How much energy is exhusted y het in the time intervl Dt? 19. Wht is the coefficient of performnce of refrigertor tht opertes with Crnot efficiency etween tempertures C nd C? 20. An idel refrigertor or idel het pump is equivlent to Crnot engine running in reverse. Tht is, energy Q c is tken in from cold reservoir nd energy Q h is rejected to hot reservoir. () how tht the work tht must e supplied to run the refrigertor or het pump is W 5 T h 2 T c 0 Q T c 0 c () how tht the coefficient of performnce (COP) of the idel refrigertor is COP 5 T h 2 T c 21. Wht is the mximum possile coefficient of performnce of het pump tht rings energy from outdoors t C into 22.0 C house? Note: The work done to run the het pump is lso ville to wrm the house. 22. How much work does n idel Crnot refrigertor M require to remove 1.00 J of energy from liquid helium t 4.00 K nd expel this energy to room-temperture (293-K) environment? 23. If 35.0%-efficient Crnot het engine (Fig. 22.2) is run in reverse so s to form refrigertor (Fig. 22.4), wht would e this refrigertor s coefficient of performnce? 24. A power plnt opertes t 32.0% efficiency during the summer when the sewter used for cooling is t 20.0 C. The plnt uses 350 C stem to drive turines. If the plnt s efficiency chnges in the sme proportion s the idel efficiency, wht would e the plnt s efficiency in the winter, when the sewter is t 10.0 C? 25. A het engine is eing designed to hve Crnot efficiency of 65.0% when operting etween two energy Q/C reservoirs. () If the temperture of the cold reservoir is 20.0 C, wht must e the temperture of the hot res- T c

26 Prolems 683 ervoir? () Cn the ctul efficiency of the engine e equl to 65.0%? Explin. 26. A Crnot het engine opertes etween tempertures Q/C T h nd T c. () If T h K nd T c K, wht is the efficiency of the engine? () Wht is the chnge in its efficiency for ech degree of increse in T h ove 500 K? (c) Wht is the chnge in its efficiency for ech degree of chnge in T c? (d) Does the nswer to prt (c) depend on T c? Explin. 27. An idel gs is tken through Crnot cycle. The isotherml expnsion occurs t 250 C, nd the isother- M ml compression tkes plce t 50.0 C. The gs tkes in J of energy from the hot reservoir during the isotherml expnsion. Find () the energy expelled to the cold reservoir in ech cycle nd () the net work done y the gs in ech cycle. 28. An electric power plnt tht would mke use of the W temperture grdient in the ocen hs een proposed. Q/C The system is to operte etween 20.0 C (surfcewter temperture) nd 5.00 C (wter temperture t depth of out 1 km). () Wht is the mximum efficiency of such system? () If the electric power output of the plnt is 75.0 MW, how much energy is tken in from the wrm reservoir per hour? (c) In view of your nswer to prt (), explin whether you think such system is worthwhile. Note tht the fuel is free. 29. A het engine opertes in Crnot cycle etween M 80.0 C nd 350 C. It sors J of energy per cycle from the hot reservoir. The durtion of ech cycle is 1.00 s. () Wht is the mechnicl power output of this engine? () How much energy does it expel in ech cycle y het? 30. uppose you uild two-engine device with the exhust Q/C energy output from one het engine supplying the input energy for second het engine. We sy tht the two engines re running in series. Let e 1 nd e 2 represent the efficiencies of the two engines. () The overll efficiency of the two-engine device is defined s the totl work output divided y the energy put into the first engine y het. how tht the overll efficiency e is given y e 5 e 1 1 e 2 2 e 1 e 2 Wht If? For prts () through (e) tht follow, ssume the two engines re Crnot engines. Engine 1 opertes etween tempertures T h nd T i. The gs in engine 2 vries in temperture etween T i nd T c. In terms of the tempertures, () wht is the efficiency of the comintion engine? (c) Does n improvement in net efficiency result from the use of two engines insted of one? (d) Wht vlue of the intermedite temperture T i results in equl work eing done y ech of the two engines in series? (e) Wht vlue of T i results in ech of the two engines in series hving the sme efficiency? 31. Argon enters turine t rte of 80.0 kg/min, W temperture of 800 C, nd pressure of 1.50 MP. It expnds diticlly s it pushes on the turine ldes nd exits t pressure 300 kp. () Clculte its tem- perture t exit. () Clculte the (mximum) power output of the turning turine. (c) The turine is one component of model closed-cycle gs turine engine. Clculte the mximum efficiency of the engine. 32. At point A in Crnot cycle, 2.34 mol of montomic idel gs hs pressure of kp, volume of 10.0 L, nd temperture of 720 K. The gs expnds isothermlly to point B nd then expnds diticlly to point C, where its volume is 24.0 L. An isotherml compression rings it to point D, where its volume is 15.0 L. An ditic process returns the gs to point A. () Determine ll the unknown pressures, volumes, nd tempertures s you fill in the following tle: P V T A kp 10.0 L 720 K B C 24.0 L D 15.0 L () Find the energy dded y het, the work done y the engine, nd the chnge in internl energy for ech of the steps A B, B C, C D, nd D A. (c) Clculte the efficiency W net / Q h. (d) how tht the efficiency is equl to 1 2 T C /T A, the Crnot efficiency. 33. An electric generting sttion is designed to hve Q/C n electric output power of 1.40 MW using turine with two-thirds the efficiency of Crnot engine. The exhust energy is trnsferred y het into cooling tower t 110 C. () Find the rte t which the sttion exhusts energy y het s function of the fuel comustion temperture T h. () If the fireox is modified to run hotter y using more dvnced comustion technology, how does the mount of energy exhust chnge? (c) Find the exhust power for T h C. (d) Find the vlue of T h for which the exhust power would e only hlf s lrge s in prt (c). (e) Find the vlue of T h for which the exhust power would e one-fourth s lrge s in prt (c). 34. An idel (Crnot) freezer in kitchen hs constnt temperture of 260 K, wheres the ir in the kitchen hs constnt temperture of 300 K. uppose the insultion for the freezer is not perfect ut rther conducts energy into the freezer t rte of W. Determine the verge power required for the freezer s motor to mintin the constnt temperture in the freezer. 35. A het pump used for heting shown in Figure P22.35 is essentilly n ir conditioner instlled ckwrd. It Het pump Q c Outside T c Figure P22.35 Q h Inside T h

27 684 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics extrcts energy from colder ir outside nd deposits it in wrmer room. uppose the rtio of the ctul energy entering the room to the work done y the device s motor is 10.0% of the theoreticl mximum rtio. Determine the energy entering the room per joule of work done y the motor given tht the inside temperture is 20.0 C nd the outside temperture is C. ection 22.5 Gsoline nd Diesel Engines Note: For prolems in this section, ssume the gs in the engine is ditomic with g A gsoline engine hs compression rtio of () Wht is the efficiency of the engine if it opertes in n idelized Otto cycle? () Wht If? If the ctul efficiency is 15.0%, wht frction of the fuel is wsted s result of friction nd energy trnsfers y het tht could e voided in reversile engine? Assume complete comustion of the ir fuel mixture. 37. In cylinder of n utomoile engine, immeditely M fter comustion the gs is confined to volume of 50.0 cm 3 nd hs n initil pressure of P. The piston moves outwrd to finl volume of 300 cm 3, nd the gs expnds without energy trnsfer y het. () Wht is the finl pressure of the gs? () How much work is done y the gs in expnding? 38. An idelized diesel engine opertes in cycle known s the ir-stndrd diesel cycle shown in Figure P Fuel is spryed into the cylinder t the point of mximum compression, B. Comustion occurs during the expnsion B C, which is modeled s n isoric process. how tht the efficiency of n engine operting in this idelized diesel cycle is P e g T D 2 T A T C 2 T B B V 2 V B Q h C V C Aditic processes D A V 1 V A Q c V neously nd then record the results of your tosses in terms of the numers of heds (H) nd tils (T) tht result. For exmple, HHTH nd HTHH re two possile wys in which three heds nd one til cn e chieved. () On the sis of your tle, wht is the most prole result recorded for toss? 41. If you roll two dice, wht is the totl numer of wys in which you cn otin () 12 nd () 7? ection 22.7 Chnges in Entropy for Thermodynmic ystems ection 22.8 Entropy nd the econd Lw 42. An ice try contins 500 g of liquid wter t 0 C. Clculte the chnge in entropy of the wter s it freezes W slowly nd completely t 0 C. 43. A tyrofom cup holding 125 g of hot wter t 100 C cools to room temperture, 20.0 C. Wht is the chnge in entropy of the room? Neglect the specific het of the cup nd ny chnge in temperture of the room. 44. A 1.00-kg iron horseshoe is tken from forge t 900 C nd dropped into 4.00 kg of wter t 10.0 C. Assuming tht no energy is lost y het to the surroundings, determine the totl entropy chnge of the horseshoeplus-wter system. (uggestion: Note tht dq 5 mc dt.) 45. A kg cr is moving t 20.0 m/s. The driver rkes AMT to stop. The rkes cool off to the temperture of the surrounding ir, which is nerly constnt t 20.0 C. Wht is the totl entropy chnge? 46. Two kg crs oth trveling t 20.0 m/s AMT undergo hed-on collision nd stick together. Find the chnge in entropy of the surrounding ir resulting from the collision if the ir temperture is 23.0 C. Ignore the energy crried wy from the collision y sound. 47. A 70.0-kg log flls from height of 25.0 m into lke. If AMT the log, the lke, nd the ir re ll t 300 K, find the chnge in entropy of the ir during this process. 48. A 1.00-mol smple of H 2 gs is contined in the left side of the continer shown in Figure P22.48, which hs equl volumes on the left nd right. The right side is evcuted. When the vlve is opened, the gs strems into the right side. () Wht is the entropy chnge of the gs? () Does the temperture of the gs chnge? Assume the continer is so lrge tht the hydrogen ehves s n idel gs. Figure P22.38 ection 22.6 Entropy 39. Prepre tle like Tle 22.1 y using the sme procedure () for the cse in which you drw three mrles from your g rther thn four nd () for the cse in which you drw five mrles rther thn four. 40. () Prepre tle like Tle 22.1 for the following occurrence. You toss four coins into the ir simult- H 2 Vlve Figure P22.48 Vcuum 49. A 2.00-L continer hs center prtition tht divides it into two equl prts s shown in Figure P The left side contins mol of H 2 gs, nd the right side contins mol of O 2 gs. Both gses re t

28 Prolems 685 room temperture nd t tmospheric pressure. The prtition is removed, nd the gses re llowed to mix. Wht is the entropy increse of the system? mol H mol O 2 Figure P Wht chnge in entropy occurs when 27.9-g ice cue t 212 C is trnsformed into stem t 115 C? 51. Clculte the chnge in entropy of 250 g of wter wrmed slowly from 20.0 C to 80.0 C. 52. How fst re you personlly mking the entropy of the Universe increse right now? Compute n order-ofmgnitude estimte, stting wht quntities you tke s dt nd the vlues you mesure or estimte for them. 53. When n luminum r is connected etween hot reservoir t 725 K nd cold reservoir t 310 K, 2.50 kj of energy is trnsferred y het from the hot reservoir to the cold reservoir. In this irreversile process, clculte the chnge in entropy of () the hot reservoir, () the cold reservoir, nd (c) the Universe, neglecting ny chnge in entropy of the luminum rod. 54. When metl r is connected etween hot reservoir t T h nd cold reservoir t T c, the energy trnsferred y het from the hot reservoir to the cold reservoir is Q. In this irreversile process, find expressions for the chnge in entropy of () the hot reservoir, () the cold reservoir, nd (c) the Universe, neglecting ny chnge in entropy of the metl rod. 55. The temperture t the surfce of the un is pproximtely K, nd the temperture t the surfce of M W the Erth is pproximtely 290 K. Wht entropy chnge of the Universe occurs when J of energy is trnsferred y rdition from the un to the Erth? Additionl Prolems 56. Clculte the increse in entropy of the Universe when you dd 20.0 g of 5.00 C crem to 200 g of 60.0 C coffee. Assume tht the specific hets of crem nd coffee re oth 4.20 J/g? C. 57. How much work is required, using n idel Crnot refrigertor, to chnge kg of tp wter t 10.0 C into ice t C? Assume tht the freezer comprtment is held t C nd tht the refrigertor exhusts energy into room t 20.0 C. 58. A stem engine is operted in cold climte where the exhust temperture is 0 C. () Clculte the theoreticl mximum efficiency of the engine using n intke stem temperture of 100 C. () If, insted, superheted stem t 200 C is used, find the mximum possile efficiency. 59. The energy sored y n engine is three times greter thn the work it performs. () Wht is its therml efficiency? () Wht frction of the energy sored is expelled to the cold reservoir? 60. Every second t Nigr Flls, some m 3 of AMT wter flls distnce of 50.0 m. Wht is the increse in entropy of the Universe per second due to the flling wter? Assume the mss of the surroundings is so gret tht its temperture nd tht of the wter sty nerly constnt t 20.0 C. Also ssume negligile mount of wter evportes. 61. Find the mximum (Crnot) efficiency of n engine tht sors energy from hot reservoir t 545 C nd exhusts energy to cold reservoir t 185 C. 62. In 1993, the U.. government instituted requirement Q/C tht ll room ir conditioners sold in the United ttes must hve n energy efficiency rtio (EER) of 10 or higher. The EER is defined s the rtio of the cooling cpcity of the ir conditioner, mesured in British therml units per hour, or Btu/h, to its electricl power requirement in wtts. () Convert the EER of 10.0 to dimensionless form, using the conversion 1 Btu J. () Wht is the pproprite nme for this dimensionless quntity? (c) In the 1970s, it ws common to find room ir conditioners with EERs of 5 or lower. tte how the operting costs compre for Btu/h ir conditioners with EERs of 5.00 nd Assume ech ir conditioner opertes for h during the summer in city where electricity costs 17.0 per kwh. 63. Energy trnsfers y het through the exterior wlls nd M roof of house t rte of J/s kw when the interior temperture is 22.0 C nd the outside temperture is C. () Clculte the electric power required to mintin the interior temperture t 22.0 C if the power is used in electric resistnce heters tht convert ll the energy trnsferred in y electricl trnsmission into internl energy. () Wht If? Clculte the electric power required to mintin the interior temperture t 22.0 C if the power is used to drive n electric motor tht opertes the compressor of het pump tht hs coefficient of performnce equl to 60.0% of the Crnot-cycle vlue. 64. One mole of neon gs is heted from 300 K to 420 K M t constnt pressure. Clculte () the energy Q trnsferred to the gs, () the chnge in the internl energy of the gs, nd (c) the work done on the gs. Note tht neon hs molr specific het of C P J/mol? K for constnt-pressure process. 65. An irtight freezer holds n moles of ir t 25.0 C nd 1.00 tm. The ir is then cooled to C. () Wht is the chnge in entropy of the ir if the volume is held constnt? () Wht would the entropy chnge e if the pressure were mintined t 1.00 tm during the cooling? 66. uppose n idel (Crnot) het pump could e constructed for use s n ir conditioner. () Otin Q/C n

29 686 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics expression for the coefficient of performnce (COP) for such n ir conditioner in terms of T h nd T c. () Would such n ir conditioner operte on smller energy input if the difference in the operting tempertures were greter or smller? (c) Compute the COP for such n ir conditioner if the indoor temperture is 20.0 C nd the outdoor temperture is 40.0 C. 67. In 1816, Roert tirling, cottish clergymn, ptented the tirling engine, which hs found wide vri- GP ety of pplictions ever since, including current use in solr energy collectors to trnsform sunlight into electricity. Fuel is urned externlly to wrm one of the engine s two cylinders. A fixed quntity of inert gs moves cycliclly etween the cylinders, expnding in the hot one nd contrcting in the cold one. Figure P22.67 represents model for its thermodynmic cycle. Consider n moles of n idel mon tomic gs eing tken once through the cycle, consisting of two isotherml processes t tempertures 3T i nd T i nd two constnt- volume processes. Let us find the efficiency of this engine. () Find the energy trnsferred y het into the gs during the isovolumetric process AB. () Find the energy trnsferred y het into the gs during the isotherml process BC. (c) Find the energy trnsferred y het into the gs during the isovolumetric process CD. (d) Find the energy trnsferred y het into the gs during the isotherml process DA. (e) Identify which of the results from prts () through (d) re positive nd evlute the energy input to the engine y het. (f) From the first lw of thermodynmics, find the work done y the engine. (g) From the results of prts (e) nd (f), evlute the efficiency of the engine. A tirling engine is esier to mnufcture thn n internl comustion engine or turine. It cn run on urning grge. It cn run on the energy trnsferred y sunlight nd produce no mteril exhust. tirling engines re not currently used in utomoiles due to long strtup times nd poor ccelertion response. P B A V i Isotherml processes D C 3T i 2V i Figure P A fireox is t 750 K, nd the mient temperture is Q/C 300 K. The efficiency of Crnot engine doing 150 J of work s it trnsports energy etween these constnttemperture ths is 60.0%. The Crnot engine must tke in energy 150 J/ J from the hot reservoir nd must put out 100 J of energy y het into the T i V environment. To follow Crnot s resoning, suppose some other het engine could hve n efficiency of 70.0%. () Find the energy input nd exhust energy output of engine s it does 150 J of work. () Let engine operte s in prt () nd run the Crnot engine in reverse etween the sme reservoirs. The output work of engine is the input work for the Crnot refrigertor. Find the totl energy trnsferred to or from the fireox nd the totl energy trnsferred to or from the environment s oth engines operte together. (c) Explin how the results of prts () nd () show tht the Clusius sttement of the second lw of thermodynmics is violted. (d) Find the energy input nd work output of engine s it puts out exhust energy of 100 J. Let engine operte s in prt (c) nd contriute 150 J of its work output to running the Crnot engine in reverse. Find (e) the totl energy the fireox puts out s oth engines operte together, (f) the totl work output, nd (g) the totl energy trnsferred to the environment. (h) Explin how the results show tht the Kelvin Plnck sttement of the second lw is violted. Therefore, our ssumption out the efficiency of engine must e flse. (i) Let the engines operte together through one cycle s in prt (d). Find the chnge in entropy of the Universe. (j) Explin how the result of prt (i) shows tht the entropy sttement of the second lw is violted. 69. Review. This prolem complements Prolem 88 in Chpter 10. In the opertion of single-cylinder internl comustion piston engine, one chrge of fuel explodes to drive the piston outwrd in the power stroke. Prt of its energy output is stored in turning flywheel. This energy is then used to push the piston inwrd to compress the next chrge of fuel nd ir. In this compression process, ssume n originl volume of L of ditomic idel gs t tmospheric pressure is compressed diticlly to one-eighth of its originl volume. () Find the work input required to compress the gs. () Assume the flywheel is solid disk of mss 5.10 kg nd rdius 8.50 cm, turning freely without friction etween the power stroke nd the compression stroke. How fst must the flywheel turn immeditely fter the power stroke? This sitution represents the minimum ngulr speed t which the engine cn operte without stlling. (c) When the engine s opertion is well ove the point of stlling, ssume the flywheel puts 5.00% of its mximum energy into compressing the next chrge of fuel nd ir. Find its mximum ngulr speed in this cse. 70. A iology lortory is mintined t constnt temperture of 7.00 C y n ir conditioner, which is vented to the ir outside. On typicl hot summer dy, the outside temperture is 27.0 C nd the ir-conditioning unit emits energy to the outside t rte of 10.0 kw. Model the unit s hving coefficient of performnce (COP) equl to 40.0% of the COP of n idel Crnot device. () At wht rte does the ir conditioner remove energy from the lortory? () Clculte the power required for the work input. (c) Find the chnge

30 Prolems 687 in entropy of the Universe produced y the ir conditioner in 1.00 h. (d) Wht If? The outside temperture increses to 32.0 C. Find the frctionl chnge in the COP of the ir conditioner. 71. A power plnt, hving Crnot efficiency, produces 1.00 GW of electricl power from turines tht tke in stem t 500 K nd reject wter t 300 K into flowing river. The wter downstrem is 6.00 K wrmer due to the output of the power plnt. Determine the flow rte of the river. 72. A power plnt, hving Crnot efficiency, produces electric power P from turines tht tke in energy from stem t temperture T h nd dischrge energy t temperture T c through het exchnger into flowing river. The wter downstrem is wrmer y DT due to the output of the power plnt. Determine the flow rte of the river. 73. A 1.00-mol smple of n idel montomic gs is tken Q/C through the cycle shown in Figure P The process A B is reversile isotherml expnsion. Clculte () the net work done y the gs, () the energy dded to the gs y het, (c) the energy exhusted from the gs y het, nd (d) the efficiency of the cycle. (e) Explin how the efficiency compres with tht of Crnot engine operting etween the sme temperture extremes. P (tm) 5 1 C A Isotherml process B Figure P22.73 V (liters) 74. A system consisting of n moles of n idel gs with Q/C molr specific het t constnt pressure C P undergoes two reversile processes. It strts with pressure P i nd volume V i, expnds isothermlly, nd then contrcts diticlly to rech finl stte with pressure P i nd volume 3V i. () Find its chnge in entropy in the isotherml process. (The entropy does not chnge in the ditic process.) () Wht If? Explin why the nswer to prt () must e the sme s the nswer to Prolem 77. (You do not need to solve Prolem 77 to nswer this question.) 75. A het engine opertes etween two reservoirs t T K nd T K. It tkes in J of energy from the higher-temperture reservoir nd performs 250 J of work. Find () the entropy chnge of the Universe D U for this process nd () the work W tht could hve een done y n idel Crnot engine operting etween these two reservoirs. (c) how tht the difference etween the mounts of work done in prts () nd () is T 1 D U. 76. A 1.00-mol smple of montomic idel gs is tken Q/C through the cycle shown in Figure P At point A, the pressure, volume, nd temperture re P i, V i, nd T i, respectively. In terms of R nd T i, find () the totl energy entering the system y het per cycle, () the totl energy leving the system y het per cycle, nd (c) the efficiency of n engine operting in this cycle. (d) Explin how the efficiency compres with tht of n engine operting in Crnot cycle etween the sme temperture extremes. 3P i 2P i P i P Q 1 B Q 2 C Q 3 D A Q 4 V V i 2V i Figure P A smple consisting of n moles of n idel gs undergoes reversile isoric expnsion from volume V i to volume 3V i. Find the chnge in entropy of the gs y clculting e f i dq /T, where dq 5 nc P dt. 78. An thlete whose mss is 70.0 kg drinks 16.0 ounces BIO (454 g) of refrigerted wter. The wter is t temperture of 35.0 F. () Ignoring the temperture chnge of Q/C the ody tht results from the wter intke (so tht the ody is regrded s reservoir lwys t 98.6 F), find the entropy increse of the entire system. () Wht If? Assume the entire ody is cooled y the drink nd the verge specific het of person is equl to the specific het of liquid wter. Ignoring ny other energy trnsfers y het nd ny metolic energy relese, find the thlete s temperture fter she drinks the cold wter given n initil ody temperture of 98.6 F. (c) Under these ssumptions, wht is the entropy increse of the entire system? (d) tte how this result compres with the one you otined in prt (). 79. A smple of n idel gs expnds isothermlly, douling in volume. () how tht the work done on the Q/C gs in expnding is W 5 2nRT ln 2. () Becuse the internl energy E int of n idel gs depends solely on its temperture, the chnge in internl energy is zero during the expnsion. It follows from the first lw tht the energy input to the gs y het during the expnsion is equl to the energy output y work. Does this process hve 100% efficiency in converting energy input y het into work output? (c) Does this conversion violte the second lw? Explin. 80. Why is the following sitution impossile? Two smples of wter re mixed t constnt pressure inside n insulted continer: 1.00 kg of wter t 10.0 C nd 1.00 kg of wter t 30.0 C. Becuse the continer is insulted, there is no exchnge of energy y het etween the wter nd the

31 688 Chpter 22 Het Engines, Entropy, nd the econd Lw of Thermodynmics environment. Furthermore, the mount of energy tht leves the wrm wter y het is equl to the mount tht enters the cool wter y het. Therefore, the entropy chnge of the Universe is zero for this process. Chllenge Prolems 81. A 1.00-mol smple of n idel gs (g ) is crried through the Crnot cycle descried in Figure At point A, the pressure is 25.0 tm nd the temperture is 600 K. At point C, the pressure is 1.00 tm nd the temperture is 400 K. () Determine the pressures nd volumes t points A, B, C, nd D. () Clculte the net work done per cycle. 82. The compression rtio of n Otto cycle s shown in Figure is V A /V B At the eginning A of the compression process, 500 cm 3 of gs is t 100 kp nd 20.0 C. At the eginning of the ditic expnsion, the temperture is T C C. Model the working fluid s n idel gs with g () Fill in this tle to follow the sttes of the gs: T (K) P (kp) V (cm 3 ) A B C D () Fill in this tle to follow the processes: A B B C C D D A ABCDA Q W DE int (c) Identify the energy input Q h, (d) the energy exhust Q c, nd (e) the net output work W eng. (f) Clculte the therml efficiency. (g) Find the numer of crnkshft revolutions per minute required for onecylinder engine to hve n output power of 1.00 kw hp. Note: The thermodynmic cycle involves four piston strokes.

32 Electricity nd Mgnetism p r t 4 A Trnsrpid mglev trin pulls into sttion in hnghi, Chin. The word mglev is n revited form of mgnetic levittion. This trin mkes no physicl contct with its rils; its weight is totlly supported y electromgnetic forces. In this prt of the ook, we will study these forces. (OTHK/Asi Imges/Jupiterimges) We now study the rnch of physics concerned with electric nd mgnetic phenomen. The lws of electricity nd mgnetism ply centrl role in the opertion of such devices s smrtphones, televisions, electric motors, computers, high-energy ccelertors, nd other electronic devices. More fundmentlly, the intertomic nd intermoleculr forces responsile for the formtion of solids nd liquids re electric in origin. Evidence in Chinese documents suggests mgnetism ws oserved s erly s 2000 BC. The ncient Greeks oserved electric nd mgnetic phenomen possily s erly s 700 BC. The Greeks knew out mgnetic forces from oservtions tht the nturlly occurring stone mgnetite (Fe 3 O 4 ) is ttrcted to iron. (The word electric comes from elecktron, the Greek word for mer. The word mgnetic comes from Mgnesi, the nme of the district of Greece where mgnetite ws first found.) Not until the erly prt of the nineteenth century did scientists estlish tht electricity nd mgnetism re relted phenomen. In 1819, Hns Oersted discovered tht compss needle is deflected when plced ner circuit crrying n electric current. In 1831, Michel Frdy nd, lmost simultneously, Joseph Henry showed tht when wire is moved ner mgnet (or, equivlently, when mgnet is moved ner wire), n electric current is estlished in the wire. In 1873, Jmes Clerk Mxwell used these oservtions nd other experimentl fcts s sis for formulting the lws of electromgnetism s we know them tody. (Electromgnetism is nme given to the comined study of electricity nd mgnetism.) Mxwell s contriutions to the field of electromgnetism were especilly significnt ecuse the lws he formulted re sic to ll forms of electromgnetic phenomen. His work is s importnt s Newton s work on the lws of motion nd the theory of grvittion. 689

33 c h p t e r 23 Electric Fields 23.1 Properties of Electric Chrges 23.2 Chrging Ojects y Induction 23.3 Coulom s Lw 23.4 Anlysis Model: Prticle in Field (Electric) 23.5 Electric Field of Continuous Chrge Distriution 23.6 Electric Field Lines 23.7 Motion of Chrged Prticle in Uniform Electric Field This young womn is enjoying the effects of electriclly chrging her ody. Ech individul hir on her hed ecomes chrged nd exerts repulsive force on the other hirs, resulting in the stnd-up hirdo seen here. (Ted Kinsmn / Photo Reserchers, Inc.) In this chpter, we egin the study of electromgnetism. The first link tht we will mke to our previous study is through the concept of force. The electromgnetic force etween chrged prticles is one of the fundmentl forces of nture. We egin y descriing some sic properties of one mnifesttion of the electromgnetic force, the electric force. We then discuss Coulom s lw, which is the fundmentl lw governing the electric force etween ny two chrged prticles. Next, we introduce the concept of n electric field ssocited with chrge distriution nd descrie its effect on other chrged prticles. We then show how to use Coulom s lw to clculte the electric field for given chrge distriution. The chpter concludes with discussion of the motion of chrged prticle in uniform electric field. The second link etween electromgnetism nd our previous study is through the concept of energy. We will discuss tht connection in Chpter Properties of Electric Chrges 690 A numer of simple experiments demonstrte the existence of electric forces. For exmple, fter ruing lloon on your hir on dry dy, you will find tht the lloon ttrcts its of pper. The ttrctive force is often strong enough to suspend the pper from the lloon.

34 23.1 Properties of Electric Chrges 691 A negtively chrged ruer rod suspended y string is ttrcted to positively chrged glss rod. A negtively chrged ruer rod is repelled y nother negtively chrged ruer rod. Figure 23.1 The electric force etween () oppositely chrged ojects nd () like-chrged ojects. F F Glss Ruer F Ruer F When mterils ehve in this wy, they re sid to e electrified or to hve ecome electriclly chrged. You cn esily electrify your ody y vigorously ruing your shoes on wool rug. Evidence of the electric chrge on your ody cn e detected y lightly touching (nd strtling) friend. Under the right conditions, you will see sprk when you touch nd oth of you will feel slight tingle. (Experiments such s these work est on dry dy ecuse n excessive mount of moisture in the ir cn cuse ny chrge you uild up to lek from your ody to the Erth.) In series of simple experiments, it ws found tht there re two kinds of electric chrges, which were given the nmes positive nd negtive y Benjmin Frnklin ( ). Electrons re identified s hving negtive chrge, nd protons re positively chrged. To verify tht there re two types of chrge, suppose hrd ruer rod tht hs een rued on fur is suspended y string s shown in Figure When glss rod tht hs een rued on silk is rought ner the ruer rod, the two ttrct ech other (Fig. 23.1). On the other hnd, if two chrged ruer rods (or two chrged glss rods) re rought ner ech other s shown in Figure 23.1, the two repel ech other. This oservtion shows tht the ruer nd glss hve two different types of chrge on them. On the sis of these oservtions, we conclude tht chrges of the sme sign repel one nother nd chrges with opposite signs ttrct one nother. Using the convention suggested y Frnklin, the electric chrge on the glss rod is clled positive nd tht on the ruer rod is clled negtive. Therefore, ny chrged oject ttrcted to chrged ruer rod (or repelled y chrged glss rod) must hve positive chrge, nd ny chrged oject repelled y chrged ruer rod (or ttrcted to chrged glss rod) must hve negtive chrge. Another importnt spect of electricity tht rises from experimentl oservtions is tht electric chrge is lwys conserved in n isolted system. Tht is, when one oject is rued ginst nother, chrge is not creted in the process. The electrified stte is due to trnsfer of chrge from one oject to the other. One oject gins some mount of negtive chrge while the other gins n equl mount of positive chrge. For exmple, when glss rod is rued on silk s in Figure 23.2, the silk otins negtive chrge equl in mgnitude to the positive chrge on the glss rod. We now know from our understnding of tomic structure tht electrons re trnsferred in the ruing process from the glss to the silk. imilrly, when ruer is rued on fur, electrons re trnsferred from the fur to the ruer, giving the ruer net negtive chrge nd the fur net positive chrge. This process works ecuse neutrl, unchrged mtter contins s mny positive chrges (protons within tomic nuclei) Ruer Becuse of conservtion of chrge, ech electron dds negtive chrge to the silk nd n equl positive chrge is left on the glss rod. Figure 23.2 When glss rod is rued with silk, electrons re trnsferred from the glss to the silk. WWElectric chrge is conserved

35 692 Chpter 23 Electric Fields The neutrl sphere hs equl numers of positive nd negtive chrges. Electrons redistriute when chrged rod is rought close. ome electrons leve the grounded sphere through the ground wire. The excess positive chrge is nonuniformly distriuted. The remining electrons redistriute uniformly, nd there is net uniform distriution of positive chrge on the sphere. c d Figure 23.3 Chrging metllic oject y induction. () A neutrl metllic sphere. () A chrged ruer rod is plced ner the sphere. (c) The sphere is grounded. (d) The ground connection is removed. (e) The rod is removed. e s negtive chrges (electrons). Conservtion of electric chrge for n isolted system is like conservtion of energy, momentum, nd ngulr momentum, ut we don t identify n nlysis model for this conservtion principle ecuse it is not used often enough in the mthemticl solution to prolems. In 1909, Roert Millikn ( ) discovered tht electric chrge lwys occurs s integrl multiples of fundmentl mount of chrge e (see ection 25.7). In modern terms, the electric chrge q is sid to e quntized, where q is the stndrd symol used for chrge s vrile. Tht is, electric chrge exists s discrete pckets, nd we cn write q 5 6Ne, where N is some integer. Other experiments in the sme period showed tht the electron hs chrge 2e nd the proton hs chrge of equl mgnitude ut opposite sign 1e. ome prticles, such s the neutron, hve no chrge. Q uick Quiz 23.1 Three ojects re rought close to ech other, two t time. When ojects A nd B re rought together, they repel. When ojects B nd C re rought together, they lso repel. Which of the following re true? () Ojects A nd C possess chrges of the sme sign. () Ojects A nd C possess chrges of opposite sign. (c) All three ojects possess chrges of the sme sign. (d) One oject is neutrl. (e) Additionl experiments must e performed to determine the signs of the chrges Chrging Ojects y Induction It is convenient to clssify mterils in terms of the ility of electrons to move through the mteril: Electricl conductors re mterils in which some of the electrons re free electrons 1 tht re not ound to toms nd cn move reltively freely through the mteril; electricl insultors re mterils in which ll electrons re ound to toms nd cnnot move freely through the mteril. Mterils such s glss, ruer, nd dry wood fll into the ctegory of electricl insultors. When such mterils re chrged y ruing, only the re rued ecomes chrged nd the chrged prticles re unle to move to other regions of the mteril. In contrst, mterils such s copper, luminum, nd silver re good electricl conductors. When such mterils re chrged in some smll region, the chrge redily distriutes itself over the entire surfce of the mteril. emiconductors re third clss of mterils, nd their electricl properties re somewhere etween those of insultors nd those of conductors. ilicon nd germnium re well-known exmples of semiconductors commonly used in the friction of vriety of electronic chips used in computers, cellulr telephones, nd home theter systems. The electricl properties of semiconductors cn e chnged over mny orders of mgnitude y the ddition of controlled mounts of certin toms to the mterils. To understnd how to chrge conductor y process known s induction, consider neutrl (unchrged) conducting sphere insulted from the ground s shown in Figure There re n equl numer of electrons nd protons in the sphere if the chrge on the sphere is exctly zero. When negtively chrged ruer rod is rought ner the sphere, electrons in the region nerest the rod experience repulsive force nd migrte to the opposite side of the sphere. This migrtion leves 1 A metl tom contins one or more outer electrons, which re wekly ound to the nucleus. When mny toms comine to form metl, the free electrons re these outer electrons, which re not ound to ny one tom. These electrons move out the metl in mnner similr to tht of gs molecules moving in continer.

36 23.2 Chrging Ojects y Induction 693 The chrged lloon induces chrge seprtion on the surfce of the wll due to relignment of chrges in the molecules of the wll. The chrged rod ttrcts the pper ecuse chrge seprtion is induced in the molecules of the pper. Figure 23.4 () A chrged lloon is rought ner n insulting wll. () A chrged rod is rought close to its of pper. Chrged lloon Wll Induced chrge seprtion the side of the sphere ner the rod with n effective positive chrge ecuse of the diminished numer of electrons s in Figure (The left side of the sphere in Figure 23.3 is positively chrged s if positive chrges moved into this region, ut rememer tht only electrons re free to move.) This process occurs even if the rod never ctully touches the sphere. If the sme experiment is performed with conducting wire connected from the sphere to the Erth (Fig. 23.3c), some of the electrons in the conductor re so strongly repelled y the presence of the negtive chrge in the rod tht they move out of the sphere through the wire nd into the Erth. The symol t the end of the wire in Figure 23.3c indictes tht the wire is connected to ground, which mens reservoir, such s the Erth, tht cn ccept or provide electrons freely with negligile effect on its electricl chrcteristics. If the wire to ground is then removed (Fig. 23.3d), the conducting sphere contins n excess of induced positive chrge ecuse it hs fewer electrons thn it needs to cncel out the positive chrge of the protons. When the ruer rod is removed from the vicinity of the sphere (Fig. 23.3e), this induced positive chrge remins on the ungrounded sphere. Notice tht the ruer rod loses none of its negtive chrge during this process. Chrging n oject y induction requires no contct with the oject inducing the chrge. Tht is in contrst to chrging n oject y ruing (tht is, y conduction), which does require contct etween the two ojects. A process similr to induction in conductors tkes plce in insultors. In most neutrl molecules, the center of positive chrge coincides with the center of negtive chrge. In the presence of chrged oject, however, these centers inside ech molecule in n insultor my shift slightly, resulting in more positive chrge on one side of the molecule thn on the other. This relignment of chrge within individul molecules produces lyer of chrge on the surfce of the insultor s shown in Figure The proximity of the positive chrges on the surfce of the oject nd the negtive chrges on the surfce of the insultor results in n ttrctive force etween the oject nd the insultor. Your knowledge of induction in insultors should help you explin why chrged rod ttrcts its of electriclly neutrl pper s shown in Figure Q uick Quiz 23.2 Three ojects re rought close to one nother, two t time. When ojects A nd B re rought together, they ttrct. When ojects B nd C re rought together, they repel. Which of the following re necessrily true? () Ojects A nd C possess chrges of the sme sign. () Ojects A nd C possess chrges of opposite sign. (c) All three ojects possess chrges of the sme sign. (d) One oject is neutrl. (e) Additionl experiments must e performed to determine informtion out the chrges on the ojects.. Cengge Lerning/Chrles D. Winters

37 694 Chpter 23 Electric Fields B A uspension hed Fier Figure 23.5 Coulom s lnce, used to estlish the inversesqure lw for the electric force. Coulom s lw Coulom constnt Chrles Coulom French physicist ( ) Coulom s mjor contriutions to science were in the res of electrosttics nd mgnetism. During his lifetime, he lso investigted the strengths of mterils, therey contriuting to the field of structurl mechnics. In ergonomics, his reserch provided n understnding of the wys in which people nd nimls cn est do work. INTERFOTO/Almy 23.3 Coulom s Lw Chrles Coulom mesured the mgnitudes of the electric forces etween chrged ojects using the torsion lnce, which he invented (Fig. 23.5). The operting principle of the torsion lnce is the sme s tht of the pprtus used y Cvendish to mesure the density of the Erth (see ection 13.1), with the electriclly neutrl spheres replced y chrged ones. The electric force etween chrged spheres A nd B in Figure 23.5 cuses the spheres to either ttrct or repel ech other, nd the resulting motion cuses the suspended fier to twist. Becuse the restoring torque of the twisted fier is proportionl to the ngle through which the fier rottes, mesurement of this ngle provides quntittive mesure of the electric force of ttrction or repulsion. Once the spheres re chrged y ruing, the electric force etween them is very lrge compred with the grvittionl ttrction, nd so the grvittionl force cn e neglected. From Coulom s experiments, we cn generlize the properties of the electric force (sometimes clled the electrosttic force) etween two sttionry chrged prticles. We use the term point chrge to refer to chrged prticle of zero size. The electricl ehvior of electrons nd protons is very well descried y modeling them s point chrges. From experimentl oservtions, we find tht the mgnitude of the electric force (sometimes clled the Coulom force) etween two point chrges is given y Coulom s lw. F e 5 k e 0 q q 2 0 r 2 (23.1) where k e is constnt clled the Coulom constnt. In his experiments, Coulom ws le to show tht the vlue of the exponent of r ws 2 to within n uncertinty of few percent. Modern experiments hve shown tht the exponent is 2 to within n uncertinty of few prts in Experiments lso show tht the electric force, like the grvittionl force, is conservtive. The vlue of the Coulom constnt depends on the choice of units. The I unit of chrge is the coulom (C). The Coulom constnt k e in I units hs the vlue This constnt is lso written in the form k e N? m 2 /C 2 (23.2) k e 5 1 4pP 0 (23.3) where the constnt P 0 (Greek letter epsilon) is known s the permittivity of free spce nd hs the vlue P C 2 /N? m 2 (23.4) The smllest unit of free chrge e known in nture, 2 the chrge on n electron (2e) or proton (1e), hs mgnitude e C (23.5) Therefore, 1 C of chrge is pproximtely equl to the chrge of electrons or protons. This numer is very smll when compred with the numer of free electrons in 1 cm 3 of copper, which is on the order of Nevertheless, 1 C is sustntil mount of chrge. In typicl experiments in which ruer or glss rod is chrged y friction, net chrge on the order of C is otined. In other 2 No unit of chrge smller thn e hs een detected on free prticle; current theories, however, propose the existence of prticles clled qurks hving chrges 2e/3 nd 2e/3. Although there is considerle experimentl evidence for such prticles inside nucler mtter, free qurks hve never een detected. We discuss other properties of qurks in Chpter 46.

38 23.3 Coulom's Lw 695 Tle 23.1 Chrge nd Mss of the Electron, Proton, nd Neutron Prticle Chrge (C) Mss (kg) Electron (e) Proton (p) Neutron (n) words, only very smll frction of the totl ville chrge is trnsferred etween the rod nd the ruing mteril. The chrges nd msses of the electron, proton, nd neutron re given in Tle Notice tht the electron nd proton re identicl in the mgnitude of their chrge ut vstly different in mss. On the other hnd, the proton nd neutron re similr in mss ut vstly different in chrge. Chpter 46 will help us understnd these interesting properties. Exmple 23.1 The Hydrogen Atom The electron nd proton of hydrogen tom re seprted (on the verge) y distnce of pproximtely m. Find the mgnitudes of the electric force nd the grvittionl force etween the two prticles. olution Conceptulize Think out the two prticles seprted y the very smll distnce given in the prolem sttement. In Chpter 13, we mentioned tht the grvittionl force etween n electron nd proton is very smll compred to the electric force etween them, so we expect this to e the cse with the results of this exmple. Ctegorize The electric nd grvittionl forces will e evluted from universl force lws, so we ctegorize this exmple s sustitution prolem. Use Coulom s lw to find the mgnitude of the electric force: Use Newton s lw of universl grvittion nd Tle 23.1 (for the prticle msses) to find the mgnitude of the grvittionl force: F e 5 k 0 e 0 0 2e 0 e N # m 2 /C C2 2 r m N F g 5 G m em p r N # m 2 /kg kg kg m N The rtio F e /F g < Therefore, the grvittionl force etween chrged tomic prticles is negligile when compred with the electric force. Notice the similr forms of Newton s lw of universl grvittion nd Coulom s lw of electric forces. Other thn the mgnitude of the forces etween elementry prticles, wht is fundmentl difference etween the two forces? When deling with Coulom s lw, rememer tht force is vector quntity nd must e treted ccordingly. Coulom s lw expressed in vector form for the electric force exerted y chrge q 1 on second chrge q 2, written F 12, is 12 F 5 k q 1q 2 e r 2 r^12 (23.6) where r^12 is unit vector directed from q 1 towrd q 2 s shown in Figure 23.6 (pge 696). Becuse the electric force oeys Newton s third lw, the electric force exerted y q 2 on q 1 is equl in mgnitude to the force exerted y q 1 on q 2 nd in the opposite direction; tht is, F F12. Finlly, Eqution 23.6 shows tht if q 1 nd q 2 hve the WWVector form of Coulom s lw

39 696 Chpter 23 Electric Fields Figure 23.6 Two point chrges seprted y distnce r exert force on ech other tht is given y Coulom s lw. The force F 21 exerted y q 2 on q 1 is equl in mgnitude nd opposite in direction to the force F 12 exerted y q 1 on q 2. Exmple 23.2 sme sign s in Figure 23.6, the product q 1 q 2 is positive nd the electric force on one prticle is directed wy from the other prticle. If q 1 nd q 2 re of opposite sign s shown in Figure 23.6, the product q 1 q 2 is negtive nd the electric force on one prticle is directed towrd the other prticle. These signs descrie the reltive direction of the force ut not the solute direction. A negtive product indictes n ttrctive force, nd positive product indictes repulsive force. The solute direction of the force on chrge depends on the loction of the other chrge. For exmple, if n x xis lies long the two chrges in Figure 23.6, the product q 1 q 2 is positive, ut F 12 points in the positive x direction nd F 21 points in the negtive x direction. When more thn two chrges re present, the force etween ny pir of them is given y Eqution Therefore, the resultnt force on ny one of them equls the vector sum of the forces exerted y the other individul chrges. For exmple, if four chrges re present, the resultnt force exerted y prticles 2, 3, nd 4 on prticle 1 is F1 5 F21 1 F31 1 F41 Q uick Quiz 23.3 Oject A hs chrge of 12 mc, nd oject B hs chrge of 16 mc. Which sttement is true out the electric forces on the ojects? () F AB 5 23 F BA () F AB 5 2 F BA (c) 3 F AB 5 2 F BA (d) F AB 5 3 F BA (e) F AB 5 F BA (f) 3 F AB 5 F BA Find the Resultnt Force Consider three point chrges locted t the corners of right tringle s shown in Figure 23.7, where q 1 5 q mc, q mc, nd m. Find the resultnt force exerted on q 3. olution When the chrges re of the sme sign, the force is repulsive. Conceptulize Think out the net force on q 3. Becuse chrge q 3 is ner two other chrges, it will experience two electric forces. These forces re exerted in different directions s shown in Figure Bsed on the forces shown in the figure, estimte the direction of the net force vector. Ctegorize Becuse two forces re exerted on chrge q 3, we ctegorize this exmple s vector ddition prolem. Anlyze The directions of the individul forces exerted y q 1 nd q 2 on q 3 re shown in Figure The force F 23 exerted y q 2 on q 3 is ttrctive ecuse q 2 nd q 3 hve opposite signs. In the coordinte system shown in Figure 23.7, the ttrctive force F 23 is to the left (in the negtive x direction). F 21 q 1 r rˆ 12 The force F 13 exerted y q 1 on q 3 is repulsive ecuse oth chrges re positive. The repulsive force F 13 mkes n ngle of with the x xis. q 2 F 12 When the chrges re of opposite signs, the force is ttrctive. q 1 q 2 q 1 F 21 y F 12 q 2 2 F 23 q 3 F 13 Figure 23.7 (Exmple 23.2) The force exerted y q 1 on q 3 is F 13. The force exerted y q 2 on q 3 is F 23. The resultnt force F 3 exerted on q 3 is the vector sum F 13 1 F 23. x

40 23.3 Coulom's Lw continued Use Eqution 23.1 to find the mgnitude of F23: Finlize The net force on q 3 is upwrd nd towrd the left in Figure If q 3 moves in response to the net force, the distnces etween q 3 nd the other chrges chnge, so the net force chnges. Therefore, if q 3 is free to move, it cn e modeled s prticle under net force s long s it is recognized tht the force exerted on q 3 is not constnt. As reminder, we disply most numericl vlues to three significnt figures, which leds to opertions such s 7.94 N 1 (28.99 N) N ove. If you crry ll intermedite results to more significnt figures, you will see tht this opertion is correct. Wht If? for F3? F 23 5 k e 0 q q Find the mgnitude of the force F13: F 13 5 k 0 q q 3 0 e 1"2 2 2 Find the x nd y components of the force F 13: Find the components of the resultnt force cting on q 3 : Express the resultnt force cting on q 3 in unit-vector form: N # m 2 /C C C m N N # m 2 /C C C m N F 13x 5 (11.2 N) cos N F 13y 5 (11.2 N) sin N F 3x 5 F 13x 1 F 23x N 1 (28.99 N) N F 3y 5 F 13y 1 F 23y N N F i^ j^2 N Wht if the signs of ll three chrges were chnged to the opposite signs? How would tht ffect the result Answer The chrge q 3 would still e ttrcted towrd q 2 nd repelled from q 1 with forces of the sme mgnitude. Therefore, the finl result for F 3 would e the sme. Exmple 23.3 Where Is the Net Force Zero? Three point chrges lie long the x xis s shown in Figure The positive chrge q mc is t x m, the positive chrge q mc is t the origin, nd the net force cting on q 3 is zero. Wht is the x coordinte of q 3? olution Conceptulize Becuse q 3 is ner two other chrges, it experiences two electric forces. Unlike the preceding exmple, however, the forces lie long the sme line in this prolem s indicted in Figure Becuse q 3 is negtive nd q 1 nd q 2 re positive, the forces F13 nd F23 re oth ttrctive. Becuse q 2 is the smller chrge, the position of q 3 t which the force is zero should e closer to q 2 thn to q 1. Ctegorize Becuse the net force on q 3 is zero, we model the point chrge s prticle in equilirium. AM y q 2 x F m q x F 13 q 1 Figure 23.8 (Exmple 23.3) Three point chrges re plced long the x xis. If the resultnt force cting on q 3 is zero, the force F13 exerted y q 1 on q 3 must e equl in mgnitude nd opposite in direction to the force F23 exerted y q 2 on q 3. x Anlyze Write n expression for the net force on chrge q 3 when it is in equilirium: Move the second term to the right side of the eqution nd set the coefficients of the unit vector i^ equl: 3 F 5 F23 1 F13 5 2k 0 q q 3 0 e x 2 k e 0 q q 3 0 x 2 5 k e 0 q q x2 2 i^ 1 k e 0 q q x2 2 i^ 5 0 continued

41 698 Chpter 23 Electric Fields 23.3 continued Eliminte k e nd uq 3 u nd rerrnge the eqution: ( x) 2 uq 2 u 5 x 2 uq 1 u Tke the squre root of oth sides of the eqution: ( x)" 0 q x " 0 q 1 0 olve for x : x " 0 q 2 0 ustitute numericl vlues, choosing the plus sign: x 5 Exmple 23.4 Find the Chrge on the pheres Two identicl smll chrged spheres, ech hving mss of kg, hng in equilirium s shown in Figure The length L of ech string is m, nd the ngle u is Find the mgnitude of the chrge on ech sphere. olution Conceptulize Figure 23.9 helps us conceptulize this exmple. The two spheres exert repulsive forces on ech other. If they re held close to ech other nd relesed, they move outwrd from the center nd settle into the configurtion in Figure 23.9 fter the oscilltions hve vnished due to ir resistnce. Ctegorize The key phrse in equilirium helps us model ech sphere s prticle in equilirium. This exmple is similr to the prticle in equilirium prolems in Chpter 5 with the dded feture tht one of the forces on sphere is n electric force. Anlyze The force digrm for the left-hnd sphere is shown in Figure The sphere is in equilirium under the ppliction of the force T from the string, the electric force F e from the other sphere, nd the grvittionl force m g. AM q " 0 q " 0 q " C " C 1 " C m Finlize Notice tht the movle chrge is indeed closer to q 2 s we predicted in the Conceptulize step. The second solution to the eqution (if we choose the negtive sign) is x m. Tht is nother loction where the mgnitudes of the forces on q 3 re equl, ut oth forces re in the sme direction, so they do not cncel. Wht If? uppose q 3 is constrined to move only long the x xis. From its initil position t x m, it is pulled smll distnce long the x xis. When relesed, does it return to equilirium, or is it pulled frther from equilirium? Tht is, is the equilirium stle or unstle? Answer If q 3 is moved to the right, F 13 ecomes lrger nd F 23 ecomes smller. The result is net force to the right, in the sme direction s the displcement. Therefore, the chrge q 3 would continue to move to the right nd the equilirium is unstle. (ee ection 7.9 for review of stle nd unstle equiliri.) If q 3 is constrined to sty t fixed x coordinte ut llowed to move up nd down in Figure 23.8, the equilirium is stle. In this cse, if the chrge is pulled upwrd (or downwrd) nd relesed, it moves ck towrd the equilirium position nd oscilltes out this point. L u u L q F e T cos u u m g T u T sin u Figure 23.9 (Exmple 23.4) () Two identicl spheres, ech crrying the sme chrge q, suspended in equilirium. () Digrm of the forces cting on the sphere on the left prt of (). From the prticle in equilirium model, set the net force on the left-hnd sphere equl to zero for ech component: (1) o F x 5 T sin u 2 F e 5 0 T sin u 5 F e (2) o F y 5 T cos u 2 mg 5 0 T cos u 5 mg Divide Eqution (1) y Eqution (2) to find F e : (3) tn u 5 F e mg F e 5 mg tn u

42 23.4 Anlysis Model: Prticle in Field (Electric) 699 Use the geometry of the right tringle in Figure 23.9 to find reltionship etween, nd (4) sin u 5 sin olve Coulom s lw (Eq. 23.1) for the chrge on ech sphere nd sustitute from Equtions (3) nd (4): mg tn sin ustitute numericl vlues: kg m tn m sin N C Finlize If the sign of the chrges were not given in Figure 23.9, we could not determine them. In fct, the sign of the chrge is not importnt. The sitution is the sme whether oth spheres re positively chrged or negtively chrged. Wht If? uppose your roommte proposes solving this prolem without the ssumption tht the chrges re of equl mgnitude. he clims the symmetry of the prolem is destroyed if the chrges re not equl, so the strings would mke two different ngles with the verticl nd the prolem would e much more complicted. How would you respond? Answer The symmetry is not destroyed nd the ngles re not different. Newton s third lw requires the mgnitudes of the electric forces on the two spheres to e the sme, regrdless of the equlity or nonequlity of the chrges. The solu tion to the exmple remins the sme with one chnge: the vlue of in the solution is replced y in the new sitution, where nd re the vlues of the chrges on the two spheres. The symmetry of the prolem would e destroyed if the msses of the spheres were not the sme. In this cse, the strings would mke different ngles with the verticl nd the prolem would e more complicted Anlysis Model: Prticle in Field (Electric) In ection 5.1, we discussed the differences etween contct forces nd field forces. Two field forces the grvittionl force in Chpter 13 nd the electric force here hve een introduced into our discussions so fr. As pointed out erlier, field forces cn ct through spce, producing n effect even when no physicl contct occurs etween intercting ojects. uch n interction cn e modeled s two-step pro cess: source prticle estlishes field, nd then chrged prticle intercts with the field nd experiences force. The grvittionl field t point in spce due to source prticle ws defined in ection 13.4 to e equl to the grvittionl force cting on test prticle of mss divided y tht mss: Then the force exerted y the field is (Eq. 5.5). The concept of field ws developed y Michel Frdy ( ) in the con text of electric forces nd is of such prcticl vlue tht we shll devote much tten tion to it in the next severl chpters. In this pproch, n electric field is sid to exist in the region of spce round chrged oject, the source chrge. The presence of the electric field cn e detected y plcing test chrge in the field nd noting the electric force on it. As n exmple, consider Figure 23.10, which shows smll positive test chrge plced ner second oject crrying much greter positive chrge We define the electric field due to the source chrge t the loction of the test chrge to e the electric force on the test chrge per unit chrge, or, to e more specific, the electric field vector t point in spce is defined s the electric force ct ing on positive test chrge plced t tht point divided y the test chrge: igure A smll positive test chrge plced t point ner n oject crrying much lrger positive chrge expe riences n electric field t point estlished y the source chrge Q. We will lwys ssume tht the test chrge is so smll tht the field of the source chrge is unffected y its presence. (23.7) WWDefinition of electric field When using Eqution 23.7, we must ssume the test chrge is smll enough tht it does not distur the chrge distri ution responsile for the electric field. If the test chrge is gret enough, the chrge on the metllic sphere is redistri uted nd the electric field it sets up is different from the field it sets up in the presence of the much smller test chrge.

43 700 Chpter 23 Electric Fields This drmtic photogrph cptures lightning olt striking tree ner some rurl homes. Lightning is ssocited with very strong electric fields in the tmosphere. Pitfll Prevention 23.1 Prticles Only Eqution 23.8 is vlid only for prticle of chrge q, tht is, n oject of zero size. For chrged oject of finite size in n electric field, the field my vry in mgnitude nd direction over the size of the oject, so the corresponding force eqution my e more complicted. Courtesy Johnny Autery The vector E hs the I units of newtons per coulom (N/C). The direction of E s shown in Figure is the direction of the force positive test chrge experiences when plced in the field. Note tht E is the field produced y some chrge or chrge distriution seprte from the test chrge; it is not the field produced y the test chrge itself. Also note tht the existence of n electric field is property of its source; the presence of the test chrge is not necessry for the field to exist. The test chrge serves s detector of the electric field: n electric field exists t point if test chrge t tht point experiences n electric force. If n ritrry chrge q is plced in n electric field E, it experiences n electric force given y Fe 5 q E (23.8) This eqution is the mthemticl representtion of the electric version of the prticle in field nlysis model. If q is positive, the force is in the sme direction s the field. If q is negtive, the force nd the field re in opposite directions. Notice the similrity etween Eqution 23.8 nd the corresponding eqution from the grvittionl version of the prticle in field model, F g 5 m g (ection 5.5). Once the mgnitude nd direction of the electric field re known t some point, the electric force exerted on ny chrged prticle plced t tht point cn e clculted from Eqution To determine the direction of n electric field, consider point chrge q s source chrge. This chrge cretes n electric field t ll points in spce surrounding it. A test chrge q 0 is plced t point P, distnce r from the source chrge, s in Figure We imgine using the test chrge to determine the direction of the electric force nd therefore tht of the electric field. According to Coulom s lw, the force exerted y q on the test chrge is e F 5 k qq 0 e r r^ 2 where r^ is unit vector directed from q towrd q 0. This force in Figure is directed wy from the source chrge q. Becuse the electric field t P, the position of the test chrge, is defined y E 5 Fe /q 0, the electric field t P creted y q is E q 5 ke 2 r^ (23.9) r If the source chrge q is positive, Figure shows the sitution with the test chrge removed: the source chrge sets up n electric field t P, directed wy from q. If q is If q is positive, the force on the test chrge q 0 is directed wy from q. q rˆ r q 0 P F e q rˆ F e q 0 P If q is negtive, the force on the test chrge q 0 is directed towrd q. c Figure (), (c) When test chrge q 0 is plced ner source chrge q, the test chrge experiences force. (), (d) At point P ner source chrge q, there exists n electric field. For positive source chrge, the electric field t P points rdilly outwrd from q. q rˆ P E q d rˆ E P For negtive source chrge, the electric field t P points rdilly inwrd towrd q.

44 23.4 Anlysis Model: Prticle in Field (Electric) 701 negtive s in Figure 23.11c, the force on the test chrge is towrd the source chrge, so the electric field t P is directed towrd the source chrge s in Figure 23.11d. To clculte the electric field t point P due to smll numer of point chrges, we first clculte the electric field vectors t P individully using Eqution 23.9 nd then dd them vectorilly. In other words, t ny point P, the totl electric field due to group of source chrges equls the vector sum of the electric fields of ll the chrges. This superposition principle pplied to fields follows directly from the vector ddition of electric forces. Therefore, the electric field t point P due to group of source chrges cn e expressed s the vector sum E 5 ke i q i r i 2 r^i (23.10) where r i is the distnce from the ith source chrge q i to the point P nd r^i is unit vector directed from q i towrd P. In Exmple 23.6, we explore the electric field due to two chrges using the superposition principle. Prt (B) of the exmple focuses on n electric dipole, which is defined s positive chrge q nd negtive chrge 2q seprted y distnce 2. The electric dipole is good model of mny molecules, such s hydrochloric cid (HCl). Neutrl toms nd molecules ehve s dipoles when plced in n externl electric field. Furthermore, mny molecules, such s HCl, re permnent dipoles. The effect of such dipoles on the ehvior of mterils sujected to electric fields is discussed in Chpter 26. Q uick Quiz 23.4 A test chrge of 13 mc is t point P where n externl electric field is directed to the right nd hs mgnitude of N/C. If the test chrge is replced with nother test chrge of 23 mc, wht hppens to the externl electric field t P? () It is unffected. () It reverses direction. (c) It chnges in wy tht cnnot e determined. Anlysis Model Imgine n oject with E q chrge tht we cll source chrge. The source chrge estlishes n electric field E throughout spce. Now imgine F e qe prticle with chrge q is plced in tht field. The prticle intercts with the electric field so tht the prticle experiences n electric force given y Exmple 23.5 Prticle in Field (Electric) Fe 5 q E (23.8) Exmples: A uspended Wter Droplet W WElectric field due to finite numer of point chrges n electron moves etween the deflection pltes of cthode ry oscilloscope nd is deflected from its originl pth chrged ions experience n electric force from the electric field in velocity selector efore entering mss spectrometer (Chpter 29) n electron moves round the nucleus in the electric field estlished y the proton in hydrogen tom s modeled y the Bohr theory (Chpter 42) hole in semiconducting mteril moves in response to the electric field estlished y pplying voltge to the mteril (Chpter 43) AM A wter droplet of mss kg is locted in the ir ner the ground during stormy dy. An tmospheric electric field of mgnitude N/C points verticlly downwrd in the vicinity of the wter droplet. The droplet remins suspended t rest in the ir. Wht is the electric chrge on the droplet? olution Conceptulize Imgine the wter droplet hovering t rest in the ir. This sitution is not wht is normlly oserved, so something must e holding the wter droplet up. continued

45 702 Chpter 23 Electric Fields 23.5 continued Ctegorize The droplet cn e modeled s prticle nd is descried y two nlysis models ssocited with fields: the prticle in field (grvittionl) nd the prticle in field (electric). Furthermore, ecuse the droplet is suject to forces ut remins t rest, it is lso descried y the prticle in equilirium model. Anlyze Write Newton s second lw from the prticle in equilirium model in the verticl direction: (1) F y 5 0 F e 2 F g 5 0 Using the two prticle in field models mentioned in the Ctegorize step, sustitute for the forces in Eqution (1), recognizing tht the verticl component of the electric field is negtive: olve for the chrge on the wter droplet: ustitute numericl vlues: Exmple 23.6 Electric Field Due to Two Chrges Chrges q 1 nd q 2 re locted on the x xis, t distnces nd, respectively, from the origin s shown in Figure (A) Find the components of the net electric field t the point P, which is t position (0, y). olution Conceptulize Compre this exmple with Exmple There, we dd vector forces to find the net force on chrged prticle. Here, we dd electric field vectors to find the net electric field t point in spce. If chrged prticle were plced t P, we could use the prticle in field model to find the electric force on the prticle. Ctegorize We hve two source chrges nd wish to find the resultnt electric field, so we ctegorize this exmple s one in which we cn use the superposition principle represented y Eqution Anlyze Find the mgnitude of the electric field t P due to chrge q 1 : q 5 2 mg E E 1 5 k 0 q 1 0 e r 5 k 2 e 1 q 12E 2 2 mg 5 0 q kg m/s C N/C Finlize Noting the smllest unit of free chrge in Eqution 23.5, the chrge on the wter droplet is lrge numer of these units. Notice tht the electric force is upwrd to lnce the downwrd grvittionl force. The prolem sttement clims tht the electric field is in the downwrd direction. Therefore, the chrge found ove is negtive so tht the electric force is in the direction opposite to the electric field. Figure (Exmple 23.6) The totl electric field E t P equls the vector sum E 1 1 E 2, where E 1 is the field due to the positive chrge q 1 nd E 2 is the field due to the negtive chrge q 2. 0 q y 2 r 1 q 1 P f y E 1 f u E 2 E u r 2 q 2 x Find the mgnitude of the electric field t P due to chrge q 2 : Write the electric field vectors for ech chrge in unit-vector form: E 2 5 k 0 q 2 0 e r 5 k 2 e 2 E 1 5 k e E 2 5 k e 0 q y 2 0 q y cos f i^ 0 q k 2 e sin f j^ y 0 q y cos u i^ 0 q k 2 e sin u j^ y

46 23.4 Anlysis Model: Prticle in Field (Electric) continued Write the components of the net electric field vector: 0 q 1 0 (1) E x 5 E 1x 1 E 2x 5 k e 2 1 y cos f 1 k 0 q e 2 1 y cos u 2 0 q 1 0 (2) E y 5 E 1y 1 E 2y 5 k e 2 1 y sin f 2 k 0 q e 2 1 y sin u 2 (B) Evlute the electric field t point P in the specil cse tht uq 1 u 5 uq 2 u nd 5. olution Conceptulize Figure shows the sitution in this specil cse. Notice the symmetry in the sitution nd tht the chrge distriution is now n electric dipole. Ctegorize Becuse Figure is specil cse of the generl cse shown in Figure 23.12, we cn ctegorize this exmple s one in which we cn tke the result of prt (A) nd sustitute the pproprite vlues of the vriles. Anlyze Bsed on the symmetry in Figure 23.13, evlute Equtions (1) nd (2) from prt (A) with 5, uq 1 u 5 uq 2 u 5 q, nd f 5 u: From the geometry in Figure 23.13, evlute cos u: ustitute Eqution (4) into Eqution (3): Figure (Exmple 23.6) When the chrges in Figure re of equl mgnitude nd equidistnt from the origin, the sitution ecomes symmetric s shown here. q r u P y u u E 1 E 2 E u x q q (3) E x 5 k e 2 1 y cos u 1 k q 2 e 2 1 y cos u 5 2k q 2 e 2 1 y cos u 2 q E y 5 k e 2 1 y sin u 2 k q 2 e 2 1 y sin u (4) cos u 5 r y 2 2 1/2 q E x 5 2k e 2 1 y c y 2 2 d 5 k 1/2 e 2q y 2 2 3/2 (C) Find the electric field due to the electric dipole when point P is distnce y.. from the origin. olution In the solution to prt (B), ecuse y.., neglect 2 compred with y 2 nd write the expression for E in this cse: (5) E < k e 2q y 3 Finlize From Eqution (5), we see tht t points fr from dipole ut long the perpendiculr isector of the line joining the two chrges, the mgnitude of the electric field creted y the dipole vries s 1/r 3, wheres the more slowly vrying field of point chrge vries s 1/r 2 (see Eq. 23.9). Tht is ecuse t distnt points, the fields of the two chrges of equl mgnitude nd opposite sign lmost cncel ech other. The 1/r 3 vrition in E for the dipole lso is otined for distnt point long the x xis nd for ny generl distnt point.

47 704 Chpter 23 Electric Fields ˆr 2 r 2 q 2 r 1 P E E 3 2 E 1 rˆ 1 r3 ˆr 3 q 1 q 3 Figure The electric field t P due to continuous chrge distriution is the vector sum of the fields D E i due to ll the elements Dq i of the chrge distriution. Three smple elements re shown. Electric field due to continuous chrge distriution Volume chrge density 23.5 Electric Field of Continuous Chrge Distriution Eqution is useful for clculting the electric field due to smll numer of chrges. In mny cses, we hve continuous distriution of chrge rther thn collection of discrete chrges. The chrge in these situtions cn e descried s continuously distriuted long some line, over some surfce, or throughout some volume. To set up the process for evluting the electric field creted y continuous chrge distriution, let s use the following procedure. First, divide the chrge distriution into smll elements, ech of which contins smll chrge Dq s shown in Figure Next, use Eqution 23.9 to clculte the electric field due to one of these elements t point P. Finlly, evlute the totl electric field t P due to the chrge distriution y summing the contriutions of ll the chrge elements (tht is, y pplying the superposition principle). The electric field t P due to one chrge element crrying chrge Dq is D E 5 k Dq e r r^ 2 where r is the distnce from the chrge element to point P nd r^ is unit vector directed from the element towrd P. The totl electric field t P due to ll elements in the chrge distriution is pproximtely Dq i E < ke i 2 r r^i i where the index i refers to the ith element in the distriution. Becuse the numer of elements is very lrge nd the chrge distriution is modeled s continuous, the totl field t P in the limit Dq i 0 is Dq i E 5 ke lim r r^ 2 i 5 k dq e 3 2 r^ (23.11) i r Dq i 0 i where the integrtion is over the entire chrge distriution. The integrtion in Eqution is vector opertion nd must e treted ppropritely. Let s illustrte this type of clcultion with severl exmples in which the chrge is distriuted on line, on surfce, or throughout volume. When performing such clcultions, it is convenient to use the concept of chrge density long with the following nottions: If chrge Q is uniformly distriuted throughout volume V, the volume chrge density r is defined y r ; Q V where r hs units of couloms per cuic meter (C/m 3 ). If chrge Q is uniformly distriuted on surfce of re A, the surfce chrge density s (Greek letter sigm) is defined y urfce chrge density s ; Q A where s hs units of couloms per squre meter (C/m 2 ). If chrge Q is uniformly distriuted long line of length,, the liner chrge density l is defined y Liner chrge density l ; Q, where l hs units of couloms per meter (C/m).

48 23.5 Electric Field of Continuous Chrge Distriution 705 If the chrge is nonuniformly distriuted over volume, surfce, or line, the mounts of chrge dq in smll volume, surfce, or length element re dq 5 r dv dq 5 s da dq 5 l d, Prolem-olving trtegy Clculting the Electric Field The following procedure is recommended for solving prolems tht involve the determintion of n electric field due to individul chrges or chrge distriution. 1. Conceptulize. Estlish mentl representtion of the prolem: think crefully out the individul chrges or the chrge distriution nd imgine wht type of electric field it would crete. Appel to ny symmetry in the rrngement of chrges to help you visulize the electric field. 2. Ctegorize. Are you nlyzing group of individul chrges or continuous chrge distriution? The nswer to this question tells you how to proceed in the Anlyze step. 3. Anlyze. () If you re nlyzing group of individul chrges, use the superposition principle: when severl point chrges re present, the resultnt field t point in spce is the vector sum of the individul fields due to the individul chrges (Eq ). Be very creful in the mnipultion of vector quntities. It my e useful to review the mteril on vector ddition in Chpter 3. Exmple 23.6 demonstrted this procedure. () If you re nlyzing continuous chrge distriution, the superposition principle is pplied y replcing the vector sums for evluting the totl electric field from individul chrges y vector integrls. The chrge distriution is divided into infinitesiml pieces, nd the vector sum is crried out y integrting over the entire chrge distriution (Eq ). Exmples 23.7 through 23.9 demonstrte such procedures. Consider symmetry when deling with either distriution of point chrges or continuous chrge distriution. Tke dvntge of ny symmetry in the system you oserved in the Conceptulize step to simplify your clcultions. The cncelltion of field components perpendiculr to the xis in Exmple 23.8 is n exmple of the ppliction of symmetry. 4. Finlize. Check to see if your electric field expression is consistent with the mentl representtion nd if it reflects ny symmetry tht you noted previously. Imgine vrying prmeters such s the distnce of the oservtion point from the chrges or the rdius of ny circulr ojects to see if the mthemticl result chnges in resonle wy. Exmple 23.7 The Electric Field Due to Chrged Rod A rod of length, hs uniform positive chrge per unit length l nd totl chrge Q. Clculte the electric field t point P tht is locted long the long xis of the rod nd distnce from one end (Fig ). olution P Conceptulize The field d E t P due to ech segment of chrge on the rod is in the negtive x direction ecuse every segment Figure (Exmple 23.7) The electric field t P crries positive chrge. Figure shows the pproprite due to uniformly chrged rod lying long the x xis. geometry. In our result, we expect the electric field to ecome smller s the distnce ecomes lrger ecuse point P is frther from the chrge distriution. continued E y x dx x

49 706 Chpter 23 Electric Fields 23.7 continued Ctegorize Becuse the rod is continuous, we re evluting the field due to continuous chrge distriution rther thn group of individul chrges. Becuse every segment of the rod produces n electric field in the negtive x direction, the sum of their contriutions cn e hndled without the need to dd vectors. Anlyze Let s ssume the rod is lying long the x xis, dx is the length of one smll segment, nd dq is the chrge on tht segment. Becuse the rod hs chrge per unit length l, the chrge dq on the smll segment is dq 5 l dx. Find the mgnitude of the electric field t P due to one segment of the rod hving chrge dq: Finlize We see tht our prediction is correct; if ecomes lrger, the denomintor of the frction grows lrger, nd E ecomes smller. On the other hnd, if 0, which corresponds to sliding the r to the left until its left end is t the origin, then E `. Tht represents the condition in which the oservtion point P is t zero distnce from the chrge t the end of the rod, so the field ecomes infinite. We explore lrge vlues of elow. Wht If? de 5 k dq e x 5 k l dx 2 e x 2,1 Find the totl field t P using 4 Eqution 23.11: E 5 3 k e l dx Noting tht k e nd l 5 Q /, re constnts nd cn e removed from the integrl, evlute the integrl: uppose point P is very fr wy from the rod. Wht is the nture of the electric field t such point? Answer If P is fr from the rod (..,), then, in the denomintor of Eqution (1) cn e neglected nd E < k e Q/ 2. Tht is exctly the form you would expect for point chrge. Therefore, t lrge vlues of /,, the chrge distriution ppers to e point chrge of mgnitude Q ; the point P is so fr wy from the rod we cnnot distinguish tht it hs size. The use of the limiting technique (/, `) is often good method for checking mthemticl expression. Exmple 23.8 A ring of rdius crries uniformly distriuted positive totl chrge Q. Clculte the electric field due to the ring t point P lying distnce x from its center long the centrl xis perpendiculr to the plne of the ring (Fig ). olution Conceptulize Figure shows the electric field contriution d E t P due to single segment of chrge t the top of the ring. This field vector cn e resolved into components de x prllel to The Electric Field of Uniform Ring of Chrge dq x x 2,1 dx E 5 k e l 3 x 5 k 2 e l c2 1,1 x d (1) E 5 k Q e, 1 2 1, 1 5 k e Q 1, 1 2 r u P de x de de x 2 1 x de 2 u Figure (Exmple 23.8) A uniformly chrged ring of rdius. () The field t P on the x xis due to n element of chrge dq. () The totl electric field t P is long the x xis. The perpendiculr component of the field t P due to segment 1 is cnceled y the perpendiculr component due to segment 2. de 1 x 4 To crry out integrtions such s this one, first express the chrge element dq in terms of the other vriles in the integrl. (In this exmple, there is one vrile, x, so we mde the chnge dq 5 l dx.) The integrl must e over sclr quntities; therefore, express the electric field in terms of components, if necessry. (In this exmple, the field hs only n x component, so this detil is of no concern.) Then, reduce your expression to n integrl over single vrile (or to multiple integrls, ech over single vrile). In exmples tht hve sphericl or cylindricl symmetry, the single vrile is rdil coordinte.

50 23.5 Electric Field of Continuous Chrge Distriution continued the xis of the ring nd de perpendiculr to the xis. Figure shows the electric field contriutions from two segments on opposite sides of the ring. Becuse of the symmetry of the sitution, the perpendiculr components of the field cncel. Tht is true for ll pirs of segments round the ring, so we cn ignore the perpendiculr component of the field nd focus solely on the prllel components, which simply dd. Ctegorize Becuse the ring is continuous, we re evluting the field due to continuous chrge distriution rther thn group of individul chrges. Anlyze Evlute the prllel component of n electric field contriution from segment of chrge dq on the ring: Wht If? uppose negtive chrge is plced t the center of the ring in Figure nd displced slightly y distnce x,, long the x xis. When the chrge is relesed, wht type of motion does it exhiit? Answer In the expression for the field due to ring of chrge, let x,,, which results in E x 5 k eq x 3 (1) de x 5 k dq e r cos u 5 k dq 2 e 2 1 x cos u 2 From the geometry in Figure 23.16, evlute cos u: (2) cos u 5 x r 5 x x 2 2 1/2 ustitute Eqution (2) into Eqution (1): All segments of the ring mke the sme contriution to the field t P ecuse they re ll equidistnt from this point. Integrte over the circumference of the ring to otin the totl field t P : dq de x 5 k e 2 1 x c x x 2 2 d 5 1/2 E x 5 3 (3) E 5 k e x k dq x 2 3/2 2 k e x x 2 2 Q 3/2 e x x 2 2 3/2 3 dq k e x dq x 2 3/2 2 Finlize This result shows tht the field is zero t x 5 0. Is tht consistent with the symmetry in the prolem? Furthermore, notice tht Eqution (3) reduces to k e Q /x 2 if x.., so the ring cts like point chrge for loctions fr wy from the ring. From frwy point, we cnnot distinguish the ring shpe of the chrge. Exmple 23.9 The Electric Field of Uniformly Chrged Disk A disk of rdius R hs uniform surfce chrge density s. Clculte the electric field t point P tht lies long the centrl perpendiculr xis of the disk nd distnce x from the center of the disk (Fig ). Therefore, from Eqution 23.8, the force on chrge 2q plced ner the center of the ring is F x 5 2 k eqq x 3 Becuse this force hs the form of Hooke s lw (Eq. 15.1), the motion of the negtive chrge is descried with the prticle in simple hrmonic motion model! r P olution x x Figure (Exmple 23.9) A uniformly chrged disk of rdius R. Conceptulize If the disk is considered to e dr The electric field t n xil point P set of concentric rings, we cn use our result is directed long the centrl xis, perpendiculr to the plne of the disk. from Exmple 23.8 which gives the field creted y single ring of rdius nd sum the contriutions of ll rings mking up the disk. By symmetry, the field t n xil point must e long the centrl xis. continued R dq

51 708 Chpter 23 Electric Fields 23.9 continued Ctegorize Becuse the disk is continuous, we re evluting the field due to continuous chrge distriution rther thn group of individul chrges. Anlyze Find the mount of chrge dq on the surfce re of ring of rdius r nd width dr s shown in Figure 23.17: dq 5 s da 5 s12pr dr2 5 2psr dr Use this result in the eqution given for E x in Exmple 23.8 (with replced y r nd Q replced y dq) to find the field due to the ring: To otin the totl field t P, integrte this expression over the limits r 5 0 to r 5 R, noting tht x is constnt in this sitution: Finlize This result is vlid for ll vlues of x. 0. For lrge vlues of x, the result ove cn e evluted y series expnsion nd shown to e equivlent to the electric field of point chrge Q. We cn clculte the field close to the disk long the xis y ssuming x,, R ; in this cse, the expression in rckets reduces to unity to give us the nerfield pproximtion E 5 2pk e s 5 s 2P 0 where P 0 is the permittivity of free spce. In Chpter 24, we otin the sme result for the field creted y n infinite plne of chrge with uniform surfce chrge density. Wht If? de x 5 Wht if we let the rdius of the disk grow so tht the disk ecomes n infinite plne of chrge? Answer The result of letting R ` in the finl result of the exmple is tht the mgnitude of the electric field ecomes E 5 2pk e s 5 s 2P 0 k e x 12psr dr2 1r 2 1 x 2 3/2 2 R 2r dr E x 5 k e x ps 3 1r 2 1 x 2 2 3/2 This is the sme expression tht we otined for x,, R. If R `, everywhere is ner-field the result is independent of the position t which you mesure the electric field. Therefore, the electric field due to n infinite plne of chrge is uniform throughout spce. An infinite plne of chrge is impossile in prctice. If two plnes of chrge re plced close to ech other, however, with one plne positively chrged, nd the other negtively, the electric field etween the pltes is very close to uniform t points fr from the edges. uch configurtion will e investigted in Chpter Electric Field Lines 0 R 5 k e x ps 3 1r 2 1 x /2 d1r k e x ps c 1r 2 1 x /2 d 21/2 R 0 5 2pk e s c1 2 x 1R 2 1 x 2 2 1/2 d We hve defined the electric field in the mthemticl representtion with Eqution Let s now explore mens of visulizing the electric field in pictoril representtion. A convenient wy of visulizing electric field ptterns is to drw lines, clled electric field lines nd first introduced y Frdy, tht re relted to the electric field in region of spce in the following mnner: The electric field vector E is tngent to the electric field line t ech point. The line hs direction, indicted y n rrowhed, tht is the sme s tht

52 23.6 Electric Field Lines 709 of the electric field vector. The direction of the line is tht of the force on positive chrge plced in the field ccording to the prticle in field model. The numer of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of the electric field in tht region. Therefore, the field lines re close together where the electric field is strong nd fr prt where the field is wek. These properties re illustrted in Figure The density of field lines through surfce A is greter thn the density of lines through surfce B. Therefore, the mgnitude of the electric field is lrger on surfce A thn on surfce B. Furthermore, ecuse the lines t different loctions point in different directions, the field is nonuniform. Is this reltionship etween strength of the electric field nd the density of field lines consistent with Eqution 23.9, the expression we otined for E using Coulom s lw? To nswer this question, consider n imginry sphericl surfce of rdius r concentric with point chrge. From symmetry, we see tht the mgnitude of the electric field is the sme everywhere on the surfce of the sphere. The numer of lines N tht emerge from the chrge is equl to the numer tht penetrte the sphericl surfce. Hence, the numer of lines per unit re on the sphere is N/4pr 2 (where the surfce re of the sphere is 4pr 2 ). Becuse E is proportionl to the numer of lines per unit re, we see tht E vries s 1/r 2 ; this finding is consistent with Eqution Representtive electric field lines for the field due to single positive point chrge re shown in Figure This two-dimensionl drwing shows only the field lines tht lie in the plne contining the point chrge. The lines re ctully directed rdilly outwrd from the chrge in ll directions; therefore, insted of the flt wheel of lines shown, you should picture n entire sphericl distriution of lines. Becuse positive chrge plced in this field would e repelled y the positive source chrge, the lines re directed rdilly wy from the source chrge. The electric field lines representing the field due to single negtive point chrge re directed towrd the chrge (Fig ). In either cse, the lines re long the rdil direction nd extend ll the wy to infinity. Notice tht the lines ecome closer together s they pproch the chrge, indicting tht the strength of the field increses s we move towrd the source chrge. The rules for drwing electric field lines re s follows: The lines must egin on positive chrge nd terminte on negtive chrge. In the cse of n excess of one type of chrge, some lines will egin or end infinitely fr wy. For positive point chrge, the field lines re directed rdilly outwrd. For negtive point chrge, the field lines re directed rdilly inwrd. The mgnitude of the field is greter on surfce A thn on surfce B. A Figure Electric field lines penetrting two surfces. Pitfll Prevention 23.2 Electric Field Lines Are Not Pths of Prticles! Electric field lines represent the field t vrious loctions. Except in very specil cses, they do not represent the pth of chrged prticle moving in n electric field. B q q Figure The electric field lines for point chrge. Notice tht the figures show only those field lines tht lie in the plne of the pge.

53 710 Chpter 23 Electric Fields Pitfll Prevention 23.3 Electric Field Lines Are Not Rel Electric field lines re not mteril ojects. They re used only s pictoril representtion to provide qulittive description of the electric field. Only finite numer of lines from ech chrge cn e drwn, which mkes it pper s if the field were quntized nd exists only in certin prts of spce. The field, in fct, is continuous, existing t every point. You should void otining the wrong impression from two-dimensionl drwing of field lines used to descrie threedimensionl sitution. The numer of field lines leving the positive chrge equls the numer terminting t the negtive chrge. Figure The electric field lines for two point chrges of equl mgnitude nd opposite sign (n electric dipole). The numer of lines drwn leving positive chrge or pproching negtive chrge is proportionl to the mgnitude of the chrge. No two field lines cn cross. We choose the numer of field lines strting from ny oject with positive chrge q 1 to e Cq 1 nd the numer of lines ending on ny oject with negtive chrge q 2 to e C uq 2 u, where C is n ritrry proportionlity constnt. Once C is chosen, the numer of lines is fixed. For exmple, in two-chrge system, if oject 1 hs chrge Q 1 nd oject 2 hs chrge Q 2, the rtio of numer of lines in contct with the chrges is N 2 /N 1 5 uq 2 /Q 1 u. The electric field lines for two point chrges of equl mgnitude ut opposite signs (n electric dipole) re shown in Figure Becuse the chrges re of equl mgnitude, the numer of lines tht egin t the positive chrge must equl the numer tht terminte t the negtive chrge. At points very ner the chrges, the lines re nerly rdil, s for single isolted chrge. The high density of lines etween the chrges indictes region of strong electric field. Figure shows the electric field lines in the vicinity of two equl positive point chrges. Agin, the lines re nerly rdil t points close to either chrge, nd the sme numer of lines emerges from ech chrge ecuse the chrges re equl in mgnitude. Becuse there re no negtive chrges ville, the electric field lines end infinitely fr wy. At gret distnces from the chrges, the field is pproximtely equl to tht of single point chrge of mgnitude 2q. Finlly, in Figure 23.22, we sketch the electric field lines ssocited with positive chrge 12q nd negtive chrge 2q. In this cse, the numer of lines leving 12q is twice the numer terminting t 2q. Hence, only hlf the lines tht leve the positive chrge rech the negtive chrge. The remining hlf terminte on negtive chrge we ssume to e t infinity. At distnces much greter thn the chrge seprtion, the electric field lines re equivlent to those of single chrge 1q. Q uick Quiz 23.5 Rnk the mgnitudes of the electric field t points A, B, nd C shown in Figure (gretest mgnitude first). A C B Figure The electric field lines for two positive point chrges. (The loctions A, B, nd C re discussed in Quick Quiz 23.5.) Two field lines leve 2q for every one tht termintes on q. 2q q Figure The electric field lines for point chrge +2q nd second point chrge 2q Motion of Chrged Prticle in Uniform Electric Field When prticle of chrge q nd mss m is plced in n electric field E, the electric force exerted on the chrge is q E ccording to Eqution 23.8 in the prticle in

54 23.7 Motion of Chrged Prticle in Uniform Electric Field 711 field model. If tht is the only force exerted on the prticle, it must e the net force, nd it cuses the prticle to ccelerte ccording to the prticle under net force model. Therefore, e F 5 q E 5 m nd the ccelertion of the prticle is 5 q E (23.12) m If E is uniform (tht is, constnt in mgnitude nd direction), nd the prticle is free to move, the electric force on the prticle is constnt nd we cn pply the prticle under constnt ccelertion model to the motion of the prticle. Therefore, the prticle in this sitution is descried y three nlysis models: prticle in field, prticle under net force, nd prticle under constnt ccelertion! If the prticle hs positive chrge, its ccelertion is in the direction of the electric field. If the prticle hs negtive chrge, its ccelertion is in the direction opposite the electric field. Exmple An Accelerting Positive Chrge: Two Models A uniform electric field E is directed long the x xis etween prllel pltes of chrge seprted y distnce d s shown in Figure A positive point chrge q of mss m is relesed from rest t point A next to the positive plte nd ccelertes to point B next to the negtive plte. (A) Find the speed of the prticle t B y modeling it s prticle under constnt ccelertion. olution Pitfll Prevention 23.4 Just Another Force Electric forces nd fields my seem strct to you. Once F e is evluted, however, it cuses prticle to move ccording to our well-estlished models of forces nd motion from Chpters 2 through 6. Keeping this link with the pst in mind should help you solve prolems in this chpter. Conceptulize When the positive chrge is plced t A, it experiences n electric force towrd the right in Figure Figure (Exmple 23.10) due to the electric field directed towrd the right. As A positive point chrge q in uniform electric field E undergoes result, it will ccelerte to the right nd rrive t B with some speed. constnt ccelertion in the direction of the field. d Ctegorize Becuse the electric field is uniform, constnt electric force cts on the chrge. Therefore, s suggested in the discussion preceding the exmple nd in the prolem sttement, the point chrge cn e modeled s chrged prticle under constnt ccelertion. Anlyze Use Eqution 2.17 to express the velocity of the prticle s function of position: AM v 0 A v f 2 5 v i 2 1 2(x f 2 x i ) (d 2 0) 5 2d olve for v f nd sustitute for the mgnitude of the ccelertion from Eqution 23.12: v f 5 "2d 5 2 qe Å m d 5 2qEd Å m E v q B (B) Find the speed of the prticle t B y modeling it s nonisolted system in terms of energy. olution Ctegorize The prolem sttement tells us tht the chrge is nonisolted system for energy. The electric force, like ny force, cn do work on system. Energy is trnsferred to the system of the chrge y work done y the electric force exerted on the chrge. The initil configurtion of the system is when the prticle is t rest t A, nd the finl configurtion is when it is moving with some speed t B. continued

55 712 Chpter 23 Electric Fields continued Anlyze Write the pproprite reduction of the conservtion of energy eqution, Eqution 8.2, for the system of the chrged prticle: Replce the work nd kinetic energies with vlues pproprite for this sitution: ustitute for the mgnitude of the electric force F e from the prticle in field model nd the displcement Dx: Exmple An Accelerted Electron AM An electron enters the region of uniform electric field s shown in Figure 23.24, with v i m/s nd E N/C. The horizontl length of the pltes is, m. (A) Find the ccelertion of the electron while it is in the electric field. olution Conceptulize This exmple differs from the preceding one ecuse the velocity of the chrged prticle is initilly perpendiculr to the electric field lines. (In Exmple 23.10, the velocity of the chrged prticle is lwys prllel to the electric field lines.) As result, the electron in this exmple follows curved pth s shown in Figure The motion of the electron is the sme s tht of mssive prticle projected horizontlly in grvittionl field ner the surfce of the Erth. W 5 DK F e Dx 5 K B 2 K A 5 1 2mv f v f 5 Å 2F e Dx m v f 5 Å 21qE2 1d2 m 5 Å 2qEd m Finlize The nswer to prt (B) is the sme s tht for prt (A), s we expect. This prolem cn e solved with different pproches. We sw the sme possiilities with mechnicl prolems. Ctegorize The electron is prticle in field (electric). Becuse the electric field is uniform, constnt electric force is exerted on the electron. To find the ccelertion of the electron, we cn model it s prticle under net force. Anlyze From the prticle in field model, we know tht the direction of the electric force on the electron is downwrd in Figure 23.24, opposite the direction of the electric field lines. From the prticle under net force model, therefore, the ccelertion of the electron is downwrd. The prticle under net force model ws used to develop y 5 2 ee m Eqution in the cse in which the electric force on e prticle is the only force. Use this eqution to evlute the y component of the ccelertion of the electron: ustitute numericl vlues: y C N/C m/s kg v i î The electron undergoes downwrd ccelertion (opposite E), nd its motion is prolic while it is etween the pltes. (0, 0) (x,y) E Figure (Exmple 23.11) An electron is projected horizontlly into uniform electric field produced y two chrged pltes. (B) Assuming the electron enters the field t time t 5 0, find the time t which it leves the field. y v x olution Ctegorize Becuse the electric force cts only in the verticl direction in Figure 23.24, the motion of the prticle in the horizontl direction cn e nlyzed y modeling it s prticle under constnt velocity.

56 ummry continued Anlyze olve Eqution 2.7 for the time t which the electron rrives t the right edges of the pltes: x f 5 x i 1 v x t t 5 x f 2 x i v x ustitute numericl vlues: t 5, m 5 v x m/s s (C) Assuming the verticl position of the electron s it enters the field is y i 5 0, wht is its verticl position when it leves the field? olution Ctegorize Becuse the electric force is constnt in Figure 23.24, the motion of the prticle in the verticl direction cn e nlyzed y modeling it s prticle under constnt ccelertion. Anlyze Use Eqution 2.16 to descrie the position of the prticle t ny time t: y f 5 y i 1 v yi t y t 2 ustitute numericl vlues: y f m/s s m cm Finlize If the electron enters just elow the negtive plte in Figure nd the seprtion etween the pltes is less thn the vlue just clculted, the electron will strike the positive plte. Notice tht we hve used four nlysis models to descrie the electron in the vrious prts of this prolem. We hve neglected the grvittionl force cting on the electron, which represents good pproximtion when deling with tomic prticles. For n electric field of 200 N/C, the rtio of the mgnitude of the electric force ee to the mgnitude of the grvittionl force mg is on the order of for n electron nd on the order of 10 9 for proton. ummry Definitions The electric field E t some point in spce is defined s the electric force F e tht cts on smll positive test chrge plced t tht point divided y the mgnitude q 0 of the test chrge: E ; F e q 0 (23.7) Concepts nd Principles Electric chrges hve the following importnt properties: Chrges of opposite sign ttrct one nother, nd chrges of the sme sign repel one nother. The totl chrge in n isolted system is conserved. Chrge is quntized. Conductors re mterils in which electrons move freely. Insultors re mterils in which electrons do not move freely. continued

57 714 Chpter 23 Electric Fields Coulom s lw sttes tht the electric force exerted y point chrge q 1 on second point chrge q 2 is 12 F 5 k q 1q 2 e r 2 r^12 (23.6) where r is the distnce etween the two chrges nd r^12 is unit vector directed from q 1 towrd q 2. The constnt k e, which is clled the Coulom constnt, hs the vlue k e N? m 2 /C 2. The electric field due to group of point chrges cn e otined y using the superposition principle. Tht is, the totl electric field t some point equls the vector sum of the electric fields of ll the chrges: E 5 ke i q i r i 2 r^i (23.10) Anlysis Models for Prolem olving At distnce r from point chrge q, the electric field due to the chrge is E q 5 ke 2 r^ (23.9) r where r^ is unit vector directed from the chrge towrd the point in question. The electric field is directed rdilly outwrd from positive chrge nd rdilly inwrd towrd negtive chrge. The electric field t some point due to continuous chrge distriution is E dq 5 ke 3 2 r^ (23.11) r where dq is the chrge on one element of the chrge distriution nd r is the distnce from the element to the point in question. Prticle in Field (Electric) A source prticle with some electric chrge estlishes n electric field E throughout spce. When prticle with chrge q is plced in tht field, it experiences n electric force given y Ojective Questions 1. A free electron nd free proton re relesed in identicl electric fields. (i) How do the mgnitudes of the electric force exerted on the two prticles compre? () It is millions of times greter for the electron. () It is thousnds of times greter for the electron. (c) They re equl. (d) It is thousnds of times smller for the electron. (e) It is millions of times smller for the electron. (ii) Compre the mgnitudes of their ccelertions. Choose from the sme possiilities s in prt (i). 2. Wht prevents grvity from pulling you through the ground to the center of the Erth? Choose the est nswer. () The density of mtter is too gret. () The positive nuclei of your ody s toms repel the positive nuclei of the toms of the ground. (c) The density of the ground is greter thn the density of your ody. (d) Atoms re ound together y chemicl onds. (e) Electrons on the ground s surfce nd the surfce of your feet repel one nother. 3. A very smll ll hs mss of kg nd chrge of 4.00 mc. Wht mgnitude electric field directed upwrd will lnce the weight of the ll so tht the ll is suspended motionless ove the ground? () N/C () N/C (c) N/C (d) N/C (e) N/C 4. An electron with speed of m/s moves into uniform electric field of mgnitude N/C. Fe 5 q E (23.8) 1. denotes nswer ville in tudent olutions Mnul/tudy Guide The field lines re prllel to the electron s velocity nd pointing in the sme direction s the velocity. How fr does the electron trvel efore it is rought to rest? () 2.56 cm () 5.12 cm (c) 11.2 cm (d) 3.34 m (e) 4.24 m 5. A point chrge of nc is locted t (0, 1.00) m. Wht is the x component of the electric field due to the point chrge t (4.00, 22.00) m? () 1.15 N/C () N/C (c) 1.44 N/C (d) N/C (e) N/C 6. A circulr ring of chrge with rdius hs totl chrge q uniformly distriuted round it. Wht is the mgnitude of the electric field t the center of the ring? () 0 () k e q/ 2 (c) k e q 2 / 2 (d) k e q 2 / (e) none of those nswers 7. Wht hppens when chrged insultor is plced ner n unchrged metllic oject? () They repel ech other. () They ttrct ech other. (c) They my ttrct or repel ech other, depending on whether the chrge on the insultor is positive or negtive. (d) They exert no electrosttic force on ech other. (e) The chrged insultor lwys spontneously dischrges. 8. Estimte the mgnitude of the electric field due to the proton in hydrogen tom t distnce of m, the expected position of the electron in the tom. () N/C () 10 8 N/C (c) N/C (d) 10 6 N/C (e) N/C E q F e qe

58 Conceptul Questions (i) A metllic coin is given positive electric chrge. Does its mss () increse mesurly, () increse y n mount too smll to mesure directly, (c) remin unchnged, (d) decrese y n mount too smll to mesure directly, or (e) decrese mesurly? (ii) Now the coin is given negtive electric chrge. Wht hppens to its mss? Choose from the sme possiilities s in prt (i). 10. Assume the chrged ojects in Figure OQ23.10 re fixed. Notice tht there is no sight line from the loction of q 2 to the loction of q 1. If you were t q 1, you would e unle to see q 2 ecuse it is ehind q 3. How would you clculte the electric force exerted on the oject with chrge q 1? () Find only the force exerted y q 2 on chrge q 1. () Find only the force exerted y q 3 on chrge q 1. (c) Add the force tht q 2 would exert y itself on chrge q 1 to the force tht q 3 would exert y itself on chrge q 1. (d) Add the force tht q 3 would exert y itself to certin frction of the force tht q 2 would exert y itself. (e) There is no definite wy to find the force on chrge q 1. q 1 q3 q 2 Figure OQ Three chrged prticles () (e) re rrnged on corners of Q () squre s shown in Figure OQ23.11, with chrge (d) (c) 2Q on oth the prticle t the upper left corner nd the prticle t the lower 2Q Q right corner nd with chrge 12Q on the prticle Figure OQ23.11 t the lower left corner. (i) Wht is the direction of the electric field t the upper right corner, which is point in empty spce? () It is upwrd nd to the right. () It is stright to the right. (c) It is stright downwrd. (d) It is downwrd nd to the left. (e) It is perpendiculr to the plne of the picture nd outwrd. (ii) uppose the 12Q chrge t the lower left corner is removed. Then does the mgnitude of the field t the upper right corner () ecome lrger, () ecome smller, (c) sty the sme, or (d) chnge unpredictly? Conceptul Questions x 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 12. Two point chrges ttrct ech other with n electric force of mgnitude F. If the chrge on one of the prticles is reduced to one-third its originl vlue nd the distnce etween the prticles is douled, wht is the resulting mgnitude of the electric force etween them? () 1 12 F () 1 3 F (c) 1 6 F (d) 3 4 F (e) 3 2 F 13. Assume uniformly chrged ring of rdius R nd chrge Q produces n electric field E ring t point P on its xis, t distnce x wy from the center of the ring s in Figure OQ Now the sme chrge Q is spred uniformly over the circulr re the ring encloses, forming flt disk of chrge with the sme rdius s in Figure OQ How does the field E disk produced y the disk t P compre with the field produced y the ring t the sme point? () E disk, E ring () E disk 5 E ring (c) E disk. E ring (d) impossile to determine R R Q Q x x P P Figure OQ23.13 Ering Edisk 14. An oject with negtive chrge is plced in region of spce where the electric field is directed verticlly upwrd. Wht is the direction of the electric force exerted on this chrge? () It is up. () It is down. (c) There is no force. (d) The force cn e in ny direction. 15. The mgnitude of the electric force etween two protons is N. How fr prt re they? () m () m (c) 3.10 m (d) m (e) m x x 1. () Would life e different if the electron were positively chrged nd the proton were negtively chrged? () Does the choice of signs hve ny ering on physicl nd chemicl interctions? Explin your nswers. 2. A chrged com often ttrcts smll its of dry pper tht then fly wy when they touch the com. Explin why tht occurs. 3. A person is plced in lrge, hollow, metllic sphere tht is insulted from ground. If lrge chrge is plced on the sphere, will the person e hrmed upon touching the inside of the sphere? 4. A student who grew up in tropicl country nd is studying in the United ttes my hve no experience with sttic electricity sprks nd shocks until his or her first Americn winter. Explin. 5. If suspended oject A is ttrcted to chrged oject B, cn we conclude tht A is chrged? Explin.

59 716 Chpter 23 Electric Fields 6. Consider point A in A Figure CQ23.6 locted n ritrry distnce from two positive point chrges in otherwise empty spce. () Is it possile for n electric field to exist t point A in empty spce? Explin. () Does chrge exist t this point? Explin. (c) Does force exist t Figure CQ23.6 this point? Explin. 7. In fir wether, there is n electric field t the surfce of the Erth, pointing down into the ground. Wht is the sign of the electric chrge on the ground in this sitution? Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign ection 23.1 Properties of Electric Chrges 1. Find to three significnt digits the chrge nd the mss of the following prticles. uggestion: Begin y looking up the mss of neutrl tom on the periodic tle of the elements in Appendix C. () n ionized hydrogen tom, represented s H 1 () singly ionized sodium tom, N 1 (c) chloride ion Cl 2 (d) douly ionized clcium tom, C 11 5 C 21 (e) the center of n mmoni molecule, modeled s n N 32 ion (f) qudruply ionized nitrogen toms, N 41, found in plsm in hot str (g) the nucleus of nitrogen tom (h) the moleculr ion H 2 O 2 2. () Clculte the numer of electrons in smll, electriclly neutrl silver pin tht hs mss of 10.0 g. W ilver hs 47 electrons per tom, nd its molr mss is g/mol. () Imgine dding electrons to the pin until the negtive chrge hs the very lrge vlue 1.00 mc. How mny electrons re dded for every 10 9 electrons lredy present? ection 23.2 Chrging Ojects y Induction ection 23.3 Coulom s Lw 3. Two protons in n tomic nucleus re typiclly seprted y distnce of m. The electric repulsive force etween the protons is huge, ut the ttrctive nucler force is even stronger nd keeps the nucleus from ursting prt. Wht is the mgnitude of the electric force etween two protons seprted y m? 8. Why must hospitl personnel wer specil conducting shoes while working round oxygen in n operting room? Wht might hppen if the personnel wore shoes with ruer soles? 9. A lloon clings to wll fter it is negtively chrged y ruing. () Does tht occur ecuse the wll is positively chrged? () Why does the lloon eventully fll? 10. Consider two electric dipoles in empty spce. Ech dipole hs zero net chrge. () Does n electric force exist etween the dipoles; tht is, cn two ojects with zero net chrge exert electric forces on ech other? () If so, is the force one of ttrction or of repulsion? 11. A glss oject receives positive chrge y ruing it with silk cloth. In the ruing process, hve protons een dded to the oject or hve electrons een removed from it? BIO Q/C 4. A chrged prticle A exerts force of 2.62 mn to the right on chrged prticle B when the prticles re 13.7 mm prt. Prticle B moves stright wy from A to mke the distnce etween them 17.7 mm. Wht vector force does it then exert on A? 5. In thundercloud, there my e electric chrges of C ner the top of the cloud nd C ner the ottom of the cloud. These chrges re seprted y 2.00 km. Wht is the electric force on the top chrge? 6. () Find the mgnitude of the electric force etween Q/C N 1 ion nd Cl 2 ion seprted y 0.50 nm. () Would the nswer chnge if the sodium ion were replced y Li 1 nd the chloride ion y Br 2? Explin. 7. Review. A molecule of DNA (deoxyrionucleic cid) is BIO 2.17 mm long. The ends of the molecule ecome singly ionized: negtive on one end, positive on the other. The helicl molecule cts like spring nd compresses 1.00% upon ecoming chrged. Determine the effective spring constnt of the molecule. 8. Noel lurete Richrd Feynmn ( ) once sid tht if two persons stood t rm s length from ech other nd ech person hd 1% more electrons thn protons, the force of repulsion etween them would e enough to lift weight equl to tht of the entire Erth. Crry out n order-of-mgnitude clcultion to sustntite this ssertion. 9. A 7.50-nC point chrge is locted 1.80 m from Q/C 4.20-nC point chrge. () Find the mgnitude of the

60 Prolems 717 electric force tht one prticle exerts on the other. () Is the force ttrctive or repulsive? 10. () Two protons in molecule re m W prt. Find the mgnitude of the electric force exerted Q/C y one proton on the other. () tte how the mgnitude of this force compres with the mgnitude of the grvittionl force exerted y one proton on the other. (c) Wht If? Wht must e prticle s chrge-tomss rtio if the mgnitude of the grvittionl force etween two of these prticles is equl to the mgnitude of electric force etween them? 11. Three point chrges re rrnged s shown in Figure M P Find () the mgnitude nd () the direction of the electric force on the prticle t the origin. y 5.00 nc 6.00 nc m x m 3.00 nc Figure P23.11 Prolems 11 nd Three point chrges lie long stright line s shown in Figure P23.12, where q mc, q mc, nd q mc. The seprtion distnces re d cm nd d cm. Clculte the mgnitude nd direction of the net electric force on () q 1, () q 2, nd (c) q 3. q 1 d 1 q 2 Figure P Two smll eds hving positive chrges q 1 5 3q nd W q 2 5 q re fixed t the opposite ends of horizontl Q/C insulting rod of length d m. The ed with chrge q 1 is t the origin. As shown in Figure P23.13, third smll, chrged ed is free to slide on the rod. () At wht position x is the third ed in equilirium? () Cn the equilirium e stle? d 2 q 3 x q 1 q 2 x d Figure P23.13 Prolems 13 nd Two smll eds hving chrges q 1 nd q 2 of the sme Q/C sign re fixed t the opposite ends of horizontl insulting rod of length d. The ed with chrge q 1 is t the origin. As shown in Figure P23.13, third smll, chrged ed is free to slide on the rod. () At wht position x is the third ed in equilirium? () Cn the equilirium e stle? 15. Three chrged prticles re locted t the corners of M n equilterl tringle s shown in Figure P Clculte the totl electric force on the 7.00-mC chrge. y 7.00 mc m 60.0 x 2.00 mc 4.00 mc Figure P23.15 Prolems 15 nd Two smll metllic spheres, ech of mss m g, re suspended s pendulums y light strings of length L s shown in Figure P The L θ spheres re given the sme electric chrge of 7.2 nc, nd they come to equilirium when ech string is t m m n ngle of u with the verticl. How long re the strings? Figure P Review. In the Bohr theory of the hydrogen tom, n electron moves in circulr orit out proton, where the rdius of the orit is m. () Find the mgnitude of the electric force exerted on ech prticle. () If this force cuses the centripetl ccelertion of the electron, wht is the speed of the electron? 18. Prticle A of chrge C is t the origin, prticle B of chrge C is t (4.00 m, 0), nd GP prticle C of chrge C is t (0, 3.00 m). We wish to find the net electric force on C. () Wht is the x component of the electric force exerted y A on C? () Wht is the y component of the force exerted y A on C? (c) Find the mgnitude of the force exerted y B on C. (d) Clculte the x component of the force exerted y B on C. (e) Clculte the y component of the force exerted y B on C. (f) um the two x components from prts () nd (d) to otin the resultnt x component of the electric force cting on C. (g) imilrly, find the y component of the resultnt force vector cting on C. (h) Find the mgnitude nd direction of the resultnt electric force cting on C. 19. A point chrge 12Q is t y the origin nd point chrge 2Q is locted Q long the x xis t x 5 d s in Figure P Find d symolic expression for the net force on third x d point chrge 1Q locted 2Q Q long the y xis t y 5 d. Figure P Review. Two identicl prticles, ech hving chrge 1q, re fixed in spce nd seprted y distnce d. A third prticle with chrge 2Q is free to move nd lies initilly t rest on the

61 718 Chpter 23 Electric Fields perpendiculr isector of the two fixed chrges distnce x from the midpoint etween those chrges (Fig. P23.20). () how tht if x is smll compred with d, the motion of 2Q is simple hrmonic long the perpendiculr isector. () Determine the period of tht motion. (c) How fst will the chrge 2Q e moving when it is t the midpoint etween the two fixed chrges if initilly it is relesed t distnce,, d from the midpoint? 21. Two identicl conducting smll spheres re plced with W their centers m prt. One is given chrge of 12.0 nc nd the other chrge of nc. () Find the electric force exerted y one sphere on the other. () Wht If? The spheres re connected y conducting wire. Find the electric force ech exerts on the other fter they hve come to equilirium. 22. Why is the following sitution impossile? Two identicl dust prticles of mss 1.00 mg re floting in empty spce, fr from ny externl sources of lrge grvittionl or electric fields, nd t rest with respect to ech other. Both prticles crry electric chrges tht re identicl in mgnitude nd sign. The grvittionl nd electric forces etween the prticles hppen to hve the sme mgnitude, so ech prticle experiences zero net force nd the distnce etween the prticles remins constnt. ection 23.4 Anlysis Model: Prticle in Field (Electric) 23. Wht re the mgnitude nd direction of the electric field tht will lnce the weight of () n electron nd () proton? (You my use the dt in Tle 23.1.) 24. A smll oject of mss 3.80 g nd chrge mc is suspended motionless ove the ground when immersed in uniform electric field perpendiculr to the ground. Wht re the mgnitude nd direction of the electric field? 25. Four chrged prticles re t the corners of squre of side s shown in Figure P Determine () the electric field t the loction of chrge q nd () the totl electric force exerted on q. 2q q q d 2 d 2 y q x Q Figure P23.20 x q 27. Two equl positively Q d P chrged prticles re t opposite corners of trpezoid s shown in Figure P 2d Q P Find symolic expressions for the totl Figure P23.27 electric field t () the point P nd () the point P Consider n equl positively chrged prticles ech of Q/C mgnitude Q /n plced symmetriclly round circle of rdius. () Clculte the mgnitude of the electric field t point distnce x from the center of the circle nd on the line pssing through the center nd perpendiculr to the plne of the circle. () Explin why this result is identicl to the result of the clcultion done in Exmple In Figure P23.29, determine the point (other thn M infinity) t which the electric field is zero mc r 270 y 150 q 30 x 2q Figure P m 6.00 mc Figure P Three chrged prticles re t the corners of n equilterl tringle s shown in Figure P () Clcu- W lte the electric field t the position of the 2.00-mC chrge due to the 7.00-mC nd mC chrges. () Use your nswer to prt () to determine the force on the 2.00-mC chrge. 31. Three point chrges re locted on circulr rc s shown in Figure P () Wht is the totl electric field t P, the center of the rc? () Find the electric force tht would e exerted on nC point chrge plced t P nc 4.00 cm 3q 4q Figure P Three point chrges lie long circle of rdius r t ngles of 308, 1508, nd 2708 s shown in Figure P Find symolic expression for the resultnt electric field t the center of the circle nc cm 3.00 nc Figure P23.31 P

62 Prolems Two chrged prticles re locted on the x xis. The first Q/C is chrge 1Q t x 5 2. The second is n unknown chrge locted t x The net electric field these chrges produce t the origin hs mgnitude of 2k e Q / 2. Explin how mny vlues re possile for the unknown chrge nd find the possile vlues. 33. A smll, 2.00-g plstic ll is suspended y 20.0-cmlong string in uniform electric field s shown in Fig- AMT ure P If the ll is in equilirium when the string mkes 15.0 ngle with the verticl, wht is the net chrge on the ll? y x 15.0 E = N/C L m = 2.00 g Figure P Two 2.00-mC point chrges re locted on the x xis. One is t x m, nd the other is t x m. () Determine the electric field on the y xis t y m. () Clculte the electric force on mC chrge plced on the y xis t y m. 35. Three point chrges re rrnged s shown in Figure P () Find the vector electric field tht the 6.00-nC nd nC chrges together crete t the origin. () Find the vector force on the 5.00-nC chrge. 36. Consider the electric dipole shown in Figure P how tht the electric field t distnt point on the 1x xis is E x < 4k e q/x 3. y q q 2 Figure P23.36 ection 23.5 Electric Field of Continuous Chrge Distriution 37. A rod 14.0 cm long is uniformly chrged nd hs totl W chrge of mc. Determine () the mgnitude nd () the direction of the electric field long the xis of the rod t point 36.0 cm from its center. 38. A uniformly chrged disk of rdius 35.0 cm crries chrge with density of C/m 2. Clculte the electric field on the xis of the disk t () 5.00 cm, () 10.0 cm, (c) 50.0 cm, nd (d) 200 cm from the center of the disk. 39. A uniformly chrged ring of rdius 10.0 cm hs totl M chrge of 75.0 mc. Find the electric field on the xis of x the ring t () 1.00 cm, () 5.00 cm, (c) 30.0 cm, nd (d) 100 cm from the center of the ring. 40. The electric field long the xis of uniformly chrged disk of rdius R nd totl chrge Q ws clculted in Exmple how tht the electric field t distnces x tht re lrge compred with R pproches tht of prticle with chrge Q 5 spr 2. uggestion: First show tht x/(x 2 1 R 2 ) 1/2 5 (1 1 R 2 /x 2 ) 21/2 nd use the inomil expnsion (1 1 d) n < 1 1 nd, when d,, Exmple 23.9 derives the exct expression for the Q/C electric field t point on the xis of uniformly chrged disk. Consider disk of rdius R cm hving uniformly distriuted chrge of mc. () Using the result of Exmple 23.9, compute the electric field t point on the xis nd 3.00 mm from the center. () Wht If? Explin how the nswer to prt () compres with the field computed from the ner-field pproximtion E 5 s/2p 0. (We derived this expression in Exmple 23.9.) (c) Using the result of Exmple 23.9, compute the electric field t point on the xis nd 30.0 cm from the center of the disk. (d) Wht If? Explin how the nswer to prt (c) compres with the electric field otined y treting the disk s mC chrged prticle t distnce of 30.0 cm. 42. A uniformly chrged y Q/C rod of length L nd totl chrge Q lies long the x P xis s shown in Figure P () Find the components of the electric d field t the point P on the x O y xis distnce d from L the origin. () Wht re Figure P23.42 the pproximte vlues of the field components when d.. L? Explin why you would expect these results. 43. A continuous line of chrge lies long the x xis, W extending from x 5 1x 0 to positive infinity. The line crries positive chrge with uniform liner chrge density l 0. Wht re () the mgnitude nd () the direction of the electric field t the origin? 44. A thin rod of length, nd uniform chrge per unit length l lies long the x xis s shown in Figure P () how tht the electric field t P, distnce d from the rod long its perpendiculr isector, hs no x d y O P u 0 Figure P23.44 x

63 720 Chpter 23 Electric Fields component nd is given y E 5 2k e l sin u 0 /d. () Wht If? Using your result to prt (), show tht the field of rod of infinite length is E 5 2k e l/d. 45. A uniformly chrged insulting rod M of length 14.0 cm is ent into the shpe of semicircle s shown in Figure P The rod hs totl chrge O of mc. Find () the mgnitude nd () the direction of the electric field t O, the center of the semicircle. 46. () Consider uniformly chrged, Figure P23.45 thin-wlled, right circulr cylindricl shell hving totl chrge Q, rdius R, nd length,. Determine the electric field t point distnce d from the right side of the cylinder s shown in Figure P uggestion: Use the result of Exmple 23.8 nd tret the cylinder s collection of ring chrges. () Wht If? Consider now solid cylinder with the sme dimensions nd crrying the sme chrge, uniformly distriuted through its volume. Use the result of Exmple 23.9 to find the field it cretes t the sme point., R Figure P23.46 ection 23.6 Electric Field Lines 47. A negtively chrged rod of finite length crries chrge with uniform chrge per unit length. ketch the electric field lines in plne contining the rod. 48. A positively chrged disk hs uniform chrge per unit re s s descried in Exmple ketch the electric field lines in plne perpendiculr to the plne of the disk pssing through its center. 49. Figure P23.49 shows the electric W field lines for two chrged prticles q 2 seprted y smll distnce. () Determine the rtio q 1 /q 2. () Wht re the signs of q 1 nd q 2? q Three equl positive chrges q re t the corners of n equilterl tringle of side s shown in Figure P Assume the Figure P23.49 three chrges together crete n P q electric field. () ketch the field lines in the plne of the chrges. () Find the loction of one point (other thn `) where the electric field is zero. Wht re (c) the mgnitude nd (d) the direction q q of the electric field t P due to the two chrges t the se? Figure P23.50 Q d ection 23.7 Motion of Chrged Prticle in Uniform Electric Field 51. A proton ccelertes from rest in uniform electric AMT field of 640 N/C. At one lter moment, its speed is M 1.20 Mm/s (nonreltivistic ecuse v is much less thn the speed of light). () Find the ccelertion of the proton. () Over wht time intervl does the proton rech this speed? (c) How fr does it move in this time intervl? (d) Wht is its kinetic energy t the end of this intervl? 52. A proton is projected in the positive x direction W into region of uniform electric field E i^ N/C t t 5 0. The proton trvels 7.00 cm s it comes to rest. Determine () the ccelertion of the proton, () its initil speed, nd (c) the time intervl over which the proton comes to rest. 53. An electron nd proton re ech plced t rest in uniform electric field of mgnitude 520 N/C. Clculte the speed of ech prticle 48.0 ns fter eing relesed. 54. Protons re projected with n initil speed v i 5 GP 9.55 km/s from field-free region through plne nd into region where uniform electric field E j^ N/C is present ove the plne s shown in Figure P The initil velocity vector of the protons mkes n ngle u with the plne. The protons re to hit trget tht lies t horizontl distnce of R mm from the point where the protons cross the plne nd enter the electric field. We wish to find the ngle u t which the protons must pss through the plne to strike the trget. () Wht nlysis model descries the horizontl motion of the protons ove the plne? () Wht nlysis model descries the verticl motion of the protons ove the plne? (c) Argue tht Eqution 4.13 would e pplicle to the protons in this sitution. (d) Use Eqution 4.13 to write n expression for R in terms of v i, E, the chrge nd mss of the proton, nd the ngle u. (e) Find the two possile vlues of the ngle u. (f) Find the time intervl during which the proton is ove the plne in Figure P23.54 for ech of the two possile vlues of u. Proton em v i u R E 720j ˆ N/C Trget E 0 elow the plne Figure P The electrons in prticle em ech hve kinetic energy K. Wht re () the mgnitude nd () the direction of the electric field tht will stop these electrons in distnce d?

64 Prolems Two horizontl metl pltes, ech 10.0 cm squre, re Q/C ligned 1.00 cm prt with one ove the other. They re given equl-mgnitude chrges of opposite sign so tht uniform downwrd electric field of N/C exists in the region etween them. A prticle of mss kg nd with positive chrge of C leves the center of the ottom negtive plte with n initil speed of m/s t n ngle of ove the horizontl. () Descrie the trjectory of the prticle. () Which plte does it strike? (c) Where does it strike, reltive to its strting point? 57. A proton moves t m/s in the horizontl M direction. It enters uniform verticl electric field with mgnitude of N/C. Ignoring ny grvittionl effects, find () the time intervl required for the proton to trvel 5.00 cm horizontlly, () its verticl displcement during the time intervl in which it trvels 5.00 cm horizontlly, nd (c) the horizontl nd verticl components of its velocity fter it hs trveled 5.00 cm horizontlly. Additionl Prolems 58. Three solid plstic cylinders ll hve rdius 2.50 cm nd length 6.00 cm. Find the chrge of ech cylinder given the following dditionl informtion out ech one. Cylinder () crries chrge with uniform density 15.0 nc/m 2 everywhere on its surfce. Cylinder () crries chrge with uniform density 15.0 nc/m 2 on its curved lterl surfce only. Cylinder (c) crries chrge with uniform density 500 nc/m 3 throughout the plstic. 59. Consider n infinite numer of identicl prticles, ech with chrge q, plced long the x xis t distnces, 2, 3, 4,... from the origin. Wht is the electric field t the origin due to this distriution? uggestion: Use p A prticle with chrge nc is t the origin, nd prticle with negtive chrge of mgnitude Q is t x cm. A third prticle with positive chrge is in equilirium t x cm. Wht is Q? 61. A smll lock of mss m Q AMT nd chrge Q is plced on m n insulted, frictionless, inclined plne of ngle u s in Figure P An electric u field is pplied prllel to the incline. () Find n expression for the mgnitude of the electric field tht Figure P23.61 enles the lock to remin t rest. () If m g, Q mc, nd u , determine the mgnitude nd the direction of the electric field tht enles the lock to remin t rest on the incline. 62. A smll sphere of chrge q mc hngs from the end of spring s in Figure P When nother smll sphere of chrge q mc is held eneth the first sphere s in Figure P23.62, the spring stretches y d cm from its originl length nd reches new equilirium position with seprtion etween the chrges of r cm. Wht is the force constnt of the spring? k k q 1 q 1 q 2 Figure P A line of chrge strts t x 5 1x 0 nd extends to positive infinity. The liner chrge density is l 5 l 0 x 0 /x, where l 0 is constnt. Determine the electric field t the origin. 64. A smll sphere of mss m g nd chrge q nc is ttched to the end of string nd hngs verticlly s in Figure P A second chrge of equl mss nd chrge q nc is locted elow the first chrge distnce d cm elow the first chrge s in Figure P () Find the tension in the string. () If the string cn withstnd mximum tension of N, wht is the smllest vlue d cn hve efore the string reks? q 1 q 2 d Figure P A uniform electric field of mgnitude 640 N/C exists AMT etween two prllel pltes tht re 4.00 cm prt. A proton is relesed from rest t the positive plte t the sme instnt n electron is relesed from rest t the negtive plte. () Determine the distnce from the positive plte t which the two pss ech other. Ignore the electricl ttrction etween the proton nd electron. () Wht If? Repet prt () for sodium ion (N 1 ) nd chloride ion (Cl 2 ). 66. Two smll silver spheres, ech with mss of 10.0 g, re seprted y 1.00 m. Clculte the frction of the electrons in one sphere tht must e trnsferred to the other to produce n ttrctive force of N (out 1 ton) etween the spheres. The numer of electrons per tom of silver is 47. d r

65 722 Chpter 23 Electric Fields 67. A chrged cork ll of M mss 1.00 g is suspended on light string in the presence of uniform u y E electric field s shown in Figure P When E i^ j^ x N/C, the ll is in equilirium t q u Find () the chrge on the ll nd () the tension in the string. Figure P23.67 Prolems 67 nd A chrged cork ll of mss m is suspended on light string in the presence of uniform electric field s shown in Figure P When E 5 Ai^ 1 B j^, where A nd B re positive quntities, the ll is in equilirium t the ngle u. Find () the chrge on the ll nd () the tension in the string. 69. Three chrged prticles re ligned long the x xis s shown in Figure P Find the electric field t () the position (2.00 m, 0) nd () the position (0, 2.00 m) nc m y 5.00 nc m 3.00 nc Figure P Two point chrges q A mc nd q B mc Q/C nd third prticle with unknown chrge q C re locted on the x xis. The prticle q A is t the origin, nd q B is t x cm. The third prticle is to e plced so tht ech prticle is in equilirium under the ction of the electric forces exerted y the other two prticles. () Is this sitution possile? If so, is it possile in more thn one wy? Explin. Find () the required loction nd (c) the mgnitude nd the sign of the chrge of the third prticle. 71. A line of positive chrge is y formed into semicircle of rdius R cm s shown in Figure P The chrge per unit u length long the semicircle is descried y the x R expression l 5 l 0 cos u. P The totl chrge on the semicircle is 12.0 mc. Clculte the totl force on Figure P23.71 chrge of 3.00 mc plced t the center of curvture P. 72. Four identicl chrged prticles (q mc) re locted on the corners of rectngle s shown in Figure P The dimensions of the rectngle re L cm nd W cm. Clculte () the mgnitude nd () the direction of the totl electric force exerted on the chrge t the lower left corner y the other three chrges. x 73. Two smll spheres hng in equilirium t the ottom ends of threds, 40.0 cm long, tht hve their top ends tied to the sme fixed point. One sphere hs mss 2.40 g nd chrge 1300 nc. The other sphere hs the sme mss nd chrge 1200 nc. Find the distnce etween the centers of the spheres. 74. Why is the following sitution impossile? An electron enters region of uniform electric field etween two prllel pltes. The pltes re used in cthode-ry tue to djust the position of n electron em on distnt fluorescent screen. The mgnitude of the electric field etween the pltes is 200 N/C. The pltes re m in length nd re seprted y 1.50 cm. The electron enters the region t speed of m/s, trveling prllel to the plne of the pltes in the direction of their length. It leves the pltes heding towrd its correct loction on the fluorescent screen. 75. Review. Two identicl locks resting on frictionless, horizontl surfce re connected y light spring hving spring constnt k N/m nd n unstretched length L i m s shown in Figure P A chrge Q is slowly plced on ech lock, cusing the spring to stretch to n equilirium length L m s shown in Figure P Determine the vlue of Q, modeling the locks s chrged prticles. Q y q q L Figure P23.72 L i k L k q x q Figure P23.75 Prolems 75 nd Review. Two identicl locks resting on frictionless, horizontl surfce re connected y light spring hving spring constnt k nd n unstretched length L i s shown in Figure P A chrge Q is slowly plced on ech lock, cusing the spring to stretch to n equilirium length L s shown in Figure P Determine the vlue of Q, modeling the locks s chrged prticles. 77. Three identicl point chrges, ech of mss m kg, hng from three strings s shown in Figure W Q

66 Prolems 723 P If the lengths of the left nd right strings re ech L cm nd the ngle u is 45.08, determine the vlue of q. q m L θ q m θ Figure P23.77 L q m 78. how tht the mximum mgnitude E mx of the electric field long the xis of uniformly chrged ring occurs t x 5 /!2 (see Fig ) nd hs the vlue Q / 16!3pP Two hrd ruer spheres, ech of mss m g, re rued with fur on dry dy nd re then suspended with two insulting strings of length L cm whose support points re distnce d cm from ech other s shown in Figure P During the ruing process, one sphere receives exctly twice the chrge of the other. They re oserved to hng t equilirium, ech t n ngle of u with the verticl. Find the mount of chrge on ech sphere. m u d u L m Figure P Two identicl eds ech hve mss m nd chrge q. When plced in hemisphericl owl of rdius R with frictionless, nonconducting wlls, the eds move, nd t equilirium, they re distnce d prt (Fig. P23.80). () Determine the chrge q on ech ed. () Determine the chrge required for d to ecome equl to 2R. () Explin how u 1 nd u 2 re relted. () Assume u 1 nd u 2 re smll. how tht the distnce r etween the spheres is pproximtely r < 4k eq 2, 1/3 mg 82. Review. A negtively chrged prticle 2q is plced t the center of uniformly chrged ring, where the ring hs totl positive chrge Q s shown in Figure P The prticle, confined to move long the x xis, is moved smll distnce x long the xis (where x,, ) nd relesed. how tht the prticle oscilltes in simple hrmonic motion with frequency given y f 5 1 1/2 2p k eqq m 3 q Q Figure P Review. A 1.00-g cork ll with chrge 2.00 mc is suspended verticlly on m-long light string in the Q/C presence of uniform, downwrd-directed electric field of mgnitude E N/C. If the ll is displced slightly from the verticl, it oscilltes like simple pendulum. () Determine the period of this oscilltion. () hould the effect of grvittion e included in the clcultion for prt ()? Explin. Chllenge Prolems 84. Identicl thin rods of length 2 crry equl chrges 1Q uniformly distriuted long their lengths. The rods lie long the x xis with their centers seprted y distnce. 2 (Fig. P23.84). how tht the mgnitude of the force exerted y the left rod on the right one is F 5 k eq 2 4 ln m R d R m y x x Figure P Two smll spheres of mss m re suspended from strings Q/C of length, tht re connected t common point. One sphere hs chrge Q nd the other chrge 2Q. The strings mke ngles u 1 nd u 2 with the verticl. Figure P Eight chrged prticles, ech of mgnitude q, re locted on the corners of cue of edge s s shown in Figure P23.85 (pge 724). () Determine the x, y, nd z components of the totl force exerted y the other chrges on the chrge locted t point A. Wht re

67 724 Chpter 23 Electric Fields () the mgnitude nd (c) the direction of this totl force? x q s q s s q q z q q Point A Figure P23.85 Prolems 85 nd Consider the chrge distriution shown in Figure P () how tht the mgnitude of the electric field t the center of ny fce of the cue hs vlue of 2.18k e q /s 2. () Wht is the direction of the electric field t the center of the top fce of the cue? 87. Review. An electric dipole in uniform horizontl electric field is displced slightly from its equilirium position s shown in Figure P23.87, where u is smll. The seprtion of the chrges is 2, nd ech of the two prticles hs mss m. () Assuming the dipole is relesed from this position, show tht its ngulr orienttion exhiits simple hrmonic motion with frequency f 5 1 qe 2p Å m Wht If? () uppose the msses of the two chrged prticles in the dipole re not the sme even though ech prticle continues to hve chrge q. Let the msses of the prticles e m 1 nd m 2. how tht the frequency of the oscilltion in this cse is f 5 1 qe 1m 1 1 m 2 2 2p Å 2m 1 m 2 q 2 q u Figure P23.87 q q y E 88. Inez is putting up decortions for her sister s quinceñer (fifteenth irthdy prty). he ties three light silk rions together to the top of gtewy nd hngs ruer lloon from ech rion (Fig. P23.88). To include the effects of the grvittionl nd uoynt forces on it, ech lloon cn e modeled s prticle of mss 2.00 g, with its center 50.0 cm from the point of support. Inez rus the whole surfce of ech lloon with her woolen scrf, mking the lloons hng seprtely with gps etween them. Looking directly upwrd from elow the lloons, Inez notices tht the centers of the hnging lloons form horizontl equilterl tringle with sides 30.0 cm long. Wht is the common chrge ech lloon crries? Figure P A line of chrge with uniform density 35.0 nc/m lies long the line y cm etween the points with coordintes x 5 0 nd x cm. Find the electric field it cretes t the origin. 90. A prticle of mss m nd chrge q moves t high speed long the x xis. It is initilly ner x 5 2`, nd it ends up ner x 5 1`. A second chrge Q is fixed t the point x 5 0, y 5 2d. As the moving chrge psses the sttionry chrge, its x component of velocity does not chnge pprecily, ut it cquires smll velocity in the y direction. Determine the ngle through which the moving chrge is deflected from the direction of its initil velocity. 91. Two prticles, ech with chrge 52.0 nc, re locted on Q/C the y xis t y cm nd y cm. () Find the vector electric field t point on the x xis s function of x. () Find the field t x cm. (c) At wht loction is the field 1.00i^ kn/c? You my need computer to solve this eqution. (d) At wht loction is the field 16.0i^ kn/c?

68 Guss s Lw chpter In Chpter 23, we showed how to clculte the electric field due to given chrge distriution y integrting over the distriution. In this chpter, we descrie Guss s lw nd n lterntive procedure for clculting electric fields. Guss s lw is sed on the inversesqure ehvior of the electric force etween point chrges. Although Guss s lw is direct consequence of Coulom s lw, it is more convenient for clculting the electric fields of highly symmetric chrge distriutions nd mkes it possile to del with complicted prolems using qulittive resoning. As we show in this chpter, Guss s lw is importnt in understnding nd verifying the properties of conductors in electrosttic equilirium. 4.1 Electric Flux The concept of electric field lines ws descried qulittively in Chpter 23. We now tret electric field lines in more quntittive wy. Consider n electric field tht is uniform in oth mgnitude nd direction s shown in Figure The field lines penetrte rectngulr surfce of re whose plne is oriented perpendiculr to the field. Recll from ection 23.6 tht the numer of lines per unit re (in other words, the line density) is proportionl to the mgnitude of the electric field. Therefore, the totl numer of lines penetrt ing the surfce is proportionl to the product EA. This product of the mgnitude of the electric field nd surfce re perpendiculr to the field is clled the electric flux (uppercse Greek letter phi): 24.1 Electric Flux 24.2 Guss s Lw 24.3 Appliction of Guss s Lw to Vrious Chrge Distriutions 24.4 Conductors in Electrosttic Equilirium In tletop plsm ll, the colorful lines emnting from the sphere give evidence of strong electric fields. Using Guss s lw, we show in this chpter tht the electric field surrounding uniformly chrged sphere is identicl to tht of point chrge. (teve Cole/Getty Imges) (24.1) Figure 24.1 Field lines representing uniform electric field penetrting plne of re perpendiculr to the field. 725

69 726 Chpter 24 Guss s Lw A The numer of field lines tht go through the re A is the sme s the numer tht go through re A. w u A w Norml Figure 24.2 Field lines representing uniform electric field penetrting n re A whose norml is t n ngle u to the field. The electric field mkes n ngle u i with the vector A i, defined s eing norml to the surfce element. E i u i Definition of electric flux, u A i Figure 24.3 A smll element of surfce re DA i in n electric field. E From the I units of E nd A, we see tht F E hs units of newton meters squred per coulom (N? m 2 /C). Electric flux is proportionl to the numer of electric field lines penetrting some surfce. If the surfce under considertion is not perpendiculr to the field, the flux through it must e less thn tht given y Eqution Consider Figure 24.2, where the norml to the surfce of re A is t n ngle u to the uniform electric field. Notice tht the numer of lines tht cross this re A is equl to the numer of lines tht cross the re A, which is projection of re A onto plne oriented perpendiculr to the field. The re A is the product of the length nd the width of the surfce: A 5,w. At the left edge of the figure, we see tht the widths of the surfces re relted y w 5 w cos u. The re A is given y A 5,w 5,w cos u nd we see tht the two res re relted y A 5 A cos u. Becuse the flux through A equls the flux through A, the flux through A is F E 5 EA 5 EA cos u (24.2) From this result, we see tht the flux through surfce of fixed re A hs mximum vlue EA when the surfce is perpendiculr to the field (when the norml to the surfce is prllel to the field, tht is, when u 5 08 in Fig. 24.2); the flux is zero when the surfce is prllel to the field (when the norml to the surfce is perpendiculr to the field, tht is, when u 5 908). In this discussion, the ngle u is used to descrie the orienttion of the surfce of re A. We cn lso interpret the ngle s tht etween the electric field vector nd the norml to the surfce. In this cse, the product E cos u in Eqution 24.2 is the component of the electric field perpendiculr to the surfce. The flux through the surfce cn then e written F E 5 (E cos u)a 5 E n A, where we use E n s the component of the electric field norml to the surfce. We ssumed uniform electric field in the preceding discussion. In more generl situtions, the electric field my vry over lrge surfce. Therefore, the definition of flux given y Eqution 24.2 hs mening only for smll element of re over which the field is pproximtely constnt. Consider generl surfce divided into lrge numer of smll elements, ech of re DA i. It is convenient to define vector D A i whose mgnitude represents the re of the ith element of the lrge surfce nd whose direction is defined to e perpendiculr to the surfce element s shown in Figure The electric field E i t the loction of this element mkes n ngle u i with the vector D A i. The electric flux F E,i through this element is F E,i 5 E i DA i cos u i 5 E i? D A i where we hve used the definition of the sclr product of two vectors ( A? B ; AB cos u ; see Chpter 7). umming the contriutions of ll elements gives n pproximtion to the totl flux through the surfce: F E < E i? D A i If the re of ech element pproches zero, the numer of elements pproches infinity nd the sum is replced y n integrl. Therefore, the generl definition of electric flux is F E ; 3 E? d A (24.3) surfce Eqution 24.3 is surfce integrl, which mens it must e evluted over the surfce in question. In generl, the vlue of F E depends oth on the field pttern nd on the surfce. We re often interested in evluting the flux through closed surfce, defined s surfce tht divides spce into n inside nd n outside region so tht one cnnot move from one region to the other without crossing the surfce. The surfce of sphere, for exmple, is closed surfce. By convention, if the re element in Equ-

70 24.1 Electric Flux 727 Figure 24.4 A closed surfce in n electric field. The re vectors re, y convention, norml to the surfce nd point outwrd. A 3 u E n The electric flux through this re element is negtive. E tion 24.3 is prt of closed surfce, the direction of the re vector is chosen so tht the vector points outwrd from the surfce. If the re element is not prt of closed surfce, the direction of the re vector is chosen so tht the ngle etween the re vector nd the electric field vector is less thn or equl to 90. Consider the closed surfce in Figure The vectors D A i point in different directions for the vrious surfce elements, ut for ech element they re norml to the surfce nd point outwrd. At the element leled, the field lines re crossing the surfce from the inside to the outside nd u, 908; hence, the flux F E,1 5 E? D A 1 through this element is positive. For element, the field lines grze the surfce (perpendiculr to D A 2); therefore, u nd the flux is zero. For elements such s, where the field lines re crossing the surfce from outside to inside, u. 908 nd the flux is negtive ecuse cos u is negtive. The net flux through the surfce is proportionl to the net numer of lines leving the surfce, where the net numer mens the numer of lines leving the surfce minus the numer of lines entering the surfce. If more lines re leving thn entering, the net flux is positive. If more lines re entering thn leving, the net flux is negtive. Using the symol r to represent n integrl over closed surfce, we cn write the net flux F E through closed surfce s F E 5 C E? d A 5 C E n da (24.4) where E n represents the component of the electric field norml to the surfce. Q uick Quiz 24.1 uppose point chrge is locted t the center of sphericl surfce. The electric field t the surfce of the sphere nd the totl flux through the sphere re determined. Now the rdius of the sphere is hlved. A 2 The electric flux through this re element is zero. E E n A 1 u The electric flux through this re element is positive. E

71 728 Chpter 24 Guss s Lw Wht hppens to the flux through the sphere nd the mgnitude of the electric field t the surfce of the sphere? () The flux nd field oth increse. () The flux nd field oth decrese. (c) The flux increses, nd the field decreses. (d) The flux decreses, nd the field increses. (e) The flux remins the sme, nd the field increses. (f) The flux decreses, nd the field remins the sme. Exmple 24.1 Write the integrls for the net flux through fces nd : For fce, E is constnt nd directed inwrd ut d A 1 is directed outwrd (u ). Find the flux through this fce: For fce, E is constnt nd outwrd nd in the sme direction s d A 2 (u 5 08). Find the flux through this fce: 24.2 Guss s Lw F E 5 3 E? d A 1 3 E? d A E? d A 5 3 E 1cos da 5 2E 3 da 5 2EA 5 2E, E? d A 5 3 E 1cos 082 da 5 E 3 da 5 1EA 5 E, Find the net flux y dding the flux over ll six fces: F E 5 2E, 2 1 E, When the chrge is t the center of the sphere, the electric field is everywhere norml to the surfce nd constnt in mgnitude. phericl gussin surfce r q Flux Through Cue Consider uniform electric field E oriented in the x direction in empty spce. A cue of edge length, is plced in the field, oriented s shown in Figure Find the net electric flux through the surfce of the cue. olution Conceptulize Exmine Figure 24.5 crefully. Notice tht the electric field lines pss through two fces perpendiculrly nd re prllel to four other fces of the cue. Ctegorize We evlute the flux from its definition, so we ctegorize this exmple s sustitution prolem. The flux through four of the fces (,, nd the unnumered fces) is zero ecuse E is prllel to the four fces nd therefore perpendiculr to d A on these fces. E A i z In this section, we descrie generl reltionship etween the net electric flux through closed surfce (often clled gussin surfce) nd the chrge enclosed y the surfce. This reltionship, known s Guss s lw, is of fundmentl importnce in the study of electric fields. Consider positive point chrge q locted t the center of sphere of rdius r s shown in Figure From Eqution 23.9, we know tht the mgnitude of the electric field everywhere on the surfce of the sphere is E 5 k e q/r 2. The field lines re directed rdilly outwrd nd hence re perpendiculr to the surfce t every point on the surfce. Tht is, t ech surfce point, E is prllel to the vector D A i representing locl element of re DA i surrounding the surfce point. Therefore, da 1 E? D A i 5 E DA i y da 3 da 4 da 2 Figure 24.5 (Exmple 24.1) A closed surfce in the shpe of cue in uniform electric field oriented prllel to the x xis. ide is the ottom of the cue, nd side is opposite side. E x Figure 24.6 A sphericl gussin surfce of rdius r surrounding positive point chrge q. nd, from Eqution 24.4, we find tht the net flux through the gussin surfce is F E 5 C E? d A 5 C E da 5 E C da

72 24.2 Guss s Lw 729 where we hve moved E outside of the integrl ecuse, y symmetry, E is constnt over the surfce. The vlue of E is given y E 5 k e q/r 2. Furthermore, ecuse the surfce is sphericl, rda 5 A 5 4pr 2. Hence, the net flux through the gussin surfce is q F E 5 k e r 14pr pk 2 e q Reclling from Eqution 23.3 tht k e 5 1/4pP 0, we cn write this eqution in the form F E 5 q P 0 (24.5) Eqution 24.5 shows tht the net flux through the sphericl surfce is proportionl to the chrge inside the surfce. The flux is independent of the rdius r ecuse the re of the sphericl surfce is proportionl to r 2, wheres the electric field is proportionl to 1/r 2. Therefore, in the product of re nd electric field, the dependence on r cncels. Now consider severl closed surfces surrounding chrge q s shown in Figure urfce 1 is sphericl, ut surfces 2 nd 3 re not. From Eqution 24.5, the flux tht psses through 1 hs the vlue q/p 0. As discussed in the preceding section, flux is proportionl to the numer of electric field lines pssing through surfce. The construction shown in Figure 24.7 shows tht the numer of lines through 1 is equl to the numer of lines through the nonsphericl surfces 2 nd 3. Therefore, the net flux through ny closed surfce surrounding point chrge q is given y q/p 0 nd is independent of the shpe of tht surfce. Now consider point chrge locted outside closed surfce of ritrry shpe s shown in Figure As cn e seen from this construction, ny electric field line entering the surfce leves the surfce t nother point. The numer of electric field lines entering the surfce equls the numer leving the surfce. Therefore, the net electric flux through closed surfce tht surrounds no chrge is zero. Applying this result to Exmple 24.1, we see tht the net flux through the cue is zero ecuse there is no chrge inside the cue. Let s extend these rguments to two generlized cses: (1) tht of mny point chrges nd (2) tht of continuous distriution of chrge. We once gin use the superposition principle, which sttes tht the electric field due to mny chrges is The net electric flux is the sme through ll surfces. 2 3 The numer of field lines entering the surfce equls the numer leving the surfce. Krl Friedrich Guss Germn mthemticin nd stronomer ( ) Guss received doctorl degree in mthemtics from the University of Helmstedt in In ddition to his work in electromgnetism, he mde contriutions to mthemtics nd science in numer theory, sttistics, non-eucliden geometry, nd cometry oritl mechnics. He ws founder of the Germn Mgnetic Union, which studies the Erth s mgnetic field on continul sis. Photo Reserchers/Almy 1 q Figure 24.7 Closed surfces of vrious shpes surrounding positive chrge. Figure 24.8 A point chrge locted outside closed surfce.

73 730 Chpter 24 Guss s Lw Chrge q 4 does not contriute to the flux through ny surfce ecuse it is outside ll surfces. q 1 q 4 q 3 q 2 Figure 24.9 The net electric flux through ny closed surfce depends only on the chrge inside tht surfce. The net flux through surfce is q 1 /P 0, the net flux through surfce 9 is (q 2 1 q 3 )/P 0, nd the net flux through surfce 0 is zero. Pitfll Prevention 24.1 Zero Flux Is Not Zero Field In two situtions, there is zero flux through closed surfce: either (1) there re no chrged prticles enclosed y the surfce or (2) there re chrged prticles enclosed, ut the net chrge inside the surfce is zero. For either sitution, it is incorrect to conclude tht the electric field on the surfce is zero. Guss s lw sttes tht the electric flux is proportionl to the enclosed chrge, not the electric field. Conceptul Exmple 24.2 the vector sum of the electric fields produced y the individul chrges. Therefore, the flux through ny closed surfce cn e expressed s C E? d A 5 C 1 E 1 1 E 2 1 c 2? d A where E is the totl electric field t ny point on the surfce produced y the vector ddition of the electric fields t tht point due to the individul chrges. Consider the system of chrges shown in Figure The surfce surrounds only one chrge, q 1 ; hence, the net flux through is q 1 /P 0. The flux through due to chrges q 2, q 3, nd q 4 outside it is zero ecuse ech electric field line from these chrges tht enters t one point leves it t nother. The surfce 9 surrounds chrges q 2 nd q 3 ; hence, the net flux through it is (q 2 1 q 3 )/P 0. Finlly, the net flux through surfce 0 is zero ecuse there is no chrge inside this surfce. Tht is, ll the electric field lines tht enter 0 t one point leve t nother. Chrge q 4 does not contriute to the net flux through ny of the surfces. The mthemticl form of Guss s lw is generliztion of wht we hve just descried nd sttes tht the net flux through ny closed surfce is F E 5 C E? d A 5 q in P 0 (24.6) where E represents the electric field t ny point on the surfce nd q in represents the net chrge inside the surfce. When using Eqution 24.6, you should note tht lthough the chrge q in is the net chrge inside the gussin surfce, E represents the totl electric field, which includes contriutions from chrges oth inside nd outside the surfce. In principle, Guss s lw cn e solved for E to determine the electric field due to system of chrges or continuous distriution of chrge. In prctice, however, this type of solution is pplicle only in limited numer of highly symmetric situtions. In the next section, we use Guss s lw to evlute the electric field for chrge distriutions tht hve sphericl, cylindricl, or plnr symmetry. If one chooses the gussin surfce surrounding the chrge distriution crefully, the integrl in Eqution 24.6 cn e simplified nd the electric field determined. Q uick Quiz 24.2 If the net flux through gussin surfce is zero, the following four sttements could e true. Which of the sttements must e true? () There re no chrges inside the surfce. () The net chrge inside the surfce is zero. (c) The electric field is zero everywhere on the surfce. (d) The numer of electric field lines entering the surfce equls the numer leving the surfce. Flux Due to Point Chrge A sphericl gussin surfce surrounds point chrge q. Descrie wht hppens to the totl flux through the surfce if (A) the chrge is tripled, (B) the rdius of the sphere is douled, (C) the surfce is chnged to cue, nd (D) the chrge is moved to nother loction inside the surfce. olution (A) The flux through the surfce is tripled ecuse flux is proportionl to the mount of chrge inside the surfce. (B) The flux does not chnge ecuse ll electric field lines from the chrge pss through the sphere, regrdless of its rdius. (C) The flux does not chnge when the shpe of the gussin surfce chnges ecuse ll electric field lines from the chrge pss through the surfce, regrdless of its shpe. (D) The flux does not chnge when the chrge is moved to nother loction inside tht surfce ecuse Guss s lw refers to the totl chrge enclosed, regrdless of where the chrge is locted inside the surfce.

74 24.3 Appliction of Guss s Lw to Vrious Chrge Distriutions Appliction of Guss s Lw to Vrious Chrge Distriutions As mentioned erlier, Guss s lw is useful for determining electric fields when the chrge distriution is highly symmetric. The following exmples demonstrte wys of choosing the gussin surfce over which the surfce integrl given y Eqution 24.6 cn e simplified nd the electric field determined. In choosing the surfce, lwys tke dvntge of the symmetry of the chrge distriution so tht E cn e removed from the integrl. The gol in this type of clcultion is to determine surfce for which ech portion of the surfce stisfies one or more of the following conditions: 1. The vlue of the electric field cn e rgued y symmetry to e constnt over the portion of the surfce. 2. The dot product in Eqution 24.6 cn e expressed s simple lgeric product E da ecuse E nd d A re prllel. 3. The dot product in Eqution 24.6 is zero ecuse E nd d A re perpendiculr. 4. The electric field is zero over the portion of the surfce. Different portions of the gussin surfce cn stisfy different conditions s long s every portion stisfies t lest one condition. All four conditions re used in exmples throughout the reminder of this chpter nd will e identified y numer. If the chrge distriution does not hve sufficient symmetry such tht gussin surfce tht stisfies these conditions cn e found, Guss s lw is still true, ut is not useful for determining the electric field for tht chrge distriution. Exmple 24.3 A phericlly ymmetric Chrge Distriution An insulting solid sphere of rdius hs uniform volume chrge density r nd crries totl positive chrge Q (Fig ). (A) Clculte the mgnitude of the electric field t point outside the sphere. olution Conceptulize Notice how this prolem differs from our previous discussion of Guss s lw. The electric field due to point chrges ws discussed in ection Now we re considering the electric field due to distriution of chrge. We found the field for vrious distriutions of chrge in Chpter 23 y integrting over the distriution. This exmple demonstrtes difference from our discussions in Chpter 23. In this chpter, we find the electric field using Guss s lw. For points outside the sphere, lrge, sphericl gussin surfce is drwn concentric with the sphere. Ctegorize Becuse the chrge is distriuted uniformly throughout the sphere, the chrge distriution hs sphericl symmetry nd we cn pply Guss s lw to find the electric field. Gussin sphere For points inside the sphere, sphericl gussin surfce smller thn the sphere is drwn. Gussin sphere Anlyze To reflect the sphericl symmetry, let s choose sphericl gussin surfce of rdius r, concentric with the sphere, s shown in Figure For this choice, condition (2) is stisfied everywhere on the surfce nd E? d A 5 E da. continued r Q Pitfll Prevention 24.2 Gussin urfces Are Not Rel A gussin surfce is n imginry surfce you construct to stisfy the conditions listed here. It does not hve to coincide with physicl surfce in the sitution. Figure (Exmple 24.3) A uniformly chrged insulting sphere of rdius nd totl chrge Q. In digrms such s this one, the dotted line represents the intersection of the gussin surfce with the plne of the pge. r

75 732 Chpter 24 Guss s Lw 24.3 continued Replce E? d A in Guss s lw with E da: F E 5 C E? d A 5 C E da 5 Q P 0 By symmetry, E hs the sme vlue everywhere on the surfce, which stisfies condition (1), so we cn remove E from the integrl: C E da 5 E C da 5 E 14pr Q P 0 olve for E: (1) E 5 Q 4pP 0 r 5 k Q 2 e r 2 1for r. 2 Finlize This field is identicl to tht for point chrge. Therefore, the electric field due to uniformly chrged sphere in the region externl to the sphere is equivlent to tht of point chrge locted t the center of the sphere. (B) Find the mgnitude of the electric field t point inside the sphere. olution Anlyze In this cse, let s choose sphericl gussin surfce hving rdius r,, concentric with the insulting sphere (Fig ). Let V 9 e the volume of this smller sphere. To pply Guss s lw in this sitution, recognize tht the chrge q in within the gussin surfce of volume V 9 is less thn Q. Clculte q in y using q in 5 rv 9: q in 5 rv r 5 r1 4 3pr 3 2 Notice tht conditions (1) nd (2) re stisfied everywhere on the gussin surfce in Figure Apply Guss s lw in the region r, : C E da 5 E C da 5 E 14pr q in olve for E nd sustitute for q in : E 5 q in 4pP 0 r 5 r1 4 3pr pP 0 r 5 r r 2 3P 0 ustitute r 5 Q / 4 3p 3 nd P 0 5 1/4pk e : (2) E 5 Q / 4 3 p 3 Finlize This result for E differs from the one otined in prt (A). It shows tht E 0 s r 0. Therefore, the result elimintes the prolem tht would exist t r 5 0 if E vried s 1/r 2 inside the sphere s it does outside the sphere. Tht is, if E ~ 1/r 2 for r,, the field would e infinite t r 5 0, which is physiclly impossile. Wht If? uppose the rdil position r 5 is pproched from inside the sphere nd from outside. Do we otin the sme vlue of the electric field from oth directions? Answer Eqution (1) shows tht the electric field pproches vlue from the outside given y From the inside, Eqution (2) gives E 5 lim k Q e r r 5 k Q 2 e 2 E 5 lim k Q e r r 5 k Q 3 e 5 k Q 3 e 2 Therefore, the vlue of the field is the sme s the surfce is pproched from oth directions. A plot of E versus r is shown in Figure Notice tht the mgnitude of the field is continuous. 311/4pk e 2 r 5 k Q e r 1for r, 2 3 P 0 E k E e Q 3 r k E e Q r 2 Figure (Exmple 24.3) A plot of E versus r for uniformly chrged insulting sphere. The electric field inside the sphere (r, ) vries linerly with r. The field outside the sphere (r. ) is the sme s tht of point chrge Q locted t r 5 0. r

76 24.3 Appliction of Guss s Lw to Vrious Chrge Distriutions 733 Exmple 24.4 A Cylindriclly ymmetric Chrge Distriution Find the electric field distnce r from line of positive chrge of infinite length nd constnt chrge per unit length l (Fig ). Gussin surfce r olution Conceptulize The line of chrge is infinitely long. Therefore, the field is the sme t ll points equidistnt from the line, regrdless of the verticl position of the point in Figure We expect the field to ecome weker s we move frther wy from the line of chrge. Ctegorize Becuse the chrge is distriuted uniformly long the line, the chrge distriution hs cylindricl symmetry nd we cn pply Guss s lw to find the electric field. Anlyze The symmetry of the chrge distriution requires tht E e perpendiculr to the line chrge nd directed outwrd s shown in Figure To reflect the symmetry of the chrge distriution, let s choose cylindricl gussin surfce of rdius r nd length, tht is coxil with the line chrge. For the curved prt of this surfce, E is constnt in mgnitude nd perpendiculr to the surfce t ech point, stisfying conditions (1) nd (2). Furthermore, the flux through the ends of the gussin cylinder is zero ecuse E is prllel to these surfces. Tht is the first ppliction we hve seen of condition (3). We must tke the surfce integrl in Guss s lw over the entire gussin surfce. Becuse E? d A is zero for the flt ends of the cylinder, however, we restrict our ttention to only the curved surfce of the cylinder. Apply Guss s lw nd conditions (1) nd (2) for the curved surfce, noting tht the totl chrge inside our gussin surfce is l,: ustitute the re A 5 2pr, of the curved surfce: E 12pr,2 5 l, P 0 olve for the mgnitude of the electric field: E 5 l 2pP 0 r 5 2k e l r da Figure (Exmple 24.4) () An infinite line of chrge surrounded y cylindricl gussin surfce concentric with the line. () An end view shows tht the electric field t the cylindricl surfce is constnt in mgnitude nd perpendiculr to the surfce. E F E 5 C E? d A 5 E C da 5 EA 5 q in P 0 5 l, P 0 Finlize This result shows tht the electric field due to cylindriclly symmetric chrge distriution vries s 1/r, wheres the field externl to sphericlly symmetric chrge distriution vries s 1/r 2. Eqution 24.7 cn lso e derived y direct integrtion over the chrge distriution. (ee Prolem 44 in Chpter 23.) Wht If? Wht if the line segment in this exmple were not infinitely long? Answer If the line chrge in this exmple were of finite length, the electric field would not e given y Eqution A finite line chrge does not possess sufficient symmetry to mke use of Guss s lw ecuse the mgnitude of the electric field is no longer constnt over the surfce of the gussin cylinder: the field ner the ends of the line would e different from tht fr from the ends. Therefore, condition (1) would not e stisfied in this sitution. Furthermore, E is not perpendiculr to the cylindricl surfce t ll points: the field vectors ner the ends would hve component prllel to the line. Therefore, condition (2) would not e stisfied. For points close to finite line chrge nd fr from the ends, Eqution 24.7 gives good pproximtion of the vlue of the field. It is left for you to show (see Prolem 33) tht the electric field inside uniformly chrged rod of finite rdius nd infinite length is proportionl to r. (24.7) E

77 734 Chpter 24 Guss s Lw Exmple 24.5 A Plne of Chrge Find the electric field due to n infinite plne of positive chrge with uniform surfce chrge density s. olution Conceptulize Notice tht the plne of chrge is infinitely lrge. Therefore, the electric field should e the sme t ll points equidistnt from the plne. How would you expect the electric field to depend on the distnce from the plne? Ctegorize Becuse the chrge is distriuted uniformly on the plne, the chrge distriution is symmetric; hence, we cn use Guss s lw to find the electric field. Gussin Anlyze By symmetry, surfce E must e perpendiculr to the plne t ll points. The direction of E is wy from positive chrges, indicting tht the direction of E Figure (Exmple 24.5) A cylindricl gussin surfce penetrting n infinite plne of chrge. on one side of the plne must e opposite its direction on the other side s shown in Figure A gussin surfce tht reflects the symmetry is smll cylinder The flux is EA through ech end whose xis is perpendiculr to the plne nd whose ends ech hve n re A nd re equidistnt from the plne. Becuse of the gussin surfce nd zero E is prllel to the curved surfce of the cylinder nd therefore perpendiculr to d through its curved surfce. A t ll points on this surfce condition (3) is stisfied nd there is no contriution to the surfce integrl from this surfce. For the flt ends of the cylinder, conditions (1) nd (2) re stisfied. The flux through ech end of the cylinder is EA; hence, the totl flux through the entire gussin surfce is just tht through the ends, F E 5 2EA. Write Guss s lw for this surfce, noting tht the enclosed chrge is q in 5 sa: olve for E: E 5 Finlize Becuse the distnce from ech flt end of the cylinder to the plne does not pper in Eqution 24.8, we conclude tht E 5 s/2p 0 t ny distnce from the plne. Tht is, the field is uniform everywhere. Figure shows this uniform field due to n infinite plne of chrge, seen edge-on. Wht If? uppose two infinite plnes of chrge re prllel to ech other, one positively chrged nd the other negtively chrged. The surfce chrge densities of oth plnes re of the sme mgnitude. Wht does the electric field look like in this sitution? Answer We first ddressed this configurtion in the Wht If? section of Exmple The electric fields due to the two plnes dd in the region etween the plnes, resulting in uniform field of mgnitude s/p 0, nd cncel elsewhere to give field of zero. Figure shows the field lines for such configurtion. This method is prcticl wy to chieve uniform electric fields with finite-sized plnes plced close to ech other. F E 5 2EA 5 q in 5 sa P 0 P 0 s (24.8) 2P 0 Figure (Exmple 24.5) The electric field lines due to n infinite plne of positive chrge. E E Figure (Exmple 24.5) The electric field lines etween two infinite plnes of chrge, one positive nd one negtive. In prctice, the field lines ner the edges of finite-sized sheets of chrge will curve outwrd. A Conceptul Exmple 24.6 Don t Use Guss s Lw Here! Explin why Guss s lw cnnot e used to clculte the electric field ner n electric dipole, chrged disk, or tringle with point chrge t ech corner.

78 24.4 Conductors in Electrosttic Equilirium continued olution The chrge distriutions of ll these configurtions do not hve sufficient symmetry to mke the use of Guss s lw prcticl. We cnnot find closed surfce surrounding ny of these distriutions for which ll portions of the surfce stisfy one or more of conditions (1) through (4) listed t the eginning of this section Conductors in Electrosttic Equilirium As we lerned in ection 23.2, good electricl conductor contins chrges (electrons) tht re not ound to ny tom nd therefore re free to move out within the mteril. When there is no net motion of chrge within conductor, the conductor is in electrosttic equilirium. A conductor in electrosttic equilirium hs the following properties: 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolted nd crries chrge, the chrge resides on its surfce. 3. The electric field t point just outside chrged conductor is perpendiculr to the surfce of the conductor nd hs mgnitude s/p 0, where s is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest t loctions where the rdius of curvture of the surfce is smllest. We verify the first three properties in the discussion tht follows. The fourth property is presented here (ut not verified until we hve studied the pproprite mteril in Chpter 25) to provide complete list of properties for conductors in electrosttic equilirium. We cn understnd the first property y considering conducting sl plced in n externl field E (Fig ). The electric field inside the conductor must e zero, ssuming electrosttic equilirium exists. If the field were not zero, free electrons in the conductor would experience n electric force ( F 5 q E) nd would ccelerte due to this force. This motion of electrons, however, would men tht the conductor is not in electrosttic equilirium. Therefore, the existence of electrosttic equilirium is consistent only with zero field in the conductor. Let s investigte how this zero field is ccomplished. Before the externl field is pplied, free electrons re uniformly distriuted throughout the conductor. When the externl field is pplied, the free electrons ccelerte to the left in Figure 24.16, cusing plne of negtive chrge to ccumulte on the left surfce. The movement of electrons to the left results in plne of positive chrge on the right surfce. These plnes of chrge crete n dditionl electric field inside the conductor tht opposes the externl field. As the electrons move, the surfce chrge densities on the left nd right surfces increse until the mgnitude of the internl field equls tht of the externl field, resulting in net field of zero inside the conductor. The time it tkes good conductor to rech equilirium is on the order of s, which for most purposes cn e considered instntneous. If the conductor is hollow, the electric field inside the conductor is lso zero, whether we consider points in the conductor or in the cvity within the conductor. The zero vlue of the electric field in the cvity is esiest to rgue with the concept of electric potentil, so we will ddress this issue in ection Guss s lw cn e used to verify the second property of conductor in electrosttic equilirium. Figure shows n ritrrily shped conductor. A gussin W WProperties of conductor in electrosttic equilirium E E Figure A conducting sl in n externl electric field E. The chrges induced on the two surfces of the sl produce n electric field tht opposes the externl field, giving resultnt field of zero inside the sl. Gussin surfce Figure A conductor of ritrry shpe. The roken line represents gussin surfce tht cn e just inside the conductor s surfce.

79 736 Chpter 24 Guss s Lw The flux through the gussin surfce is EA. A E Figure A gussin surfce in the shpe of smll cylinder is used to clculte the electric field immeditely outside chrged conductor. surfce is drwn inside the conductor nd cn e very close to the conductor s surfce. As we hve just shown, the electric field everywhere inside the conductor is zero when it is in electrosttic equilirium. Therefore, the electric field must e zero t every point on the gussin surfce, in ccordnce with condition (4) in ection 24.3, nd the net flux through this gussin surfce is zero. From this result nd Guss s lw, we conclude tht the net chrge inside the gussin surfce is zero. Becuse there cn e no net chrge inside the gussin surfce (which is ritrrily close to the conductor s surfce), ny net chrge on the conductor must reside on its surfce. Guss s lw does not indicte how this excess chrge is distriuted on the conductor s surfce, only tht it resides exclusively on the surfce. To verify the third property, let s egin with the perpendiculrity of the field to the surfce. If the field vector E hd component prllel to the conductor s surfce, free electrons would experience n electric force nd move long the surfce; in such cse, the conductor would not e in equilirium. Therefore, the field vector must e perpendiculr to the surfce. To determine the mgnitude of the electric field, we use Guss s lw nd drw gussin surfce in the shpe of smll cylinder whose end fces re prllel to the conductor s surfce (Fig ). Prt of the cylinder is just outside the conductor, nd prt is inside. The field is perpendiculr to the conductor s surfce from the condition of electrosttic equilirium. Therefore, condition (3) in ection 24.3 is stisfied for the curved prt of the cylindricl gussin surfce: there is no flux through this prt of the gussin surfce ecuse E is prllel to the surfce. There is no flux through the flt fce of the cylinder inside the conductor ecuse here E 5 0, which stisfies condition (4). Hence, the net flux through the gussin surfce is equl to tht through only the flt fce outside the conductor, where the field is perpendiculr to the gussin surfce. Using conditions (1) nd (2) for this fce, the flux is EA, where E is the electric field just outside the conductor nd A is the re of the cylinder s fce. Applying Guss s lw to this surfce gives F E 5 C E da 5 EA 5 q in 5 sa where we hve used q in 5 sa. olving for E gives for the electric field immeditely outside chrged conductor: P 0 P 0 E 5 s P 0 (24.9) Q uick Quiz 24.3 Your younger rother likes to ru his feet on the crpet nd then touch you to give you shock. While you re trying to escpe the shock tretment, you discover hollow metl cylinder in your sement, lrge enough to clim inside. In which of the following cses will you not e shocked? () You clim inside the cylinder, mking contct with the inner surfce, nd your chrged rother touches the outer metl surfce. () Your chrged rother is inside touching the inner metl surfce nd you re outside, touching the outer metl surfce. (c) Both of you re outside the cylinder, touching its outer metl surfce ut not touching ech other directly. Exmple 24.7 A phere Inside phericl hell A solid insulting sphere of rdius crries net positive chrge Q uniformly distriuted throughout its volume. A conducting sphericl shell of inner rdius nd outer rdius c is concentric with the solid sphere nd crries net chrge 22Q. Using Guss s lw, find the electric field in the regions leled y, x,, nd in Figure nd the chrge distriution on the shell when the entire system is in electrosttic equilirium.

80 24.4 Conductors in Electrosttic Equilirium continued olution Conceptulize Notice how this prolem differs from Exmple The chrged sphere in Figure ppers in Figure 24.19, ut it is now surrounded y shell crrying chrge 22Q. Think out how the presence of the shell will ffect the electric field of the sphere. Ctegorize The chrge is distriuted uniformly throughout the sphere, nd we know tht the chrge on the conducting shell distriutes itself uniformly on the surfces. Therefore, the system hs sphericl symmetry nd we cn pply Guss s lw to find the electric field in the vrious regions. Anlyze In region etween the surfce of the solid sphere nd the inner surfce of the shell we construct sphericl gussin surfce of rdius r, where, r,, noting tht the chrge inside this surfce is 1Q (the chrge on the solid sphere). Becuse of the sphericl symmetry, the electric field lines must e directed rdilly outwrd nd e constnt in mgnitude on the gussin surfce. The chrge on the conducting shell cretes zero electric field in the region r,, so the shell hs no effect on the field in region due to the sphere. Therefore, write n expression for the field in region s tht due to the sphere from prt (A) of Exmple 24.3: Becuse the conducting shell cretes zero field inside itself, it lso hs no effect on the field inside the sphere. Therefore, write n expression for the field in region s tht due to the sphere from prt (B) of Exmple 24.3: In region, where r. c, construct sphericl gussin surfce; this surfce surrounds totl chrge q in 5 Q 1 (22Q ) 5 2Q. Therefore, model the chrge distriution s sphere with chrge 2Q nd write n expression for the field in region from prt (A) of Exmple 24.3: In region, the electric field must e zero ecuse the sphericl shell is conductor in equilirium: Construct gussin surfce of rdius r in region, where, r, c, nd note tht q in must e zero ecuse E Find the mount of chrge q inner on the inner surfce of the shell: E 2 5 k Q e 1for, r, 2 2 r E 1 5 k e Q 3 r E 4 5 2k e Q r 2 E for r, 2 1for r. c2 1for, r, c2 q in 5 q sphere 1 q inner q inner 5 q in 2 q sphere Q 5 2Q Finlize The chrge on the inner surfce of the sphericl shell must e 2Q to cncel the chrge 1Q on the solid sphere nd give zero electric field in the mteril of the shell. Becuse the net chrge on the shell is 22Q, its outer surfce must crry chrge 2Q. Wht If? r Q How would the results of this prolem differ if the sphere were conducting insted of insulting? Answer The only chnge would e in region, where r,. Becuse there cn e no chrge inside conductor in electrosttic equilirium, q in 5 0 for gussin surfce of rdius r, ; therefore, on the sis of Guss s lw nd symmetry, E In regions,, nd, there would e no wy to determine from oservtions of the electric field whether the sphere is conducting or insulting. 2Q c Figure (Exmple 24.7) An insulting sphere of rdius nd crrying chrge Q surrounded y conducting sphericl shell crrying chrge 22Q.

81 738 Chpter 24 Guss s Lw ummry Definition Electric flux is proportionl to the numer of electric field lines tht penetrte surfce. If the electric field is uniform nd mkes n ngle u with the norml to surfce of re A, the electric flux through the surfce is In generl, the electric flux through surfce is Concepts nd Principles Guss s lw sys tht the net electric flux F E through ny closed gussin surfce is equl to the net chrge q in inside the surfce divided y P 0 : F E 5 C E? d A 5 q in P 0 (24.6) Using Guss s lw, you cn clculte the electric field due to vrious symmetric chrge distriutions. Ojective Questions F E 5 EA cos u (24.2) F E ; 3 E? d A surfce 1. A cuicl gussin surfce surrounds long, stright, chrged filment tht psses perpendiculrly through two opposite fces. No other chrges re nery. (i) Over how mny of the cue s fces is the electric field zero? () 0 () 2 (c) 4 (d) 6 (ii) Through how mny of the cue s fces is the electric flux zero? Choose from the sme possiilities s in prt (i). 2. A coxil cle consists of long, stright filment surrounded y long, coxil, cylindricl conducting shell. Assume chrge Q is on the filment, zero net chrge is on the shell, nd the electric field is E 1 i^ t prticulr point P midwy etween the filment nd the inner surfce of the shell. Next, you plce the cle into uniform externl field 2E i^. Wht is the x component of the electric field t P then? () 0 () etween 0 nd E 1 (c) E 1 (d) etween 0 nd 2E 1 (e) 2E 1 3. In which of the following contexts cn Guss s lw not e redily pplied to find the electric field? () ner long, uniformly chrged wire () ove lrge, uniformly chrged plne (c) inside uniformly chrged ll (d) outside uniformly chrged sphere (e) Guss s lw cn e redily pplied to find the electric field in ll these contexts. (24.3) A conductor in electrosttic equilirium hs the following properties: 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolted nd crries chrge, the chrge resides on its surfce. 3. The electric field t point just outside chrged conductor is perpendiculr to the surfce of the conductor nd hs mgnitude s/p 0, where s is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest t loctions where the rdius of curvture of the surfce is smllest. 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 4. A prticle with chrge q is locted inside cuicl gussin surfce. No other chrges re nery. (i) If the prticle is t the center of the cue, wht is the flux through ech one of the fces of the cue? () 0 () q/2p 0 (c) q/6p 0 (d) q/8p 0 (e) depends on the size of the cue (ii) If the prticle cn e moved to ny point within the cue, wht mximum vlue cn the flux through one fce pproch? Choose from the sme possiilities s in prt (i). 5. Chrges of 3.00 nc, nc, nc, nd 1.00 nc re contined inside rectngulr ox with length 1.00 m, width 2.00 m, nd height 2.50 m. Outside the ox re chrges of 1.00 nc nd 4.00 nc. Wht is the electric flux through the surfce of the ox? () 0 () N? m 2 /C (c) N? m 2 /C (d) N? m 2 /C (e) N? m 2 /C 6. A lrge, metllic, sphericl shell hs no net chrge. It is supported on n insulting stnd nd hs smll hole t the top. A smll tck with chrge Q is lowered on silk thred through the hole into the interior of the shell. (i) Wht is the chrge on the inner surfce of the shell, () Q () Q/2 (c) 0 (d) 2Q/2 or (e) 2Q? Choose your nswers to the following questions from

82 conceptul Questions 739 the sme possiilities. (ii) Wht is the chrge on the outer surfce of the shell? (iii) The tck is now llowed to touch the interior surfce of the shell. After this contct, wht is the chrge on the tck? (iv) Wht is the chrge on the inner surfce of the shell now? (v) Wht is the chrge on the outer surfce of the shell now? 7. Two solid spheres, oth of rdius 5 cm, crry identicl totl chrges of 2 mc. phere A is good conductor. phere B is n insultor, nd its chrge is distriuted uniformly throughout its volume. (i) How do the mgnitudes of the electric fields they seprtely crete t rdil distnce of 6 cm compre? () E A. E B 5 0 () E A. E B. 0 (c) E A 5 E B. 0 (d) 0, E A, E B (e) 0 5 E A, E B (ii) How do the mgnitudes of the electric fields they seprtely crete t rdius 4 cm compre? Choose from the sme possiilities s in prt (i). 8. A uniform electric field of 1.00 N/C is set up y uniform distriution of chrge in the xy plne. Wht is the electric field inside metl ll plced m ove the xy plne? () 1.00 N/C () N/C (c) 0 (d) N/C (e) vries depending on the position inside the ll 9. A solid insulting sphere of rdius 5 cm crries electric chrge uniformly distriuted throughout its volume. Concentric with the sphere is conducting sphericl shell with no net chrge s shown in Figure OQ24.9. The inner rdius of the shell is 10 cm, nd the outer rdius is 15 cm. No other chrges re nery. () Rnk Conceptul Questions 1. Consider n electric field tht is uniform in direction throughout certin volume. Cn it e uniform in mgnitude? Must it e uniform in mgnitude? Answer these questions () ssuming the volume is filled with n insulting mteril crrying chrge descried y volume chrge density nd () ssuming the volume is empty spce. tte resoning to prove your nswers. 2. A cuicl surfce surrounds point chrge q. Descrie wht hppens to the totl flux through the surfce if () the chrge is douled, () the volume of the cue is douled, (c) the surfce is chnged to sphere, (d) the chrge is moved to nother loction inside the surfce, nd (e) the chrge is moved outside the surfce. 3. A uniform electric field exists in region of spce contining no chrges. Wht cn you conclude out the net electric flux through gussin surfce plced in this region of spce? 4. If the totl chrge inside closed surfce is known ut the distriution of the chrge is unspecified, cn you use Guss s lw to find the electric field? Explin. 5. Explin why the electric flux through closed surfce with given enclosed chrge is independent of the size or shpe of the surfce. the mgnitude of the electric field t points A (t rdius 4 cm), B (rdius 8 cm), C (rdius 12 cm), nd D (rdius 16 cm) from lrgest to smllest. Disply ny cses of equlity in your rnking. () imilrly rnk the electric flux through A B C D concentric sphericl surfces Figure OQ24.9 through points A, B, C, nd D. 10. A cuicl gussin surfce is isected y lrge sheet of chrge, prllel to its top nd ottom fces. No other chrges re nery. (i) Over how mny of the cue s fces is the electric field zero? () 0 () 2 (c) 4 (d) 6 (ii) Through how mny of the cue s fces is the electric flux zero? Choose from the sme possiilities s in prt (i). 11. Rnk the electric fluxes through ech gussin surfce shown in Figure OQ24.11 from lrgest to smllest. Disply ny cses of equlity in your rnking. Q Q 3Q 4Q c d 1. denotes nswer ville in tudent olutions Mnul/tudy Guide Figure OQ If more electric field lines leve gussin surfce thn enter it, wht cn you conclude out the net chrge enclosed y tht surfce? 7. A person is plced in lrge, hollow, metllic sphere tht is insulted from ground. () If lrge chrge is plced on the sphere, will the person e hrmed upon touching the inside of the sphere? () Explin wht will hppen if the person lso hs n initil chrge whose sign is opposite tht of the chrge on the sphere. 8. Consider two identicl conducting spheres whose surfces re seprted y smll distnce. One sphere is given lrge net positive chrge, nd the other is given smll net positive chrge. It is found tht the force etween the spheres is ttrctive even though they oth hve net chrges of the sme sign. Explin how this ttrction is possile. 9. A common demonstrtion involves chrging ruer lloon, which is n insultor, y ruing it on your hir nd then touching the lloon to ceiling or wll, which is lso n insultor. Becuse of the electricl ttrction etween the chrged lloon nd the neutrl wll, the lloon sticks to the wll. Imgine now tht we hve two infinitely lrge, flt sheets of insulting

83 740 Chpter 24 Guss s Lw mteril. One is chrged, nd the other is neutrl. If these sheets re rought into contct, does n ttrctive force exist etween them s there ws for the lloon nd the wll? 10. On the sis of the repulsive nture of the force etween like chrges nd the freedom of motion of chrge within conductor, explin why excess chrge on n isolted conductor must reside on its surfce. 11. The un is lower in the sky during the winter thn it is during the summer. () How does this chnge ffect the flux of sunlight hitting given re on the surfce of the Erth? () How does this chnge ffect the wether? Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign ection 24.1 Electric Flux 1. A flt surfce of re 3.20 m 2 is rotted in uniform electric field of mgnitude E N/C. Determine the electric flux through this re () when the electric field is perpendiculr to the surfce nd () when the electric field is prllel to the surfce. 2. A verticl electric field of mgnitude N/C W exists ove the Erth s surfce on dy when thunderstorm is rewing. A cr with rectngulr size of 6.00 m y 3.00 m is trveling long dry grvel rodwy sloping downwrd t Determine the electric flux through the ottom of the cr. 3. A 40.0-cm-dimeter circulr loop is rotted in uniform electric field until the position of mximum elec- M tric flux is found. The flux in this position is mesured to e N? m 2 /C. Wht is the mgnitude of the electric field? 4. Consider closed tringulr ox resting within horizontl electric field of mgnitude E N/C W s shown in Figure P24.4. Clculte the electric flux through () the verticl rectngulr surfce, () the slnted surfce, nd (c) the entire surfce of the ox cm cm Figure P An electric field of mgnitude 3.50 kn/c is pplied M long the x xis. Clculte the electric flux through rectngulr plne m wide nd m long () if the plne is prllel to the yz plne, () if the plne is prllel to the xy plne, nd (c) if the plne contins the y xis nd its norml mkes n ngle of with the x xis. E BIO Q/C 6. A nonuniform electric field is given y the expression E 5 y i^ 1 z j^ 1 cx k^ where,, nd c re constnts. Determine the electric flux through rectngulr surfce in the xy plne, extending from x 5 0 to x 5 w nd from y 5 0 to y 5 h. ection 24.2 Guss s Lw 7. An unchrged, nonconducting, hollow sphere of rdius 10.0 cm surrounds 10.0-mC chrge locted t the origin of Crtesin coordinte system. A drill with rdius of 1.00 mm is ligned long the z xis, nd hole is drilled in the sphere. Clculte the electric flux through the hole. 8. Find the net electric flux through the sphericl closed surfce shown in Figure P24.8. The two chrges on the right re inside the sphericl surfce nc Figure P nc 3.00 nc 9. The following chrges re locted inside sumrine: M 5.00 mc, mc, 27.0 mc, nd mc. () Clculte the net electric flux through the hull of the sumrine. () Is the numer of electric field lines leving the sumrine greter thn, equl to, or less thn the numer entering it? 10. The electric field everywhere on the surfce of W thin, sphericl shell of rdius m is of mgnitude 890 N/C nd points rdilly towrd the center of the sphere. () Wht is the net chrge within the sphere s surfce? () Wht is the distriution of the chrge inside the sphericl shell?

84 Prolems Four closed surfces, 1 W through 4, together with 1 the chrges 22Q, Q, nd 2Q re sketched in Figure 2Q P (The colored lines 4 3 re the intersections of the Q surfces with the pge.) Find the electric flux Q through ech surfce. 12. A chrge of 170 mc is t the 2 Q/C center of cue of edge 80.0 cm. No other chrges Figure P24.11 re nery. () Find the flux through ech fce of the cue. () Find the flux through the whole surfce of the cue. (c) Wht If? Would your nswers to either prt () or prt () chnge if the chrge were not t the center? Explin. 13. In the ir over prticulr region t n ltitude of 500 m ove the ground, the electric field is 120 N/C directed downwrd. At 600 m ove the ground, the electric field is 100 N/C downwrd. Wht is the verge volume chrge density in the lyer of ir etween these two elevtions? Is it positive or negtive? 14. A prticle with chrge of 12.0 mc is plced t the center of sphericl shell of rdius 22.0 cm. Wht is the Q/C totl electric flux through () the surfce of the shell nd () ny hemisphericl surfce of the shell? (c) Do the results depend on the rdius? Explin. 15. () Find the net electric Q/C flux through the cue shown in Figure P () Cn you use Guss s lw to find the electric 8.00 nc 3.00 nc field on the surfce of this cue? Explin. 16. () A prticle with chrge Figure P24.15 Q/C q is locted distnce d from n infinite plne. Determine the electric flux through the plne due to the chrged prticle. () Wht If? A prticle with chrge q is locted very smll distnce from the center of very lrge squre on the line perpendiculr to the squre nd going through its center. Determine the pproximte electric flux through the squre due to the chrged prticle. (c) How do the nswers to prts () nd () compre? Explin. 17. An infinitely long line chrge hving uniform chrge per unit length l lies distnce d from point O s shown in Figure P Determine the totl electric flux through the surfce of sphere of rdius R cen- tered t O resulting from this line chrge. Consider oth cses, where () R, d nd () R. d. 18. Find the net electric flux through () the closed sphericl surfce in uniform electric field shown in Figure Q/C P24.18 nd () the closed cylindricl surfce shown in Figure P (c) Wht cn you conclude out the chrges, if ny, inside the cylindricl surfce? R E Figure P A prticle with chrge Q mc is locted L t the center of cue of edge L m. In q ddition, six other identicl q q chrged prticles q Q q hving q mc L re positioned symmetriclly q round Q s shown in Figure P Determine the electric flux through one fce of the cue. L Figure P24.19 Prolems 19 nd A prticle with chrge Q is locted t the center of cue of edge L. In ddition, six other identicl chrged prticles q re positioned symmetriclly round Q s shown in Figure P For ech of these prticles, q is negtive numer. Determine the electric flux through one fce of the cue. 21. A prticle with chrge Q is locted smll distnce d immeditely ove the center of the flt fce of hemisphere of rdius R s shown in Figure P Wht is the electric flux () through the curved surfce nd () through the flt fce s d 0? l d Figure P24.17 O R 22. Figure P24.22 (pge 742) represents the top view of cuic gussin surfce in uniform electric field E oriented prllel to the top nd ottom fces of the cue. The field mkes n ngle u with side, nd the re of ech fce is A. In symolic form, find the electric flux through () fce, () fce, (c) fce, (d) fce, nd (e) the top nd ottom fces of the cue. (f) Wht 2R d Q R Figure P24.21 E

85 742 Chpter 24 Guss s Lw is the net electric flux through the cue? (g) How much chrge is enclosed within the gussin surfce? u Figure P24.22 ection 24.3 Appliction of Guss s Lw to Vrious Chrge Distriutions 23. In nucler fission, nucleus of urnium-238, which contins 92 protons, cn divide into two smller spheres, ech hving 46 protons nd rdius of m. Wht is the mgnitude of the repulsive electric force pushing the two spheres prt? 24. The chrge per unit length on long, stright filment W is mc/m. Find the electric field () 10.0 cm, () 20.0 cm, nd (c) 100 cm from the filment, where distnces re mesured perpendiculr to the length of the filment. 25. A 10.0-g piece of tyrofom crries net chrge of AMT mc nd is suspended in equilirium ove the center of lrge, horizontl sheet of plstic tht hs uniform chrge density on its surfce. Wht is the chrge per unit re on the plstic sheet? 26. Determine the mgnitude of the electric field t the surfce of led-208 nucleus, which contins 82 protons nd 126 neutrons. Assume the led nucleus hs volume 208 times tht of one proton nd consider proton to e sphere of rdius m. 27. A lrge, flt, horizontl sheet of chrge hs chrge M per unit re of 9.00 mc/m 2. Find the electric field just ove the middle of the sheet. 28. uppose you fill two ruer lloons with ir, suspend oth of them from the sme point, nd let them hng down on strings of equl length. You then ru ech with wool or on your hir so tht the lloons hng prt with noticele seprtion etween them. Mke order-of-mgnitude estimtes of () the force on ech, () the chrge on ech, (c) the field ech cretes t the center of the other, nd (d) the totl flux of electric field creted y ech lloon. In your solution, stte the quntities you tke s dt nd the vlues you mesure or estimte for them. 29. Consider thin, sphericl shell of rdius 14.0 cm with M totl chrge of 32.0 mc distriuted uniformly on its surfce. Find the electric field () 10.0 cm nd () 20.0 cm from the center of the chrge distriution. 30. A nonconducting wll crries chrge with uniform W density of 8.60 mc/cm 2. () Wht is the electric field Q/C 7.00 cm in front of the wll if 7.00 cm is smll compred E with the dimensions of the wll? () Does your result chnge s the distnce from the wll vries? Explin. 31. A uniformly chrged, stright filment 7.00 m in M length hs totl positive chrge of 2.00 mc. An unchrged crdord cylinder 2.00 cm in length nd 10.0 cm in rdius surrounds the filment t its center, with the filment s the xis of the cylinder. Using resonle pproximtions, find () the electric field t the surfce of the cylinder nd () the totl electric flux through the cylinder. 32. Assume the mgnitude of the electric field on ech Q/C fce of the cue of edge L m in Figure P24.32 is uniform nd the directions of the fields on ech fce re s indicted. Find () the net electric flux through the cue nd () the net chrge inside the cue. (c) Could the net chrge e single point chrge? 20.0 N/C L 25.0 N/C 20.0 N/C 20.0 N/C 35.0 N/C 15.0 N/C Figure P Consider long, cylindricl chrge distriution of rdius R with uniform chrge density r. Find the electric field t distnce r from the xis, where r, R. 34. A cylindricl shell of rdius 7.00 cm nd length 2.40 m W hs its chrge uniformly distriuted on its curved surfce. The mgnitude of the electric field t point 19.0 cm rdilly outwrd from its xis (mesured from the midpoint of the shell) is 36.0 kn/c. Find () the net chrge on the shell nd () the electric field t point 4.00 cm from the xis, mesured rdilly outwrd from the midpoint of the shell. 35. A solid sphere of rdius 40.0 cm hs totl positive W chrge of 26.0 mc uniformly distriuted throughout its volume. Clculte the mgnitude of the electric field () 0 cm, () 10.0 cm, (c) 40.0 cm, nd (d) 60.0 cm from the center of the sphere. 36. Review. A prticle with chrge of nc is plced AMT t the center of nonconducting sphericl shell of inner rdius 20.0 cm nd outer rdius 25.0 cm. The sphericl shell crries chrge with uniform density of mc/m 3. A proton moves in circulr orit just outside the sphericl shell. Clculte the speed of the proton. ection 24.4 Conductors in Electrosttic Equilirium 37. A long, stright metl rod hs rdius of 5.00 cm nd M chrge per unit length of 30.0 nc/m. Find the electric field () 3.00 cm, () 10.0 cm, nd (c) 100 cm from the

86 Prolems 743 xis of the rod, where distnces re mesured perpendiculr to the rod s xis. 38. Why is the following E (kn/c) sitution impossile? A 8 solid copper sphere 6 of rdius 15.0 cm is in electrosttic equilirium nd crries 4 2 chrge of 40.0 nc. Figure P24.38 shows 0 r (m) the mgnitude of the electric field s function of rdil position Figure P24.38 r mesured from the center of the sphere. 39. A solid metllic sphere of rdius crries totl chrge W Q. No other chrges re nery. The electric field Q/C just outside its surfce is k e Q / 2 rdilly outwrd. At this close point, the uniformly chrged surfce of the sphere looks exctly like uniform flt sheet of chrge. Is the electric field here given y s/p 0 or y s/2p 0? 40. A positively chrged prticle is t distnce R/2 from the center of n unchrged thin, conducting, sphericl shell of rdius R. ketch the electric field lines set up y this rrngement oth inside nd outside the shell. 41. A very lrge, thin, flt plte of luminum of re A hs totl chrge Q uniformly distriuted over its surfces. Assuming the sme chrge is spred uniformly over the upper surfce of n otherwise identicl glss plte, compre the electric fields just ove the center of the upper surfce of ech plte. 42. In certin region of spce, the electric field is E x 2 i^, where E is in newtons per coulom nd x is in meters. Electric chrges in this region re t rest nd remin t rest. () Find the volume density of electric chrge t x m. uggestion: Apply Guss s lw to ox etween x m nd x m 1 dx. () Could this region of spce e inside conductor? 43. Two identicl conducting spheres ech hving rdius AMT of cm re connected y light, 2.00-m-long conducting wire. A chrge of 60.0 mc is plced on one of the conductors. Assume the surfce distriution of chrge on ech sphere is uniform. Determine the tension in the wire. 44. A squre plte of copper with 50.0-cm sides hs no net chrge nd is plced in region of uniform electric field of 80.0 kn/c directed perpendiculrly to the plte. Find () the chrge density of ech fce of the plte nd () the totl chrge on ech fce. 45. A long, stright wire is surrounded y hollow metl cylinder whose xis coincides with tht of the wire. The wire hs chrge per unit length of l, nd the cylinder hs net chrge per unit length of 2l. From this informtion, use Guss s lw to find () the chrge per unit length on the inner surfce of the cylinder, () the chrge per unit length on the outer surfce of the cylinder, nd (c) the electric field outside the cylinder distnce r from the xis. 46. A thin, squre, conducting plte 50.0 cm on side lies M in the xy plne. A totl chrge of C is plced on the plte. Find () the chrge density on ech fce of the plte, () the electric field just ove the plte, nd (c) the electric field just elow the plte. You my ssume the chrge density is uniform. 47. A solid conducting sphere of rdius 2.00 cm hs M chrge of 8.00 mc. A conducting sphericl shell of inner rdius 4.00 cm nd outer rdius 5.00 cm is concentric with the solid sphere nd hs chrge of mc. Find the electric field t () r cm, () r cm, (c) r cm, nd (d) r cm from the center of this chrge configurtion. Additionl Prolems 48. Consider plne surfce in uniform electric field s E in Figure P24.48, where d 5 u 15.0 cm nd u If the net flux through the surfce is d 6.00 N? m 2 /C, find the mgnitude of the electric field. d Figure P Find the electric flux through Prolems 48 nd 49. the plne surfce shown in Figure P24.48 if u , E N/C, nd d cm. The electric field is uniform over the entire re of the surfce. 50. A hollow, metllic, sphericl shell hs exterior rdius Q/C m, crries no net chrge, nd is supported on n insulting stnd. The electric field everywhere just outside its surfce is 890 N/C rdilly towrd the center of the sphere. Explin wht you cn conclude out () the mount of chrge on the exterior surfce of the sphere nd the distriution of this chrge, () the mount of chrge on the interior surfce of the sphere nd its distriution, nd (c) the mount of chrge inside the shell nd its distriution. 51. A sphere of rdius R m surrounds prticle with chrge Q mc locted t its center s shown in Figure P Find the electric flux through circulr cp of hlf-ngle u A sphere of rdius R surrounds prticle with chrge Q locted t its center s shown in Figure P Find the electric flux through circulr cp of hlfngle u. 53. A very lrge conducting plte lying in the xy plne crries chrge per unit re of s. A second such plte locted ove the first plte t z 5 z 0 nd oriented prllel to the xy plne crries chrge per unit re of 22s. Find the electric field for () z, 0, () 0, z, z 0, nd (c) z. z A solid, insulting sphere of rdius hs uniform GP chrge density throughout its volume nd totl chrge Q. Concentric with this sphere is n unchrged, conducting, hollow sphere whose inner nd outer rdii re nd c s shown in Figure P24.54 (pge 744). We wish to R Q u Figure P24.51 Prolems 51 nd 52.

87 744 Chpter 24 Guss s Lw understnd completely the chrges nd electric fields t ll loctions. () Find the chrge contined within sphere of rdius r,. () From this vlue, find the mgnitude of the electric field for r,. (c) Wht chrge is contined within sphere of rdius r when, r,? (d) From this vlue, find the mgnitude of the electric field for r when, r,. (e) Now consider r when, r, c. Wht is the mgnitude of the electric field for this rnge of vlues of r? (f) From this vlue, wht must e the chrge on the inner surfce of the hollow sphere? (g) From prt (f), wht must e the chrge on the outer surfce of the hollow sphere? (h) Consider the three sphericl surfces of rdii,, nd c. Which of these surfces hs the lrgest mgnitude of surfce chrge density? 55. A solid insulting sphere of rdius cm crries net positive chrge of Q mc uniformly distriuted throughout its volume. Concentric with this sphere is conducting sphericl shell with inner rdius cm nd outer rdius c cm s shown in Figure P24.54, hving net chrge q mc. Prepre grph of the mgnitude of the electric field due to this configurtion versus r for 0, r, 25.0 cm. 56. Two infinite, nonconducting sheets of chrge re prllel to ech other s shown in Figure P The sheet on the left hs uniform surfce chrge density s, nd the one on the right hs uniform chrge density 2s. Clculte the electric s field t points () to the left of, () in s etween, nd (c) to the right of the Figure P24.56 two sheets. (d) Wht If? Find the electric fields in ll three regions if oth sheets hve positive uniform surfce chrge densities of vlue s. 57. For the configurtion shown in Figure P24.54, suppose cm, cm, nd c cm. Fur- W thermore, suppose the electric field t point 10.0 cm from the center is mesured to e N/C rdilly inwrd nd the electric field t point 50.0 cm from the center is of mgnitude 200 N/C nd points rdilly outwrd. From this informtion, find () the chrge on the insulting sphere, () the net chrge on the hollow conducting sphere, (c) the chrge on the inner surfce of the hollow conducting sphere, nd (d) the chrge on the outer surfce of the hollow conducting sphere. 58. An insulting solid sphere of rdius hs uniform volume chrge density nd crries totl positive chrge Q. A sphericl gussin surfce of rdius r, which shres common center with the insulting sphere, is inflted strting from r 5 0. () Find n expression for the electric flux pssing through the surfce of the gussin sphere s function of r for r,. () Find n expression for the electric flux for r.. (c) Plot the flux versus r. c Insultor Conductor Figure P24.54 Prolems 54, 55, nd A uniformly chrged sphericl shell with positive surfce chrge density s contins circulr hole in its sur- fce. The rdius r of the hole is smll compred with the rdius R of the sphere. Wht is the electric field t the center of the hole? uggestion: This prolem cn e solved y using the principle of superposition. 60. An infinitely long, cylindricl, insulting shell of inner rdius nd outer rdius hs uniform volume chrge density r. A line of uniform liner chrge density l is plced long the xis of the shell. Determine the electric field for () r,, (), r,, nd (c) r.. Chllenge Prolems 61. A sl of insulting mteril hs y nonuniform positive chrge density r 5 Cx 2, where x is mesured from the center of the sl s shown in Figure P24.61 nd C is constnt. The sl is infinite x in the y nd z directions. Derive O expressions for the electric field d in () the exterior regions (uxu. d/2) nd () the interior region of the sl (2d/2, x, d/2). Figure P Review. An erly (incorrect) Prolems 61 nd 69. AMT model of the hydrogen tom, suggested y J. J. Thomson, proposed tht positive cloud of chrge 1e ws uniformly distriuted throughout the volume of sphere of rdius R, with the electron (n equl-mgnitude negtively chrged prticle 2e) t the center. () Using Guss s lw, show tht the electron would e in equilirium t the center nd, if displced from the center distnce r, R, would experience restoring force of the form F 5 2Kr, where K is constnt. () how tht K 5 k e e 2 /R 3. (c) Find n expression for the frequency f of simple hrmonic oscilltions tht n electron of mss m e would undergo if displced smll distnce (, R) from the center nd relesed. (d) Clculte numericl vlue for R tht would result in frequency of Hz, the frequency of the light rdited in the most intense line in the hydrogen spectrum. 63. A closed surfce with dimensions m nd c m is locted s shown in Figure P The left edge of the closed surfce is locted t position x 5. The electric field throughout the region is nonuniform nd is given y E x 2 2 i^ N/C, where x is in meters. () Clculte the net electric flux z y x c Figure P24.63 E x

88 Prolems 745 leving the closed surfce. () Wht net chrge is enclosed y the surfce? 64. A sphere of rdius 2 is mde of y nonconducting mteril tht hs uniform volume chrge density r. Assume the mteril does not ffect the elec- x tric field. A sphericl cvity of 2 rdius is now removed from the sphere s shown in Figure P how tht the electric Figure P24.64 field within the cvity is uniform nd is given y E x 5 0 nd E y 5 r/3p A sphericlly symmetric chrge distriution hs chrge density given y r 5 /r, where is constnt. Find the electric field within the chrge distriution s function of r. Note: The volume element dv for sphericl shell of rdius r nd thickness dr is equl to 4pr 2 dr. 66. A solid insulting sphere of rdius R hs nonuniform chrge density tht vries with r ccording to the expression r 5 Ar 2, where A is constnt nd r, R is mesured from the center of the sphere. () how tht the mgnitude of the electric field outside (r. R) the sphere is E 5 AR 5 /5P 0 r 2. () how tht the mgnitude of the electric field inside (r, R) the sphere is E 5 Ar 3 /5P 0. Note: The volume element dv for sphericl shell of rdius r nd thickness dr is equl to 4pr 2 dr. 67. An infinitely long insulting cylinder of rdius R hs volume chrge density tht vries with the rdius s r 5 r 0 2 r where r 0,, nd re positive constnts nd r is the distnce from the xis of the cylinder. Use Guss s lw to determine the mgnitude of the electric field t rdil distnces () r, R nd () r. R. 68. A prticle with chrge Q is locted on the xis of circle of rdius R t distnce from the plne of the circle (Fig. P24.68). how tht if R one-fourth of the electric flux from the chrge psses through the circle, then R 5! Review. A sl of insulting mteril Q (infinite in the y nd z direc- tions) hs thickness d nd uniform Figure P24.68 positive chrge density r. An edge view of the sl is shown in Figure P () how tht the mgnitude of the electric field distnce x from its center nd inside the sl is E 5 rx/p 0. () Wht If? uppose n electron of chrge 2e nd mss m e cn move freely within the sl. It is relesed from rest t distnce x from the center. how tht the electron exhiits simple hrmonic motion with frequency f 5 1 re 2p Å m e P 0

89 chpter 25 Electric Potentil 25.1 Electric Potentil nd Potentil Difference 25.2 Potentil Difference in Uniform Electric Field 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges 25.4 Otining the Vlue of the Electric Field from the Electric Potentil 25.5 Electric Potentil Due to Continuous Chrge Distriutions 25.6 Electric Potentil Due to Chrged Conductor 25.7 The Millikn Oil-Drop Experiment 25.8 Applictions of Electrosttics Processes occurring during thunderstorms cuse lrge differences in electric potentil etween thundercloud nd the ground. The result of this potentil difference is n electricl dischrge tht we cll lightning, such s this disply. Notice t the left tht downwrd chnnel of lightning ( stepped leder) is out to mke contct with chnnel coming up from the ground ( return stroke). (Costzzurr/hutterstock.com) In Chpter 23, we linked our new study of electromgnetism to our erlier studies of force. Now we mke new link to our erlier investigtions into energy. The concept of potentil energy ws introduced in Chpter 7 in connection with such conservtive forces s the grvittionl force nd the elstic force exerted y spring. By using the lw of conservtion of energy, we could solve vrious prolems in mechnics tht were not solvle with n pproch using forces. The concept of potentil energy is lso of gret vlue in the study of electricity. Becuse the electrosttic force is conservtive, electrosttic phenomen cn e conveniently descried in terms of n electric potentil energy. This ide enles us to define quntity known s electric potentil. Becuse the electric potentil t ny point in n electric field is sclr quntity, we cn use it to descrie electrosttic phenomen more simply thn if we were to rely only on the electric field nd electric forces. The concept of electric potentil is of gret prcticl vlue in the opertion of electric circuits nd devices tht we will study in lter chpters Electric Potentil nd Potentil Difference 746 When chrge q is plced in n electric field E creted y some source chrge distriution, the prticle in field model tells us tht there is n electric force q E

90 25.1 Electric Potentil nd Potentil Difference 747 cting on the chrge. This force is conservtive ecuse the force etween chrges descried y Coulom s lw is conservtive. Let us identify the chrge nd the field s system. If the chrge is free to move, it will do so in response to the electric force. Therefore, the electric field will e doing work on the chrge. This work is internl to the system. This sitution is similr to tht in grvittionl system: When n oject is relesed ner the surfce of the Erth, the grvittionl force does work on the oject. This work is internl to the oject Erth system s discussed in ections 7.7 nd 7.8. When nlyzing electric nd mgnetic fields, it is common prctice to use the nottion d s to represent n infinitesiml displcement vector tht is oriented tngent to pth through spce. This pth my e stright or curved, nd n integrl performed long this pth is clled either pth integrl or line integrl (the two terms re synonymous). For n infinitesiml displcement d s of point chrge q immersed in n electric field, the work done within the chrge field system y the electric field on the chrge is W int 5 F e? d s 5 q E? d s. Recll from Eqution 7.26 tht internl work done in system is equl to the negtive of the chnge in the potentil energy of the system: W int 5 2DU. Therefore, s the chrge q is displced, the electric potentil energy of the chrge field system is chnged y n mount du 5 2W int 5 2q E? d s. For finite displcement of the chrge from some point A in spce to some other point B, the chnge in electric potentil energy of the system is B DU 5 2q 3 E? d s A (25.1) The integrtion is performed long the pth tht q follows s it moves from A to B. Becuse the force q E is conservtive, this line integrl does not depend on the pth tken from A to B. For given position of the chrge in the field, the chrge field system hs potentil energy U reltive to the configurtion of the system tht is defined s U 5 0. Dividing the potentil energy y the chrge gives physicl quntity tht depends only on the source chrge distriution nd hs vlue t every point in n electric field. This quntity is clled the electric potentil (or simply the potentil) V : V 5 U q (25.2) Becuse potentil energy is sclr quntity, electric potentil lso is sclr quntity. The potentil difference DV 5 V B 2 V A etween two points A nd B in n electric field is defined s the chnge in electric potentil energy of the system when chrge q is moved etween the points (Eq. 25.1) divided y the chrge: DV ; DU q B E? d s A (25.3) In this definition, the infinitesiml displcement d s is interpreted s the displcement etween two points in spce rther thn the displcement of point chrge s in Eqution Just s with potentil energy, only differences in electric potentil re meningful. We often tke the vlue of the electric potentil to e zero t some convenient point in n electric field. Potentil difference should not e confused with difference in potentil energy. The potentil difference etween A nd B exists solely ecuse of source chrge nd depends on the source chrge distriution (consider points A nd B in the discussion ove without the presence of the chrge q). For potentil energy to exist, we must hve system of two or more chrges. The potentil W WChnge in electric potentil energy of system Pitfll Prevention 25.1 Potentil nd Potentil Energy The potentil is chrcteristic of the field only, independent of chrged prticle tht my e plced in the field. Potentil energy is chrcteristic of the chrge-field system due to n interction etween the field nd chrged prticle plced in the field. W WPotentil difference etween two points

91 748 Chpter 25 Electric Potentil Pitfll Prevention 25.2 Voltge A vriety of phrses re used to descrie the potentil difference etween two points, the most common eing voltge, rising from the unit for potentil. A voltge pplied to device, such s television, or cross device is the sme s the potentil difference cross the device. Despite populr lnguge, voltge is not something tht moves through device. Pitfll Prevention 25.3 The Electron Volt The electron volt is unit of energy, NOT of potentil. The energy of ny system my e expressed in ev, ut this unit is most convenient for descriing the emission nd sorption of visile light from toms. Energies of nucler processes re often expressed in MeV. A B Figure 25.1 (Quick Quiz 25.1) Two points in n electric field. E energy elongs to the system nd chnges only if chrge is moved reltive to the rest of the system. This sitution is similr to tht for the electric field. An electric field exists solely ecuse of source chrge. An electric force requires two chrges: the source chrge to set up the field nd nother chrge plced within tht field. Let s now consider the sitution in which n externl gent moves the chrge in the field. If the gent moves the chrge from A to B without chnging the kinetic energy of the chrge, the gent performs work tht chnges the potentil energy of the system: W 5 DU. From Eqution 25.3, the work done y n externl gent in moving chrge q through n electric field t constnt velocity is W 5 q DV (25.4) Becuse electric potentil is mesure of potentil energy per unit chrge, the I unit of oth electric potentil nd potentil difference is joules per coulom, which is defined s volt (V): 1 V ; 1 J/C Tht is, s we cn see from Eqution 25.4, 1 J of work must e done to move 1-C chrge through potentil difference of 1 V. Eqution 25.3 shows tht potentil difference lso hs units of electric field times distnce. It follows tht the I unit of electric field (N/C) cn lso e expressed in volts per meter: 1 N/C 5 1 V/m Therefore, we cn stte new interprettion of the electric field: The electric field is mesure of the rte of chnge of the electric potentil with respect to position. A unit of energy commonly used in tomic nd nucler physics is the electron volt (ev), which is defined s the energy chrge field system gins or loses when chrge of mgnitude e (tht is, n electron or proton) is moved through potentil difference of 1 V. Becuse 1 V 5 1 J/C nd the fundmentl chrge is equl to C, the electron volt is relted to the joule s follows: 1 ev C? V J (25.5) For instnce, n electron in the em of typicl dentl x-ry mchine my hve speed of m/s. This speed corresponds to kinetic energy J (using reltivistic clcultions s discussed in Chpter 39), which is equivlent to ev. uch n electron hs to e ccelerted from rest through potentil difference of 67 kv to rech this speed. Q uick Quiz 25.1 In Figure 25.1, two points A nd B re locted within region in which there is n electric field. (i) How would you descrie the potentil difference DV 5 V B 2 V A? () It is positive. () It is negtive. (c) It is zero. (ii) A negtive chrge is plced t A nd then moved to B. How would you descrie the chnge in potentil energy of the chrge field system for this process? Choose from the sme possiilities Potentil Difference in Uniform Electric Field Equtions 25.1 nd 25.3 hold in ll electric fields, whether uniform or vrying, ut they cn e simplified for the specil cse of uniform field. First, consider uniform electric field directed long the negtive y xis s shown in Figure Let s clculte the potentil difference etween two points A nd B seprted y dis-

92 25.2 Potentil Difference in Uniform Electric Field 749 When positive chrge moves from point A to point B, the electric potentil energy of the chrge field system decreses. When n oject with mss moves from point A to point B, the grvittionl potentil energy of the oject field system decreses. Figure 25.2 () When the electric field E is directed downwrd, point B is t lower electric potentil thn point A. () A grvittionl nlog to the sitution in (). A A E q B d tnce d, where the displcement s points from A towrd B nd is prllel to the field lines. Eqution 25.3 gives B B B V B 2 V A 5 DV E? d s E ds 1cos E ds A Becuse E is constnt, it cn e removed from the integrl sign, which gives A B DV 5 2E 3 ds A DV 5 2Ed (25.6) The negtive sign indictes tht the electric potentil t point B is lower thn t point A; tht is, V B, V A. Electric field lines lwys point in the direction of decresing electric potentil s shown in Figure Now suppose chrge q moves from A to B. We cn clculte the chnge in the potentil energy of the chrge field system from Equtions 25.3 nd 25.6: DU 5 q DV 5 2qEd (25.7) This result shows tht if q is positive, then DU is negtive. Therefore, in system consisting of positive chrge nd n electric field, the electric potentil energy of the system decreses when the chrge moves in the direction of the field. If positive chrge is relesed from rest in this electric field, it experiences n electric force q E in the direction of E (downwrd in Fig. 25.2). Therefore, it ccelertes downwrd, gining kinetic energy. As the chrged prticle gins kinetic energy, the electric potentil energy of the chrge field system decreses y n equl mount. This equivlence should not e surprising; it is simply conservtion of mechnicl energy in n isolted system s introduced in Chpter 8. Figure 25.2 shows n nlogous sitution with grvittionl field. When prticle with mss m is relesed in grvittionl field, it ccelertes downwrd, gining kinetic energy. At the sme time, the grvittionl potentil energy of the oject field system decreses. The comprison etween system of positive chrge residing in n electricl field nd n oject with mss residing in grvittionl field in Figure 25.2 is useful for conceptulizing electricl ehvior. The electricl sitution, however, hs one feture tht the grvittionl sitution does not: the chrge cn e negtive. If q is negtive, then DU in Eqution 25.7 is positive nd the sitution is reversed. g m B d A W WPotentil difference etween two points in uniform electric field Pitfll Prevention 25.4 The ign of DV The negtive sign in Eqution 25.6 is due to the fct tht we strted t point A nd moved to new point in the sme direction s the electric field lines. If we strted from B nd moved to A, the potentil difference would e 1Ed. In uniform electric field, the mgnitude of the potentil difference is Ed nd the sign cn e determined y the direction of trvel.

93 750 Chpter 25 Electric Potentil Figure 25.3 A uniform electric field directed long the positive x xis. Three points in the electric field re leled. Point B is t lower electric potentil thn point A. E s B Chnge in potentil etween two points in uniform electric field 9 V 8 V 7 V 6 V A C B D Figure 25.4 (Quick Quiz 25.2) Four equipotentil surfces. E A system consisting of negtive chrge nd n electric field gins electric potentil energy when the chrge moves in the direction of the field. If negtive chrge is relesed from rest in n electric field, it ccelertes in direction opposite the direction of the field. For the negtive chrge to move in the direction of the field, n externl gent must pply force nd do positive work on the chrge. Now consider the more generl cse of chrged prticle tht moves etween A nd B in uniform electric field such tht the vector s is not prllel to the field lines s shown in Figure In this cse, Eqution 25.3 gives B DV E? d s 5 2 B E? 3 d s 5 2 E? s (25.8) A A u where gin E ws removed from the integrl ecuse it is constnt. The chnge in potentil energy of the chrge field system is A DU 5 q DV 5 2q E? s (25.9) Finlly, we conclude from Eqution 25.8 tht ll points in plne perpendiculr to uniform electric field re t the sme electric potentil. We cn see tht in Figure 25.3, where the potentil difference V B 2 V A is equl to the potentil difference V C 2 V A. (Prove this fct to yourself y working out two dot products for E? s : one for s AB, where the ngle u etween E nd s is ritrry s shown in Figure 25.3, nd one for s AC, where u 5 0.) Therefore, V B 5 V C. The nme equipotentil surfce is given to ny surfce consisting of continuous distriution of points hving the sme electric potentil. The equipotentil surfces ssocited with uniform electric field consist of fmily of prllel plnes tht re ll perpendiculr to the field. Equipotentil surfces ssocited with fields hving other symmetries re descried in lter sections. Q uick Quiz 25.2 The leled points in Figure 25.4 re on series of equipotentil surfces ssocited with n electric field. Rnk (from gretest to lest) the work done y the electric field on positively chrged prticle tht moves from A to B, from B to C, from C to D, nd from D to E. d C Points B nd C re t the sme electric potentil. Exmple 25.1 The Electric Field Between Two Prllel Pltes of Opposite Chrge A ttery hs specified potentil difference DV etween its terminls nd estlishes tht potentil difference etween conductors ttched to the terminls. A 12-V ttery is connected etween two prllel pltes s shown in Figure The seprtion etween the pltes is d cm, nd we ssume the electric field etween the pltes to e uniform. (This ssumption is resonle if the plte seprtion is smll reltive to the plte dimensions nd we do not consider loctions ner the plte edges.) Find the mgnitude of the electric field etween the pltes.

94 25.2 Potentil Difference in Uniform Electric Field continued olution Use Eqution 25.6 to evlute the mgnitude of the electric field etween the pltes: E 5 0 V B 2 V A 0 5 d 12 V m V/m The configurtion of pltes in Figure 25.5 is clled prllel-plte cpcitor nd is exmined in greter detil in Chpter 26. Exmple 25.2 Motion of Proton in Uniform Electric Field A proton is relesed from rest t point A in uniform electric field tht hs mgnitude of V/m (Fig. 25.6). The proton undergoes displcement of mgnitude d m to point B in the direction of E. Find the speed of the proton fter completing the displcement. olution Figure 25.5 (Exmple 25.1) A 12-V ttery connected to two prllel pltes. The electric field etween the pltes hs mgnitude given y the potentil difference DV divided y the plte seprtion d. Conceptulize Visulize the proton in Figure 25.6 moving downwrd through the potentil difference. The sitution is nlogous to n oject flling through grvittionl field. Also compre this exmple to Exmple where positive chrge ws moving in uniform electric field. In tht exmple, we pplied the prticle under constnt ccelertion nd nonisolted system models. Now tht we hve investigted electric potentil energy, wht model cn we use here? Ctegorize The system of the proton nd the two pltes in Figure 25.6 does not interct with the environment, so we model it s n isolted system for energy. Anlyze Write the pproprite reduction of Eqution 8.2, the conservtion of energy eqution, for the isolted system of the chrge nd the electric field: DK 1 DU 5 0 AM ustitute the chnges in energy for oth terms: 1 1 2mv e DV 5 0 B d V = 12 V Conceptulize In Exmple 24.5, we illustrted the uniform electric field etween prllel pltes. The new feture to this prolem is tht the electric field is relted to the new concept of electric potentil. Ctegorize The electric field is evluted from reltionship etween field nd potentil given in this section, so we ctegorize this exmple s sustitution prolem. A A v A 0 d v B B Figure 25.6 (Exmple 25.2) A proton ccelertes from A to B in the direction of the electric field. E olve for the finl speed of the proton nd sustitute for DV from Eqution 25.6: v 5 Å 22e DV m 5 Å 22e 1 2Ed 2 m 5 Å 2e Ed m ustitute numericl vlues: v 5 Å C V m kg m/s continued

95 752 Chpter 25 Electric Potentil 25.2 continued Finlize Becuse DV is negtive for the field, DU is lso negtive for the proton field system. The negtive vlue of DU mens the potentil energy of the system decreses s the proton moves in the direction of the electric field. As the proton ccelertes in the direction of the field, it gins kinetic energy while the electric potentil energy of the system decreses t the sme time. Figure 25.6 is oriented so tht the proton moves downwrd. The proton s motion is nlogous to tht of n oject flling in grvittionl field. Although the grvittionl field is lwys downwrd t the surfce of the Erth, n electric field cn e in ny direction, depending on the orienttion of the pltes creting the field. Therefore, Figure 25.6 could e rotted 908 or 1808 nd the proton could move horizontlly or upwrd in the electric field! A r A rˆ dr u d s r q r B The two dshed circles represent intersections of sphericl equipotentil surfces with the pge. Figure 25.7 The potentil difference etween points A nd B due to point chrge q depends only on the initil nd finl rdil coordintes r A nd r B. B Pitfll Prevention 25.5 imilr Eqution Wrning Do not confuse Eqution for the electric potentil of point chrge with Eqution 23.9 for the electric field of point chrge. Potentil is proportionl to 1/r, wheres the mgnitude of the field is proportionl to 1/r 2. The effect of chrge on the spce surrounding it cn e descried in two wys. The chrge sets up vector electric field E, which is relted to the force experienced y chrge plced in the field. It lso sets up sclr potentil V, which is relted to the potentil energy of the twochrge system when chrge is plced in the field Electric Potentil nd Potentil Energy Due to Point Chrges As discussed in ection 23.4, n isolted positive point chrge q produces n electric field directed rdilly outwrd from the chrge. To find the electric potentil t point locted distnce r from the chrge, let s egin with the generl expression for potentil difference, Eqution 25.3, B V B 2 V A E? d s where A nd B re the two ritrry points shown in Figure At ny point in spce, the electric field due to the point chrge is E 5 1k e q/r 2 2 r^ (Eq. 23.9), where r^ is unit vector directed rdilly outwrd from the chrge. Therefore, the quntity E? d s cn e expressed s A E? d s 5 k q e 2 r^? d s r Becuse the mgnitude of r^ is 1, the dot product r^? d s 5 ds cos u, where u is the ngle etween r^ nd d s. Furthermore, ds cos u is the projection of d s onto r^; therefore, ds cos u 5 dr. Tht is, ny displcement d s long the pth from point A to point B produces chnge dr in the mgnitude of r, the position vector of the point reltive to the chrge creting the field. Mking these sustitutions, we find tht E? d s 5 1k e q/r 2 2dr ; hence, the expression for the potentil difference ecomes r B dr V B 2 V A 5 2k e q 3 5 k q e r ` r A r 2 V B 2 V A 5 k e q c d (25.10) r B r A Eqution shows us tht the integrl of E? d s is independent of the pth etween points A nd B. Multiplying y chrge q 0 tht moves etween points A nd B, we see tht the integrl of q 0 E? d s is lso independent of pth. This ltter integrl, which is the work done y the electric force on the chrge q 0, shows tht the electric force is conservtive (see ection 7.7). We define field tht is relted to conservtive force s conservtive field. Therefore, Eqution tells us tht the electric field of fixed point chrge q is conservtive. Furthermore, Eqution expresses the importnt result tht the potentil difference etween ny two points A nd B in field creted y point chrge depends only on the rdil coordintes r A nd r B. It is customry to choose the reference of electric potentil for point chrge to e V 5 0 t r A 5 `. With this reference choice, the electric potentil due to point chrge t ny distnce r from the chrge is V 5 k e q r r B r A (25.11)

96 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges 753 A potentil k e q 1 /r 12 exists t point P due to chrge q 1. The potentil energy of the pir of chrges is given y k e q 1 q 2 /r 12. Figure 25.8 () Chrge q 1 estlishes n electric potentil V 1 t point P. () Chrge q 2 is rought from infinity to point P. r12 q 1 r12 q 1 P q V 1 k 1 e r 12 We otin the electric potentil resulting from two or more point chrges y pplying the superposition principle. Tht is, the totl electric potentil t some point P due to severl point chrges is the sum of the potentils due to the individul chrges. For group of point chrges, we cn write the totl electric potentil t P s V 5 k e i q i r i (25.12) Figure 25.8 shows chrge q 1, which sets up n electric field throughout spce. The chrge lso estlishes n electric potentil t ll points, including point P, where the electric potentil is V 1. Now imgine tht n externl gent rings chrge q 2 from infinity to point P. The work tht must e done to do this is given y Eqution 25.4, W 5 q 2 DV. This work represents trnsfer of energy cross the oundry of the two-chrge system, nd the energy ppers in the system s potentil energy U when the prticles re seprted y distnce r 12 s in Figure From Eqution 8.2, we hve W 5 DU. Therefore, the electric potentil energy of pir of point chrges 1 cn e found s follows: DU 5 W 5 q 2 DV U q 2 k e q 1 r U 5 k q 1q 2 e (25.13) r 12 If the chrges re of the sme sign, then U is positive. Positive work must e done y n externl gent on the system to ring the two chrges ner ech other (ecuse chrges of the sme sign repel). If the chrges re of opposite sign, s in Figure 25.8, then U is negtive. Negtive work is done y n externl gent ginst the ttrctive force etween the chrges of opposite sign s they re rought ner ech other; force must e pplied opposite the displcement to prevent q 2 from ccelerting towrd q 1. If the system consists of more thn two chrged prticles, we cn otin the totl potentil energy of the system y clculting U for every pir of chrges nd summing the terms lgericlly. For exmple, the totl potentil energy of the system of three chrges shown in Figure 25.9 is U 5 k e q 1q 2 1 q 1q 3 1 q 2q 3 (25.14) r 12 r 13 r 23 Physiclly, this result cn e interpreted s follows. Imgine q 1 is fixed t the position shown in Figure 25.9 ut q 2 nd q 3 re t infinity. The work n externl gent must do to ring q 2 from infinity to its position ner q 1 is k e q 1 q 2 /r 12, which is the first term in Eqution The lst two terms represent the work required to ring q 3 from infinity to its position ner q 1 nd q 2. (The result is independent of the order in which the chrges re trnsported.) 1 The expression for the electric potentil energy of system mde up of two point chrges, Eqution 25.13, is of the sme form s the eqution for the grvittionl potentil energy of system mde up of two point msses, 2Gm 1 m 2 /r (see Chpter 13). The similrity is not surprising considering tht oth expressions re derived from n inversesqure force lw. q 2 W WElectric potentil due to severl point chrges The potentil energy of this system of chrges is given y Eqution q 1 r 13 q 2 r 12 r 23 q 3 Figure 25.9 Three point chrges re fixed t the positions shown.

97 754 Chpter 25 Electric Potentil Exmple 25.3 The Electric Potentil Due to Two Point Chrges As shown in Figure 25.10, chrge q mc is locted t the origin nd chrge q mc is locted t (0, 3.00) m. (A) Find the totl electric potentil due to these chrges t the point P, whose coordintes re (4.00, 0) m. olution Conceptulize Recognize first tht the 2.00-mC nd mC chrges re source chrges nd set up n electric field s well s potentil t ll points in spce, including point P. Ctegorize The potentil is evluted using n eqution developed in this chpter, so we ctegorize this exmple s sustitution prolem. Use Eqution for the system of two source chrges: ustitute numericl vlues: Q uick Quiz 25.3 In Figure 25.8, tke q 2 to e negtive source chrge nd q 1 to e second chrge whose sign cn e chnged. (i) If q 1 is initilly positive nd is chnged to chrge of the sme mgnitude ut negtive, wht hppens to the potentil t the position of q 1 due to q 2? () It increses. () It decreses. (c) It remins the sme. (ii) When q 1 is chnged from positive to negtive, wht hppens to the potentil energy of the two-chrge system? Choose from the sme possiilities. y 6.00 mc 3.00 m 2.00 mc P x 6.00 mc 3.00 m 2.00 mc 3.00 mc x 4.00 m 4.00 m Figure (Exmple 25.3) () The electric potentil t P due to the two chrges q 1 nd q 2 is the lgeric sum of the potentils due to the individul chrges. () A third chrge q mc is rought from infinity to point P. V P 5 k e q 1 1 q 2 r 1 r 2 V P N # m 2 /C C 4.00 m V y C 5.00 m (B) Find the chnge in potentil energy of the system of two chrges plus third chrge q mc s the ltter chrge moves from infinity to point P (Fig ). olution Assign U i 5 0 for the system to the initil configurtion in which the chrge q 3 is t infinity. Use Eqution 25.2 to evlute the potentil energy for the configurtion in which the chrge is t P: U f 5 q 3 V P ustitute numericl vlues to evlute DU: DU 5 U f 2 U i 5 q 3 V P ( C)( V) J Therefore, ecuse the potentil energy of the system hs decresed, n externl gent hs to do positive work to remove the chrge q 3 from point P ck to infinity. Wht If? You re working through this exmple with clssmte nd she sys, Wit minute! In prt (B), we ignored the potentil energy ssocited with the pir of chrges q 1 nd q 2! How would you respond? Answer Given the sttement of the prolem, it is not necessry to include this potentil energy ecuse prt (B) sks for the chnge in potentil energy of the system s q 3 is rought in from infinity. Becuse the configurtion of chrges q 1 nd q 2 does not chnge in the process, there is no DU ssocited with these chrges. Hd prt (B) sked to find the chnge in potentil energy when ll three chrges strt out infinitely fr prt nd re then rought to the positions in Figure 25.10, however, you would hve to clculte the chnge using Eqution

98 25.4 Otining the Vlue of the Electric Field from the Electric Potentil Otining the Vlue of the Electric Field from the Electric Potentil The electric field E nd the electric potentil V re relted s shown in Eqution 25.3, which tells us how to find DV if the electric field E is known. Wht if the sitution is reversed? How do we clculte the vlue of the electric field if the electric potentil is known in certin region? From Eqution 25.3, the potentil difference dv etween two points distnce ds prt cn e expressed s dv 5 2 E? d s (25.15) If the electric field hs only one component E x, then E? d s 5 E x dx. Therefore, Eqution ecomes dv 5 2E x dx, or E x 5 2 dv (25.16) dx Tht is, the x component of the electric field is equl to the negtive of the derivtive of the electric potentil with respect to x. imilr sttements cn e mde out the y nd z components. Eqution is the mthemticl sttement of the electric field eing mesure of the rte of chnge with position of the electric potentil s mentioned in ection Experimentlly, electric potentil nd position cn e mesured esily with voltmeter ( device for mesuring potentil difference) nd meterstick. Consequently, n electric field cn e determined y mesuring the electric potentil t severl positions in the field nd mking grph of the results. According to Eqution 25.16, the slope of grph of V versus x t given point provides the mgnitude of the electric field t tht point. Imgine strting t point nd then moving through displcement d s long n equipotentil surfce. For this motion, dv 5 0 ecuse the potentil is constnt long n equipotentil surfce. From Eqution 25.15, we see tht dv 5 2 E? d s 5 0; therefore, ecuse the dot product is zero, E must e perpendiculr to the displcement long the equipotentil surfce. This result shows tht the equipotentil surfces must lwys e perpendiculr to the electric field lines pssing through them. As mentioned t the end of ection 25.2, the equipotentil surfces ssocited with uniform electric field consist of fmily of plnes perpendiculr to the field lines. Figure shows some representtive equipotentil surfces for this sitution. A uniform electric field produced y n infinite sheet of chrge A sphericlly symmetric electric field produced y point chrge An electric field produced y n electric dipole q E c Figure Equipotentil surfces (the dshed lue lines re intersections of these surfces with the pge) nd electric field lines. In ll cses, the equipotentil surfces re perpendiculr to the electric field lines t every point.

99 756 Chpter 25 Electric Potentil Finding the electric field from the potentil P r 2 dq 2 r 3 r 1 dq 1 dq 3 Figure The electric potentil t point P due to continuous chrge distriution cn e clculted y dividing the chrge distriution into elements of chrge dq nd summing the electric potentil contriutions over ll elements. Three smple elements of chrge re shown. If the chrge distriution creting n electric field hs sphericl symmetry such tht the volume chrge density depends only on the rdil distnce r, the electric field is rdil. In this cse, E? d s 5 E r dr, nd we cn express dv s dv 5 2E r dr. Therefore, E r 5 2 dv (25.17) dr For exmple, the electric potentil of point chrge is V 5 k e q/r. Becuse V is function of r only, the potentil function hs sphericl symmetry. Applying Eqution 25.17, we find tht the mgnitude of the electric field due to the point chrge is E r 5 k e q/r 2, fmilir result. Notice tht the potentil chnges only in the rdil direction, not in ny direction perpendiculr to r. Therefore, V (like E r ) is function only of r, which is gin consistent with the ide tht equipotentil surfces re perpendiculr to field lines. In this cse, the equipotentil surfces re fmily of spheres concentric with the sphericlly symmetric chrge distriution (Fig ). The equipotentil surfces for n electric dipole re sketched in Figure 25.11c. In generl, the electric potentil is function of ll three sptil coordintes. If V(r) is given in terms of the Crtesin coordintes, the electric field components E x, E y, nd E z cn redily e found from V(x, y, z) s the prtil derivtives 2 E x 5 2 'V 'x E y 5 2 'V 'y E z 5 2 'V 'z (25.18) Q uick Quiz 25.4 In certin region of spce, the electric potentil is zero everywhere long the x xis. (i) From this informtion, you cn conclude tht the x component of the electric field in this region is () zero, () in the positive x direction, or (c) in the negtive x direction. (ii) uppose the electric potentil is 12 V everywhere long the x xis. From the sme choices, wht cn you conclude out the x component of the electric field now? 25.5 Electric Potentil Due to Continuous Chrge Distriutions In ection 25.3, we found how to determine the electric potentil due to smll numer of chrges. Wht if we wish to find the potentil due to continuous distriution of chrge? The electric potentil in this sitution cn e clculted using two different methods. The first method is s follows. If the chrge distriution is known, we consider the potentil due to smll chrge element dq, treting this element s point chrge (Fig ). From Eqution 25.11, the electric potentil dv t some point P due to the chrge element dq is dv 5 k dq e r (25.19) where r is the distnce from the chrge element to point P. To otin the totl potentil t point P, we integrte Eqution to include contriutions from ll elements of the chrge distriution. Becuse ech element is, in generl, different distnce from point P nd k e is constnt, we cn express V s Electric potentil due to continuous chrge distriution V 5 k e 3 dq r 2 In vector nottion, E is often written in Crtesin coordinte systems s (25.20) where = is clled the grdient opertor. E 5 2=V 5 2i^ ' 'x 1 j^ ' 'y 1 k^ ' 'z V

100 25.5 Electric Potentil Due to Continuous Chrge Distriutions 757 In effect, we hve replced the sum in Eqution with n integrl. In this expression for V, the electric potentil is tken to e zero when point P is infinitely fr from the chrge distriution. The second method for clculting the electric potentil is used if the electric field is lredy known from other considertions such s Guss s lw. If the chrge distriution hs sufficient symmetry, we first evlute E using Guss s lw nd then sustitute the vlue otined into Eqution 25.3 to determine the potentil difference DV etween ny two points. We then choose the electric potentil V to e zero t some convenient point. Prolem-olving trtegy Clculting Electric Potentil The following procedure is recommended for solving prolems tht involve the determintion of n electric potentil due to chrge distriution. 1. Conceptulize. Think crefully out the individul chrges or the chrge distriution you hve in the prolem nd imgine wht type of potentil would e creted. Appel to ny symmetry in the rrngement of chrges to help you visulize the potentil. 2. Ctegorize. Are you nlyzing group of individul chrges or continuous chrge distriution? The nswer to this question will tell you how to proceed in the Anlyze step. 3. Anlyze. When working prolems involving electric potentil, rememer tht it is sclr quntity, so there re no components to consider. Therefore, when using the superposition principle to evlute the electric potentil t point, simply tke the lgeric sum of the potentils due to ech chrge. You must keep trck of signs, however. As with potentil energy in mechnics, only chnges in electric potentil re significnt; hence, the point where the potentil is set t zero is ritrry. When deling with point chrges or finite-sized chrge distriution, we usully define V 5 0 to e t point infinitely fr from the chrges. If the chrge distriution itself extends to infinity, however, some other nery point must e selected s the reference point. () If you re nlyzing group of individul chrges: Use the superposition principle, which sttes tht when severl point chrges re present, the resultnt potentil t point P in spce is the lgeric sum of the individul potentils t P due to the individul chrges (Eq ). Exmple 25.4 elow demonstrtes this procedure. () If you re nlyzing continuous chrge distriution: Replce the sums for evluting the totl potentil t some point P from individul chrges y integrls (Eq ). The totl potentil t P is otined y integrting over the entire chrge distriution. For mny prolems, it is possile in performing the integrtion to express dq nd r in terms of single vrile. To simplify the integrtion, give creful considertion to the geometry involved in the prolem. Exmples 25.5 through 25.7 demonstrte such procedure. To otin the potentil from the electric field: Another method used to otin the potentil is to strt with the definition of the potentil difference given y Eqution If E is known or cn e otined esily (such s from Guss s lw), the line integrl of E? d s cn e evluted. 4. Finlize. Check to see if your expression for the potentil is consistent with your mentl representtion nd reflects ny symmetry you noted previously. Imgine vrying prmeters such s the distnce of the oservtion point from the chrges or the rdius of ny circulr ojects to see if the mthemticl result chnges in resonle wy.

101 758 Chpter 25 Electric Potentil Exmple 25.4 The Electric Potentil Due to Dipole An electric dipole consists of two chrges of equl mgnitude nd opposite sign seprted y distnce 2 s shown in Figure The dipole is long the x xis nd is centered t the origin. (A) Clculte the electric potentil t point P on the y xis. olution Conceptulize Compre this sitution to tht in prt (B) of Exmple It is the sme sitution, ut here we re seeking the electric potentil rther thn the electric field. q x Ctegorize We ctegorize the prolem s one in which we hve smll numer of prticles rther thn continuous distriution of chrge. The electric potentil cn e evluted y summing the potentils due to the individul chrges. Anlyze Use Eqution to find the electric potentil t P due to the two chrges: V P 5 k e i (B) Clculte the electric potentil t point R on the positive x xis. olution Use Eqution to find the electric potentil t R due to the two chrges: (C) Clculte V nd E x t point on the x xis fr from the dipole. olution For point R fr from the dipole such tht x.., neglect 2 in the denomintor of the nswer to prt (B) nd write V in this limit: Use Eqution nd this result to clculte the x component of the electric field t point on the x xis fr from the dipole: V R 5 k e i V R 5 lim x.. q y y P Figure (Exmple 25.4) An electric dipole locted on the x xis. q i q 5 k r e i " 2 1 y 1 2q 2 " 2 1 y q i 5 k r e 2q i x 2 1 q x k eq x k eq x < E x 5 2 dv dx 5 2 d dx 2 2k eq x 2 2 2k eq 1x.. 2 x 2 5 2k e q d dx 1 x k e q 1x.. 2 x 3 Finlize The potentils in prts (B) nd (C) re negtive ecuse points on the positive x xis re closer to the negtive chrge thn to the positive chrge. For the sme reson, the x component of the electric field is negtive. Notice tht we hve 1/r 3 flloff of the electric field with distnce fr from the dipole, similr to the ehvior of the electric field on the y xis in Exmple Wht If? uppose you wnt to find the electric field t point P on the y xis. In prt (A), the electric potentil ws found to e zero for ll vlues of y. Is the electric field zero t ll points on the y xis? Answer No. Tht there is no chnge in the potentil long the y xis tells us only tht the y component of the electric field is zero. Look ck t Figure in Exmple We showed there tht the electric field of dipole on the y xis hs only n x component. We could not find the x component in the current exmple ecuse we do not hve n expression for the potentil ner the y xis s function of x. R x

102 25.5 Electric Potentil Due to Continuous Chrge Distriutions 759 Exmple 25.5 Electric Potentil Due to Uniformly Chrged Ring (A) Find n expression for the electric potentil t point P locted on the perpendiculr centrl xis of uniformly chrged ring of rdius nd totl chrge Q. olution dq 2 x 2 Conceptulize tudy Figure 25.14, in which the ring is oriented so tht its plne is perpendiculr to the x xis nd its center is t the origin. Notice tht the symmetry of the sitution mens tht ll the chrges on the ring re the sme distnce from point P. Compre this exmple to Exmple Notice tht no vector considertions re necessry here ecuse electric potentil is sclr. Ctegorize Becuse the ring consists of continuous distriution of chrge rther thn set of discrete chrges, we must use the integrtion technique represented y Eqution in this exmple. Anlyze We tke point P to e t distnce x from the center of the ring s shown in Figure Use Eqution to express V in terms of the V 5 k dq e 3 r 5 k e 3 geometry: Noting tht nd x do not vry for n integrtion over the ring, ring " 2 1 x 2 in front of the integrl sign nd integrte over the ring: (B) Find n expression for the mgnitude of the electric field t point P. olution From symmetry, notice tht long the x xis E cn hve only n x component. Therefore, pply Eqution to Eqution 25.21: k e V 5 " 2 1 x 2 3 dq 5 Figure (Exmple 25.5) A uniformly chrged ring of rdius lies in plne perpendiculr to the x xis. All elements dq of the ring re the sme distnce from point P lying on the x xis. dq " 2 1 x 2 E x 5 2 dv dx 5 2k eq d dx 12 1 x /2 E x 5 5 2k e Q x /2 12x2 x k e Q (25.21) " 2 1 x 2 k e x Q (25.22) x 2 3/2 2 Finlize The only vrile in the expressions for V nd E x is x. Tht is not surprising ecuse our clcultion is vlid only for points long the x xis, where y nd z re oth zero. This result for the electric field grees with tht otined y direct integrtion (see Exmple 23.8). For prctice, use the result of prt (B) in Eqution 25.3 to verify tht the potentil is given y the expression in prt (A). Exmple 25.6 Electric Potentil Due to Uniformly Chrged Disk A uniformly chrged disk hs rdius R nd surfce chrge density s. (A) Find the electric potentil t point P long the perpendiculr centrl xis of the disk. olution Conceptulize If we consider the disk to e set of concentric rings, we cn use our result from Exmple 25.5 which gives the potentil due to ring of rdius nd sum the contriutions of ll rings mking up the disk. Figure continued P x

103 760 Chpter 25 Electric Potentil 25.6 continued shows one such ring. Becuse point P is on the centrl xis of the disk, symmetry gin tells us tht ll points in given ring re the sme distnce from P. Ctegorize Becuse the disk is continuous, we evlute the potentil due to continuous chrge distriution rther thn group of individul chrges. Anlyze Find the mount of chrge dq on ring of rdius r nd width dr s shown in Figure 25.15: Use this result in Eqution in Exmple 25.5 (with replced y the vrile r nd Q replced y the differentil dq) to find the potentil due to the ring: To otin the totl potentil t P, integrte this expression over the limits r 5 0 to r 5 R, noting tht x is constnt: This integrl is of the common form e u n du, where n nd u 5 r 2 1 x 2, nd hs the vlue u n11 /(n 1 1). Use this result to evlute the integrl: As in Exmple 25.5, use Eqution to find the electric field t ny xil point: dq 5 s da 5 s12pr dr2 5 2psr dr dv 5 k e dq "r 2 1 x 5 k e 2psr dr 2 "r 2 1 x 2 R 2r dr V 5 pk e s 3 0 "r 2 1 x 5 pk 2 e s 3 V 5 2pk e s 3 1R 2 1 x 2 2 1/2 2 x4 (25.23) (B) Find the x component of the electric field t point P long the perpendiculr centrl xis of the disk. olution E x 5 2 dv dx 5 2pk e s c1 2 x 1R 2 1 x 2 1/2 d (25.24) 2 Finlize Compre Eqution with the result of Exmple They re the sme. The clcultion of V nd E for n ritrry point off the x xis is more difficult to perform ecuse of the sence of symmetry nd we do not tret tht sitution in this ook. Exmple 25.7 Electric Potentil Due to Finite Line of Chrge A rod of length, locted long the x xis hs totl chrge Q nd uniform liner chrge density l. Find the electric potentil t point P locted on the y xis distnce from the origin (Fig ). olution Figure (Exmple 25.6) A uniformly chrged disk of rdius R lies in plne perpendiculr to the x xis. The clcultion of the electric potentil t ny point P on the x xis is simplified y dividing the disk into mny rings of rdius r nd width dr, with re 2pr dr. Conceptulize The potentil t P due to every segment of chrge on the rod is positive ecuse every segment crries positive chrge. Notice tht we hve no symmetry to ppel to here, ut the simple geometry should mke the prolem solvle. Ctegorize Becuse the rod is continuous, we evlute the potentil due to continuous chrge distriution rther thn group of individul chrges. Anlyze In Figure 25.16, the rod lies long the x xis, dx is the length of one smll segment, nd dq is the chrge on tht segment. Becuse the rod hs chrge per unit length l, the chrge dq on the smll segment is dq 5 l dx. R dr r 0 R x r 2 x 2 da 2pr dr P 1r 2 1 x /2 2r dr P O y x r Figure (Exmple 25.7) A uniform line chrge of length, locted long the x xis. To clculte the electric potentil t P, the line chrge is divided into segments ech of length dx nd ech crrying chrge dq 5 l dx. dq dx x x

104 25.6 Electric Potentil Due to Chrged Conductor continued Find the potentil t P due to one segment of the rod t n ritrry position x: Find the totl potentil t P y integrting this expression over the limits x 5 0 to x 5,: Noting tht k e nd l 5 Q /, re constnts nd cn e removed from the integrl, evlute the integrl with the help of Appendix B: Evlute the result etween the limits: V 5 k Q e, 3ln 1, 1 " 2 1, ln 4 5 k Q e, ln, 1 " 2 1, 2 (25.25) Wht If? Wht if you were sked to find the electric field t point P? Would tht e simple clcultion? Answer Clculting the electric field y mens of Eqution would e little messy. There is no symmetry to ppel to, nd the integrtion over the line of chrge would represent vector ddition of electric fields t point P. Using Eqution 25.18, you could find E y y replcing with y in Eqution nd performing the differentition with respect to y. Becuse the chrged rod in Figure V 5 3, V 5 k e l 3, 25.6 Electric Potentil Due to Chrged Conductor In ection 24.4, we found tht when solid conductor in equilirium crries net chrge, the chrge resides on the conductor s outer surfce. Furthermore, the electric field just outside the conductor is perpendiculr to the surfce nd the field inside is zero. We now generte nother property of chrged conductor, relted to electric potentil. Consider two points A nd B on the surfce of chrged conductor s shown in Figure Along surfce pth connecting these points, E is lwys Notice from the spcing of the positive signs tht the surfce chrge density is nonuniform. dv 5 k e dq r k e 0 l dx 5 k e " 2 1 x 2 0 l dx " 2 1 x 2 dx " 2 1 x 5 k Q 2 e, ln 1x 1 " 2 1 x 2 2 ` Finlize If,,,, the potentil t P should pproch tht of point chrge ecuse the rod is very short compred to the distnce from the rod to P. By using series expnsion for the nturl logrithm from Appendix B.5, it is esy to show tht Eqution ecomes V = k e Q / lies entirely to the right of x 5 0, the electric field t point P would hve n x component to the left if the rod is chrged positively. You cnnot use Eqution to find the x component of the field, however, ecuse the potentil due to the rod ws evluted t specific vlue of x (x 5 0) rther thn generl vlue of x. You would hve to find the potentil s function of oth x nd y to e le to find the x nd y components of the electric field using Eqution Pitfll Prevention 25.6 Potentil My Not Be Zero The electric potentil inside the conductor is not necessrily zero in Figure 25.17, even though the electric field is zero. Eqution shows tht zero vlue of the field results in no chnge in the potentil from one point to nother inside the conductor. Therefore, the potentil everywhere inside the conductor, including the surfce, hs the sme vlue, which my or my not e zero, depending on where the zero of potentil is defined., 0 A B E Figure An ritrrily shped conductor crrying positive chrge. When the conductor is in electrosttic equilirium, ll the chrge resides t the surfce, E 5 0 inside the conductor, nd the direction of E immeditely outside the conductor is perpendiculr to the surfce. The electric potentil is constnt inside the conductor nd is equl to the potentil t the surfce.

105 762 Chpter 25 Electric Potentil c k e Q R R V E k e Q r k e Q r 2 r R Figure () The excess chrge on conducting sphere of rdius R is uniformly distriuted on its surfce. () Electric potentil versus distnce r from the center of the chrged conducting sphere. (c) Electric field mgnitude versus distnce r from the center of the chrged conducting sphere. Exmple 25.8 r perpendiculr to the displcement d s ; therefore, E? d s 5 0. Using this result nd Eqution 25.3, we conclude tht the potentil difference etween A nd B is necessrily zero: B V B 2 V A E? d s 5 0 This result pplies to ny two points on the surfce. Therefore, V is constnt everywhere on the surfce of chrged conductor in equilirium. Tht is, the surfce of ny chrged conductor in electrosttic equilirium is n equipotentil surfce: every point on the surfce of chrged conductor in equilirium is t the sme electric potentil. Furthermore, ecuse the electric field is zero inside the conductor, the electric potentil is constnt everywhere inside the conductor nd equl to its vlue t the surfce. Becuse of the constnt vlue of the potentil, no work is required to move chrge from the interior of chrged conductor to its surfce. Consider solid metl conducting sphere of rdius R nd totl positive chrge Q s shown in Figure As determined in prt (A) of Exmple 24.3, the electric field outside the sphere is k e Q /r 2 nd points rdilly outwrd. Becuse the field outside sphericlly symmetric chrge distriution is identicl to tht of point chrge, we expect the potentil to lso e tht of point chrge, k e Q /r. At the surfce of the conducting sphere in Figure 25.18, the potentil must e k e Q /R. Becuse the entire sphere must e t the sme potentil, the potentil t ny point within the sphere must lso e k e Q /R. Figure is plot of the electric potentil s function of r, nd Figure 25.18c shows how the electric field vries with r. When net chrge is plced on sphericl conductor, the surfce chrge density is uniform s indicted in Figure If the conductor is nonsphericl s in Figure 25.17, however, the surfce chrge density is high where the rdius of curvture is smll (s noted in ection 24.4) nd low where the rdius of curvture is lrge. Becuse the electric field immeditely outside the conductor is proportionl to the surfce chrge density, the electric field is lrge ner convex points hving smll rdii of curvture nd reches very high vlues t shrp points. In Exmple 25.8, the reltionship etween electric field nd rdius of curvture is explored mthemticlly. Two Connected Chrged pheres Two sphericl conductors of rdii r 1 nd r 2 re seprted y distnce much greter thn the rdius of either sphere. The spheres re connected y conducting wire s shown in Figure The chrges on the spheres in equilirium re q 1 nd q 2, respectively, nd they re uniformly chrged. Find the rtio of the mgnitudes of the electric fields t the surfces of the spheres. olution A q 1 r 1 Conceptulize Imgine the spheres re much frther prt thn shown in Figure Becuse they re so fr prt, the field of one does not ffect the chrge distriution on the other. The conducting wire etween them ensures tht oth spheres hve the sme electric potentil. Ctegorize Becuse the spheres re so fr prt, we model the chrge distriution on them s sphericlly symmetric, nd we cn model the field nd potentil outside the spheres to e tht due to point chrges. q 2 Figure (Exmple 25.8) Two chrged sphericl conductors connected y conducting wire. The spheres re t the sme electric potentil V. r 2 Anlyze et the electric potentils t the surfces of the spheres equl to ech other: V 5 k e q 1 r 1 5 k e q 2 r 2

106 25.6 Electric Potentil Due to Chrged Conductor continued olve for the rtio of chrges on the spheres: (1) q 1 q 2 5 r 1 r 2 Write expressions for the mgnitudes of the electric fields t the surfces of the spheres: Evlute the rtio of these two fields: E 1 5 k e q 1 r 1 2 nd E 2 5 k e q 2 r 2 2 E 1 5 q 2 1 r 2 E 2 q 2 r 1 2 ustitute for the rtio of chrges from Eqution (1): (2) A Cvity Within Conductor uppose conductor of ritrry shpe contins cvity s shown in Figure Let s ssume no chrges re inside the cvity. In this cse, the electric field inside the cvity must e zero regrdless of the chrge distriution on the outside surfce of the conductor s we mentioned in ection Furthermore, the field in the cvity is zero even if n electric field exists outside the conductor. To prove this point, rememer tht every point on the conductor is t the sme electric potentil; therefore, ny two points A nd B on the cvity s surfce must e t the sme potentil. Now imgine field E exists in the cvity nd evlute the potentil difference V B 2 V A defined y Eqution 25.3: B V B 2 V A E? d s Becuse V B 2 V A 5 0, the integrl of E? d s must e zero for ll pths etween ny two points A nd B on the conductor. The only wy tht cn e true for ll pths is if E is zero everywhere in the cvity. Therefore, cvity surrounded y conducting wlls is field-free region s long s no chrges re inside the cvity. Coron Dischrge A phenomenon known s coron dischrge is often oserved ner conductor such s high-voltge power line. When the electric field in the vicinity of the conductor is sufficiently strong, electrons resulting from rndom ioniztions of ir molecules ner the conductor ccelerte wy from their prent molecules. These rpidly moving electrons cn ionize dditionl molecules ner the conductor, creting more free electrons. The oserved glow (or coron dischrge) results from the recomintion of these free electrons with the ionized ir molecules. If conductor hs n irregulr shpe, the electric field cn e very high ner shrp points or edges of the conductor; consequently, the ioniztion process nd coron dischrge re most likely to occur round such points. Coron dischrge is used in the electricl trnsmission industry to locte roken or fulty components. For exmple, roken insultor on trnsmission tower hs shrp edges where coron dischrge is likely to occur. imilrly, coron dischrge will occur t the shrp end of roken conductor strnd. Oservtion of these dischrges is difficult ecuse the visile rdition emitted is wek nd most of the rdition is in the ultrviolet. (We will discuss ultrviolet rdition nd other portions of the electromgnetic spectrum in ection 34.7.) Even use of trditionl ultrviolet cmers is of little help ecuse the rdition from the coron A E 1 5 r 1 r 2 E 2 r 2 r 5 r Finlize The field is stronger in the vicinity of the smller sphere even though the electric potentils t the surfces of oth spheres re the sme. If r 2 0, then E 2 `, verifying the sttement ove tht the electric field is very lrge t shrp points. 2 r 1 The electric field in the cvity is zero regrdless of the chrge on the conductor. B A Figure A conductor in electrosttic equilirium contining cvity.

107 764 Chpter 25 Electric Potentil Figure chemtic drwing of the Millikn oil-drop pprtus. Oil droplets Pinhole d q v With the electric field off, the droplet flls t terminl velocity v T under the influence of the grvittionl nd drg forces. v T v T mg qe F D F D q mg When the electric field is turned on, the droplet moves upwrd t terminl velocity v T under the influence of the electric, grvittionl, nd drg forces. E Figure The forces cting on negtively chrged oil droplet in the Millikn experiment. Telescope with scle in eyepiece dischrge is overwhelmed y ultrviolet rdition from the un. Newly developed dul- spectrum devices comine nrrow-nd ultrviolet cmer with visilelight cmer to show dylight view of the coron dischrge in the ctul loction on the trnsmission tower or cle. The ultrviolet prt of the cmer is designed to operte in wvelength rnge in which rdition from the un is very wek The Millikn Oil-Drop Experiment Roert Millikn performed rillint set of experiments from 1909 to 1913 in which he mesured e, the mgnitude of the elementry chrge on n electron, nd demonstrted the quntized nture of this chrge. His pprtus, digrmmed in Figure 25.21, contins two prllel metllic pltes. Oil droplets from n tomizer re llowed to pss through smll hole in the upper plte. Millikn used x-rys to ionize the ir in the chmer so tht freed electrons would dhere to the oil drops, giving them negtive chrge. A horizontlly directed light em is used to illuminte the oil droplets, which re viewed through telescope whose long xis is perpendiculr to the light em. When viewed in this mnner, the droplets pper s shining strs ginst drk ckground nd the rte t which individul drops fll cn e determined. Let s ssume single drop hving mss m nd crrying chrge q is eing viewed nd its chrge is negtive. If no electric field is present etween the pltes, the two forces cting on the chrge re the grvittionl force mg cting downwrd 3 nd viscous drg force FD cting upwrd s indicted in Figure The drg force is proportionl to the drop s speed s discussed in ection 6.4. When the drop reches its terminl speed v T the two forces lnce ech other (mg 5 F D ). Now suppose ttery connected to the pltes sets up n electric field etween the pltes such tht the upper plte is t the higher electric potentil. In this cse, third force q E cts on the chrged drop. The prticle in field model pplies twice to the prticle: it is in grvittionl field nd n electric field. Becuse q is negtive nd E is directed downwrd, this electric force is directed upwrd s shown in Figure If this upwrd force is strong enough, the drop moves upwrd nd the drg force F D r cts downwrd. When the upwrd electric force q E lnces the sum of the grvittionl force nd the downwrd drg force F D r, the drop reches new terminl speed v9 T in the upwrd direction. With the field turned on, drop moves slowly upwrd, typiclly t rtes of hundredths of centimeter per second. The rte of fll in the sence of field is comprle. Hence, one cn follow single droplet for hours, lterntely rising nd flling, y simply turning the electric field on nd off. 3 There is lso uoynt force on the oil drop due to the surrounding ir. This force cn e incorported s correction in the grvittionl force mg on the drop, so we will not consider it in our nlysis.

108 25.8 Applictions of Electrosttics 765 After recording mesurements on thousnds of droplets, Millikn nd his coworkers found tht ll droplets, to within out 1% precision, hd chrge equl to some integer multiple of the elementry chrge e : q 5 ne n 5 0, 21, 22, 23,... where e C. Millikn s experiment yields conclusive evidence tht chrge is quntized. For this work, he ws wrded the Noel Prize in Physics in Applictions of Electrosttics The prcticl ppliction of electrosttics is represented y such devices s lightning rods nd electrosttic precipittors nd y such processes s xerogrphy nd the pinting of utomoiles. cientific devices sed on the principles of electrosttics include electrosttic genertors, the field-ion microscope, nd ion-drive rocket engines. Detils of two devices re given elow. The Vn de Grff Genertor Experimentl results show tht when chrged conductor is plced in contct with the inside of hollow conductor, ll the chrge on the chrged conductor is trnsferred to the hollow conductor. In principle, the chrge on the hollow conductor nd its electric potentil cn e incresed without limit y repetition of the process. In 1929, Roert J. Vn de Grff ( ) used this principle to design nd uild n electrosttic genertor, nd schemtic representtion of it is given in Figure This type of genertor ws once used extensively in nucler physics reserch. Chrge is delivered continuously to high-potentil electrode y mens of moving elt of insulting mteril. The high-voltge electrode is hollow metl dome mounted on n insulting column. The elt is chrged t point A y mens of coron dischrge etween com-like metllic needles nd grounded grid. The needles re mintined t positive electric potentil of typiclly 10 4 V. The positive chrge on the moving elt is trnsferred to the dome y second com of needles t point B. Becuse the electric field inside the dome is negligile, the positive chrge on the elt is esily trnsferred to the conductor regrdless of its potentil. In prctice, it is possile to increse the electric potentil of the dome until electricl dischrge occurs through the ir. Becuse the rekdown electric field in ir is out V/m, sphere 1.00 m in rdius cn e rised to mximum potentil of V. The potentil cn e incresed further y incresing the dome s rdius nd plcing the entire system in continer filled with high-pressure gs. Vn de Grff genertors cn produce potentil differences s lrge s 20 million volts. Protons ccelerted through such lrge potentil differences receive enough energy to initite nucler rections etween themselves nd vrious trget nuclei. mller genertors re often seen in science clssrooms nd museums. If person insulted from the ground touches the sphere of Vn de Grff genertor, his or her ody cn e rought to high electric potentil. The person s hir cquires net positive chrge, nd ech strnd is repelled y ll the others s in the opening photogrph of Chpter 23. B A Metl dome P Ground Belt Insultor The Electrosttic Precipittor One importnt ppliction of electricl dischrge in gses is the electrosttic precipittor. This device removes prticulte mtter from comustion gses, therey reducing ir pollution. Precipittors re especilly useful in col-urning power plnts nd industril opertions tht generte lrge quntities of smoke. Current systems re le to eliminte more thn 99% of the sh from smoke. Figure (pge 766) shows schemtic digrm of n electrosttic precipittor. A high potentil difference (typiclly 40 to 100 kv) is mintined etween The chrge is deposited on the elt t point A nd trnsferred to the hollow conductor t point B. Figure chemtic digrm of Vn de Grff genertor. Chrge is trnsferred to the metl dome t the top y mens of moving elt.

109 766 Chpter 25 Electric Potentil The high negtive electric potentil mintined on the centrl wire cretes coron dischrge in the vicinity of the wire. Insultor Dirty ir in ly.c om Clen ir out ee Weight Dirt out By Courtesy of Tenov TAKRAF Bttery c s. w ic Figure () chemtic digrm of n electrosttic precipittor. Compre the ir pollution when the electrosttic precipittor is () operting nd (c) turned off. w w w. sw r ph ys wire running down the center of duct nd the wlls of the duct, which re grounded. The wire is mintined t negtive electric potentil with respect to the wlls, so the electric field is directed towrd the wire. The vlues of the field ner the wire ecome high enough to cuse coron dischrge round the wire; the ir ner the wire contins positive ions, electrons, nd such negtive ions s O22. The ir to e clened enters the duct nd moves ner the wire. As the electrons nd negtive ions creted y the dischrge re ccelerted towrd the outer wll y the electric field, the dirt prticles in the ir ecome chrged y collisions nd ion cpture. Becuse most of the chrged dirt prticles re negtive, they too re drwn to the duct wlls y the electric field. When the duct is periodiclly shken, the prticles rek loose nd re collected t the ottom. In ddition to reducing the level of prticulte mtter in the tmosphere (compre Figs nd c), the electrosttic precipittor recovers vlule mterils in the form of metl oxides. ummry Definitions The potentil difference DV etween points A nd B in n electric field E is defined s DU s 5 23 E? d q A B DV ; (25.3) where DU is given y Eqution 25.1 on pge 767. The electric potentil V 5 U/q is sclr quntity nd hs the units of joules per coulom, where 1 J/C ; 1 V. An equipotentil surfce is one on which ll points re t the sme electric potentil. Equipotentil surfces re perpendiculr to electric field lines.

110 Ojective Questions 767 Concepts nd Principles When positive chrge q is moved etween points A nd B in n electric field E, the chnge in the potentil energy of the chrge field system is B DU 5 2q 3 E? d s A (25.1) The potentil difference etween two points seprted y distnce d in uniform electric field E is DV 5 2Ed (25.6) if the direction of trvel etween the points is in the sme direction s the electric field. If we define V 5 0 t r 5 `, the electric potentil due to point chrge t ny distnce r from the chrge is V 5 k e q r (25.11) The electric potentil ssocited with group of point chrges is otined y summing the potentils due to the individul chrges. If the electric potentil is known s function of coordintes x, y, nd z, we cn otin the components of the electric field y tking the negtive derivtive of the electric potentil with respect to the coordintes. For exmple, the x component of the electric field is Ojective Questions E x 5 2 dv dx (25.16) The electric potentil energy ssocited with pir of point chrges seprted y distnce r 12 is U 5 k e q 1q 2 r 12 (25.13) We otin the potentil energy of distriution of point chrges y summing terms like Eqution over ll pirs of prticles. The electric potentil due to continuous chrge distriution is V 5 k dq e 3 (25.20) r Every point on the surfce of chrged conductor in electrosttic equilirium is t the sme electric potentil. The potentil is constnt everywhere inside the conductor nd equl to its vlue t the surfce. 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. In certin region of spce, the electric field is zero. From this fct, wht cn you conclude out the electric potentil in this region? () It is zero. () It does not vry with position. (c) It is positive. (d) It is negtive. (e) None of those nswers is necessrily true. 2. Consider the equipotentil surfces shown in Figure In this region of spce, wht is the pproximte direction of the electric field? () It is out of the pge. () It is into the pge. (c) It is towrd the top of the pge. (d) It is towrd the ottom of the pge. (e) The field is zero. 3. (i) A metllic sphere A of rdius 1.00 cm is severl centimeters wy from metllic sphericl shell B of rdius 2.00 cm. Chrge 450 nc is plced on A, with no chrge on B or nywhere nery. Next, the two ojects re joined y long, thin, metllic wire (s shown in Fig ), nd finlly the wire is removed. How is the chrge shred etween A nd B? () 0 on A, 450 nc on B () 90.0 nc on A nd 360 nc on B, with equl surfce chrge densities (c) 150 nc on A nd 300 nc on B (d) 225 nc on A nd 225 nc on B (e) 450 nc on A nd 0 on B (ii) A metllic sphere A of rdius 1 cm with chrge 450 nc hngs on n insulting thred inside n unchrged thin metllic sphericl shell B of rdius 2 cm. Next, A is mde temporrily to touch the inner surfce of B. How is the chrge then shred etween them? Choose from the sme possiilities. Arnold Arons, the only physics techer yet to hve his picture on the cover of Time mgzine, suggested the ide for this question. 4. The electric potentil t x m is 120 V, nd the electric potentil t x m is 190 V. Wht is the x component of the electric field in this region, ssuming the field is uniform? () 140 N/C () 2140 N/C (c) 35.0 N/C (d) N/C (e) 75.0 N/C 5. Rnk the potentil energies of the four systems of prticles shown in Figure OQ25.5 from lrgest to smllest. Include equlities if pproprite. Q Q 2Q r 2r c Q Q Q Figure OQ25.5 r Q 2Q 2r 6. In certin region of spce, uniform electric field is in the x direction. A prticle with negtive chrge is crried from x cm to x cm. (i) Does d

111 768 Chpter 25 Electric Potentil the electric potentil energy of the chrge field system () increse, () remin constnt, (c) decrese, or (d) chnge unpredictly? (ii) Hs the prticle moved to position where the electric potentil is () higher thn efore, () unchnged, (c) lower thn efore, or (d) unpredictle? A B 7. Rnk the electric potentils d t the four points shown in Figure OQ25.7 from lrgest to smllest. 8. An electron in n x-ry mchine is ccelerted through potentil difference C d of V D efore it hits the trget. Wht is the kinetic Q 2Q energy of the electron in Figure OQ25.7 electron volts? () ev () ev (c) ev (d) ev (e) ev 9. Rnk the electric potentil energies of the systems of chrges shown in Figure OQ25.9 from lrgest to smllest. Indicte equlities if pproprite. Q Q d Q c d Q d d d Q Q Q Q d Q d Q d Figure OQ25.9 Q d d d Q Q Q 10. Four prticles re positioned on the rim of circle. The chrges on the prticles re mc, mc, mc, nd mc. If the electric potentil t the center of the circle due to the mc chrge lone is V, wht is the totl electric potentil t the center due to the four chrges? () V () V (c) 0 (d) V (e) V 11. A proton is relesed from rest t the origin in uniform electric field in the positive x direction with mgnitude 850 N/C. Wht is the chnge in the electric potentil energy of the proton field system when the proton trvels to x m? () J () J (c) J (d) J (e) J 12. A prticle with chrge nc is on the x xis t the point with coordinte x 5 0. A second prticle, with chrge nc, is on the x xis t x m. (i) Is the point t finite distnce where the electric field is zero () to the left of x 5 0, () etween x 5 0 nd x m, or (c) to the right of x m? (ii) Is the electric potentil zero t this point? () No; it is positive. () Yes. (c) No; it is negtive. (iii) Is there point t finite distnce where the electric potentil is zero? () Yes; it is to the left of x 5 0. () Yes; it is etween x 5 0 nd x m. (c) Yes; it is to the right of x m. (d) No. 13. A filment running long the x xis from the origin to x cm crries electric chrge with uniform density. At the point P with coordintes (x cm, y cm), this filment cretes electric potentil 100 V. Now we dd nother filment long the y xis, running from the origin to y cm, crrying the sme mount of chrge with the sme uniform density. At the sme point P, is the electric potentil creted y the pir of filments () greter thn 200 V, () 200 V, (c) 100 V, (d) etween 0 nd 200 V, or (e) 0? 14. In different experimentl trils, n electron, proton, or douly chrged oxygen tom (O 22 ), is fired within vcuum tue. The prticle s trjectory crries it through point where the electric potentil is 40.0 V nd then through point t different potentil. Rnk ech of the following cses ccording to the chnge in kinetic energy of the prticle over this prt of its flight from the lrgest increse to the lrgest decrese in kinetic energy. In your rnking, disply ny cses of equlity. () An electron moves from 40.0 V to 60.0 V. () An electron moves from 40.0 V to 20.0 V. (c) A proton moves from 40.0 V to 20.0 V. (d) A proton moves from 40.0 V to 10.0 V. (e) An O 22 ion moves from 40.0 V to 60.0 V. 15. A helium nucleus (chrge 5 2e, mss kg) trveling t m/s enters n electric field, trveling from point A, t potentil of V, to point B, t V. Wht is its speed t point B? () m/s () m/s (c) m/s (d) m/s (e) m/s chnges (if ny) in () its kinetic energy nd (c) the electric potentil energy of the proton field system. 3. When chrged prticles re seprted y n infinite distnce, the electric potentil energy of the pir is zero. When the prticles re rought close, the elec Conceptul Questions 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. Wht determines the mximum electric potentil to which the dome of Vn de Grff genertor cn e rised? 2. Descrie the motion of proton () fter it is relesed from rest in uniform electric field. Descrie the

112 Prolems 769 tric potentil energy of pir with the sme sign is positive, wheres the electric potentil energy of pir with opposite signs is negtive. Give physicl explntion of this sttement. 4. tudy Figure 23.3 nd the ccompnying text discussion of chrging y induction. When the grounding wire is touched to the rightmost point on the sphere in Figure 23.3c, electrons re drined wy from the sphere to leve the sphere positively chrged. uppose the grounding wire is touched to the leftmost point on the sphere insted. () Will electrons still drin wy, moving closer to the negtively chrged rod s they do so? () Wht kind of chrge, if ny, remins on the sphere? 5. Distinguish etween electric potentil nd electric potentil energy. 6. Descrie the equipotentil surfces for () n infinite line of chrge nd () uniformly chrged sphere. Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign ection 25.1 Electric Potentil nd Potentil Difference ection 25.2 Potentil Difference in Uniform Electric Field 1. Oppositely chrged prllel pltes re seprted M y 5.33 mm. A potentil difference of 600 V exists etween the pltes. () Wht is the mgnitude of the electric field etween the pltes? () Wht is the mgnitude of the force on n electron etween the pltes? (c) How much work must e done on the electron to move it to the negtive plte if it is initilly positioned 2.90 mm from the positive plte? 2. A uniform electric field of mgnitude 250 V/m is directed in the positive x direction. A mC chrge moves from the origin to the point (x, y) 5 (20.0 cm, 50.0 cm). () Wht is the chnge in the potentil energy of the chrge field system? () Through wht potentil difference does the chrge move? 3. () Clculte the speed of proton tht is ccelerted M from rest through n electric potentil difference of 120 V. () Clculte the speed of n electron tht is ccelerted through the sme electric potentil difference. 4. How much work is done (y ttery, genertor, or W some other source of potentil difference) in moving Avogdro s numer of electrons from n initil point where the electric potentil is 9.00 V to point where the y electric potentil is V? B (The potentil in ech cse is mesured reltive to common reference point.) x 5. A uniform electric field W of mgnitude 325 V/m is directed in the negtive y A E direction in Figure P25.5. The coordintes of point Figure P25.5 BIO Q/C A re (20.200, ) m, nd those of point B re (0.400, 0.500) m. Clculte the electric potentil difference V B 2 V A using the dshed-line pth. 6. trting with the definition of work, prove tht t every Q/C point on n equipotentil surfce, the surfce must e perpendiculr to the electric field there. 7. An electron moving prllel to the x xis hs n initil speed of m/s t the origin. Its speed is AMT M reduced to m/s t the point x cm. () Clculte the electric potentil difference etween the origin nd tht point. () Which point is t the higher potentil? 8. () Find the electric potentil difference DV e required Q/C to stop n electron (clled stopping potentil ) moving with n initil speed of m/s. () Would proton trveling t the sme speed require greter or lesser mgnitude of electric potentil difference? Explin. (c) Find symolic expression for the rtio of the proton stopping potentil nd the electron stopping potentil, DV p /DV e. 9. A prticle hving chrge q mc nd mss m 5 AMT kg is connected to string tht is L m long nd tied to the pivot point P in Figure P25.9. The prticle, string, nd pivot point ll lie on frictionless, P m v 0 q u L Top view Figure P25.9 v E

113 770 Chpter 25 Electric Potentil horizontl tle. The prticle is relesed from rest when the string mkes n ngle u with uniform electric field of mgnitude E V/m. Determine the speed of the prticle when the string is prllel to the electric field. 10. Review. A lock hving GP mss m nd chrge 1Q Q/C is connected to n insulting spring hving force constnt k. The lock lies on frictionless, insulting, horizontl trck, nd the system is immersed in uniform electric field of mgnitude E directed s shown in Figure P The lock is relesed from rest when the spring is unstretched (t x 5 0). We wish to show tht the ensuing motion of the lock is simple hrmonic. () Consider the system of the lock, the spring, nd the electric field. Is this system isolted or nonisolted? () Wht kinds of potentil energy exist within this system? (c) Cll the initil configurtion of the system tht existing just s the lock is relesed from rest. The finl configurtion is when the lock momentrily comes to rest gin. Wht is the vlue of x when the lock comes to rest momentrily? (d) At some vlue of x we will cll x 5 x 0, the lock hs zero net force on it. Wht nlysis model descries the prticle in this sitution? (e) Wht is the vlue of x 0? (f) Define new coordinte system x9 such tht x9 5 x 2 x 0. how tht x9 stisfies differentil eqution for simple hrmonic motion. (g) Find the period of the simple hrmonic motion. (h) How does the period depend on the electric field mgnitude? 11. An insulting rod hving liner Q/C chrge density l mc/m nd liner mss density m kg/m is relesed from rest in uniform electric field E V/m directed E E perpendiculr to the rod (Fig. P25.11). () Determine the speed of l, m the rod fter it hs trveled 2.00 m. Figure P25.11 () Wht If? How does your nswer to prt () chnge if the electric field is not perpendiculr to the rod? Explin. ection 25.3 Electric Potentil nd Potentil Energy Due to Point Chrges Note: Unless stted otherwise, ssume the reference level of potentil is V 5 0 t r 5 `. 12. () Clculte the electric potentil cm from n Q/C electron. () Wht is the electric potentil difference etween two points tht re cm nd cm from n electron? (c) How would the nswers chnge if the electron were replced with proton? 13. Two point chrges re on the y xis. A 4.50-mC chrge is locted t y cm, nd mC chrge is locted t y cm. Find the totl electric potentil t () the origin nd () the point whose coordintes re (1.50 cm, 0). k m, Q x 0 Figure P25.10 E 14. The two chrges in Figure A P25.14 re seprted y d cm. Find the electric potentil t () point A nd d d () point B, which is hlfwy etween the chrges B d 15. Three positive chrges re 15.0 nc 27.0 nc locted t the corners of n equilterl tringle s in Figure P25.14 Figure P Find n expression Q for the electric potentil t the center of the tringle. 16. Two point chrges Q nc d d M nd Q nc re seprted Q/C y 35.0 cm. () Wht is the electric potentil t point midwy Q d 2Q etween the chrges? () Wht is Figure P25.15 the potentil energy of the pir of chrges? Wht is the significnce of the lgeric sign of your nswer? 17. Two prticles, with chrges of 20.0 nc nd nc, re plced t the points with coordintes 20.0 nc (0, 4.00 cm) nd (0, cm) s shown 4.00 cm in Figure P A prticle with chrge 10.0 nc 10.0 nc 3.00 cm 40.0 nc is locted t the origin. () Find the electric potentil energy of the 4.00 cm configurtion of the three fixed chrges nc () A fourth prticle, with mss of kg nd chrge of 40.0 nc, is relesed from Figure P25.17 rest t the point (3.00 cm, 0). Find its speed fter it hs moved freely to very lrge distnce wy. 18. The two chrges in Figure P25.18 re seprted y distnce d cm, nd Q nc. Find () the electric potentil t A, () the electric potentil t B, nd (c) the electric potentil difference etween B nd A. A d Q d B 2Q Figure P Given two prticles with 2.00-mC chrges s shown in W Figure P25.19 nd prticle with chrge q C t the origin, () wht is the net force exerted

114 Prolems 771 y the two 2.00-mC chrges on the chrge q? () Wht is the electric field t the origin due to the two 2.00-mC prticles? (c) Wht is the electric potentil t the origin due to the two 2.00-mC prticles? 2.00 mc x m y q 2.00 mc x 0 x m Figure P At certin distnce from chrged prticle, the mgnitude of the electric field is 500 V/m nd the electric M potentil is kv. () Wht is the distnce to the prticle? () Wht is the mgnitude of the chrge? 21. Four point chrges ech hving chrge Q re locted t the corners of squre hving sides of length. Find expressions for () the totl electric potentil t the center of the squre due to the four chrges nd () the work required to ring fifth chrge q from infinity to the center of the squre. 22. The three chrged prticles in q M Figure P25.22 re t the vertices of n isosceles tringle (where d cm). Tking q mc, clculte the electric potentil t 2d point A, the midpoint of the se. 23. A prticle with chrge 1q is t A the origin. A prticle with chrge q q 22q is t x m on the x xis. d () For wht finite vlue(s) of x Figure P25.22 is the electric field zero? () For wht finite vlue(s) of x is the electric potentil zero? 24. how tht the mount of work required to ssemle four identicl chrged prticles of mgnitude Q t the corners of squre of side s is 5.41k e Q 2 /s. 25. Two prticles ech with chrge mc re locted on the x xis. One is t x m, nd the other is t x m. () Determine the electric potentil on the y xis t y m. () Clculte the chnge in electric potentil energy of the system s third chrged prticle of mc is rought from infinitely fr wy to position on the y xis t y m. 26. Two chrged prticles of equl mgnitude re locted long the y xis y equl distnces ove nd elow the Q x xis s shown in Figure P () Plot grph of the electric potentil t points long the x xis x over the intervl 23, x, 3. You should plot the potentil in units Q of k e Q /. () Let the chrge of the prticle locted t y 5 2 e negtive. Plot the potentil long the y Figure P25.26 xis over the intervl 24, y, Four identicl chrged prticles (q mc) re W locted on the corners of rectngle s shown in Figure P The dimensions of the rectngle re L cm nd W cm. Clculte the chnge in electric potentil energy of the system s the prticle t the lower left corner in Figure P25.27 is rought to this position from infinitely fr wy. Assume the other three prticles in Figure P25.27 remin fixed in position. q q q 28. Three prticles with equl positive chrges q re t the corners q of n equilterl tringle of side s shown in Figure P () At wht point, if ny, in the plne of the prticles is the electric potentil zero? () Wht is the electric q q potentil t the position of one of the prticles due to the other two Figure P25.28 prticles in the tringle? 29. Five prticles with equl negtive chrges 2q re plced symmetriclly round circle of rdius R. Clculte the electric potentil t the center of the circle. 30. Review. A light, unstressed spring hs length d. Two identicl prticles, ech with chrge q, re connected to the opposite ends of the spring. The prticles re held sttionry distnce d prt nd then relesed t the sme moment. The system then oscilltes on frictionless, horizontl tle. The spring hs it of internl kinetic friction, so the oscilltion is dmped. The prticles eventully stop virting when the distnce etween them is 3d. Assume the system of the spring nd two chrged prticles is isolted. Find the increse in internl energy tht ppers in the spring during the oscilltions. 31. Review. Two insulting spheres hve rdii cm AMT nd cm, msses kg nd kg, nd uniformly distriuted chrges mc nd 3.00 mc. Q/C They re relesed from rest when their centers re seprted y 1.00 m. () How fst will ech e moving when they collide? () Wht If? If the spheres were conductors, would the speeds e greter or less thn those clculted in prt ()? Explin. 32. Review. Two insulting spheres hve rdii r 1 nd r 2, Q/C msses m 1 nd m 2, nd uniformly distriuted chrges 2q 1 nd q 2. They re relesed from rest when their centers re seprted y distnce d. () How fst is ech moving when they collide? () Wht If? If the spheres were conductors, would their speeds e greter or less thn those clculted in prt ()? Explin. 33. How much work is required to ssemle eight identicl chrged prticles, ech of mgnitude q, t the corners of cue of side s? 34. Four identicl prticles, ech hving chrge q nd mss m, re relesed from rest t the vertices of squre of side L. How fst is ech prticle moving when their distnce from the center of the squre doules? 35. In 1911, Ernest Rutherford nd his ssistnts Geiger AMT nd Mrsden conducted n experiment in which they y W x L q Figure P25.27

115 772 Chpter 25 Electric Potentil scttered lph prticles (nuclei of helium toms) from thin sheets of gold. An lph prticle, hving chrge 12e nd mss kg, is product of certin rdioctive decys. The results of the experiment led Rutherford to the ide tht most of n tom s mss is in very smll nucleus, with electrons in orit round it. (This is the plnetry model of the tom, which we ll study in Chpter 42.) Assume n lph prticle, initilly very fr from sttionry gold nucleus, is fired with velocity of m/s directly towrd the nucleus (chrge 179e). Wht is the smllest distnce etween the lph prticle nd the nucleus efore the lph prticle reverses direction? Assume the gold nucleus remins sttionry. ection 25.4 Otining the Vlue of the Electric Field from the Electric Potentil 36. Figure P25.36 represents grph of the V (V) electric potentil in 20 region of spce versus position x, where the 10 electric field is prllel to the x xis. Drw x (cm) grph of the x component of the electric field Figure P25.36 versus x in this region. 37. The potentil in region etween x 5 0 nd x m W is V 5 1 x, where V nd V/m. Determine () the potentil t x 5 0, 3.00 m, nd 6.00 m nd () the mgnitude nd direction of the electric field t x 5 0, 3.00 m, nd 6.00 m. 38. An electric field in region of spce is prllel to the x xis. The electric potentil vries with position s shown in Figure P Grph the x component of the electric field versus position in this region of spce. V (V) x (cm) Figure P Over certin region of spce, the electric potentil is W V 5 5x 2 3x 2 y 1 2yz 2. () Find the expressions for the x, y, nd z components of the electric field over this region. () Wht is the mgnitude of the field t the point P tht hs coordintes (1.00, 0, 22.00) m? 40. Figure P25.40 shows severl equipotentil lines, ech Q/C leled y its potentil in volts. The distnce etween the lines of the squre grid represents 1.00 cm. () Is the mgnitude of the field lrger t A or t B? Explin how you cn tell. () Explin wht you cn determine B Numericl vlues re in volts. Figure P25.40 out E t B. (c) Represent wht the electric field looks like y drwing t lest eight field lines. 41. The electric potentil inside chrged sphericl conductor of rdius R is given y V 5 k e Q /R, nd the potentil outside is given y V 5 k e Q /r. Using E r 5 2dV/dr, derive the electric field () inside nd () outside this chrge distriution. 42. It is shown in Exmple 25.7 tht the potentil t point P distnce ove one end of uniformly chrged rod of length, lying long the x xis is V 5 k Q e, ln, 1 " 2 1, 2 Use this result to derive n expression for the y component of the electric field t P. ection 25.5 Electric Potentil Due to Continuous Chrge Distriutions 43. Consider ring of rdius R with the totl chrge Q spred uniformly over its perimeter. Wht is the potentil difference etween the point t the center of the ring nd point on its xis distnce 2R from the center? 44. A uniformly chrged insulting rod of W length 14.0 cm is ent into the shpe of semicircle s shown in Figure P The rod hs totl chrge of O mc. Find the electric potentil t O, the center of the semicircle. 45. A rod of length L (Fig. P25.45) lies long the x xis with its left end t the Figure P25.44 origin. It hs nonuniform chrge A d y Figure P25.45 Prolems 45 nd 46. B L A x

116 Prolems 773 density l 5 x, where is positive constnt. () Wht re the units of? () Clculte the electric potentil t A. 46. For the rrngement descried in Prolem 45, clculte the electric potentil t point B, which lies on the perpendiculr isector of the rod distnce ove the x xis. 47. A wire hving uniform liner chrge density l is ent W into the shpe shown in Figure P Find the electric potentil t point O. 2R O R Figure P25.47 ection 25.6 Electric Potentil Due to Chrged Conductor 48. The electric field mgnitude on the surfce of n irregulrly shped conductor vries from 56.0 kn/c to 28.0 kn/c. Cn you evlute the electric potentil on the conductor? If so, find its vlue. If not, explin why not. 49. How mny electrons should e removed from n initilly unchrged sphericl conductor of rdius m to produce potentil of 7.50 kv t the surfce? 50. A sphericl conductor hs rdius of 14.0 cm nd M chrge of 26.0 mc. Clculte the electric field nd the electric potentil t () r cm, () r cm, nd (c) r cm from the center. 51. Electric chrge cn ccumulte on n irplne in flight. You my hve oserved needle-shped metl extensions on the wing tips nd til of n irplne. Their purpose is to llow chrge to lek off efore much of it ccumultes. The electric field round the needle is much lrger thn the field round the ody of the irplne nd cn ecome lrge enough to produce dielectric rekdown of the ir, dischrging the irplne. To model this process, ssume two chrged sphericl conductors re connected y long conducting wire nd 1.20-mC chrge is plced on the comintion. One sphere, representing the ody of the irplne, hs rdius of 6.00 cm; the other, representing the tip of the needle, hs rdius of 2.00 cm. () Wht is the electric potentil of ech sphere? () Wht is the electric field t the surfce of ech sphere? ection 25.8 Applictions of Electrosttics 52. Lightning cn e studied M with Vn de Grff genertor, which consists of sphericl dome on which chrge is continuously deposited y moving elt. Chrge cn e dded until the electric field t the surfce of the dome Figure P25.52 ecomes equl to the Dvid Evison/hutterstock.com 2R dielectric strength of ir. Any more chrge leks off in sprks s shown in Figure P Assume the dome hs dimeter of 30.0 cm nd is surrounded y dry ir with rekdown electric field of V/m. () Wht is the mximum potentil of the dome? () Wht is the mximum chrge on the dome? Additionl Prolems 53. Why is the following sitution impossile? In the Bohr model of the hydrogen tom, n electron moves in circulr orit out proton. The model sttes tht the electron cn exist only in certin llowed orits round the proton: those whose rdius r stisfies r 5 n 2 ( nm), where n 5 1, 2, 3,.... For one of the possile llowed sttes of the tom, the electric potentil energy of the system is ev. 54. Review. In fir wether, the electric field in the ir t Q/C prticulr loction immeditely ove the Erth s surfce is 120 N/C directed downwrd. () Wht is the surfce chrge density on the ground? Is it positive or negtive? () Imgine the surfce chrge density is uniform over the plnet. Wht then is the chrge of the whole surfce of the Erth? (c) Wht is the Erth s electric potentil due to this chrge? (d) Wht is the difference in potentil etween the hed nd the feet of person 1.75 m tll? (Ignore ny chrges in the tmosphere.) (e) Imgine the Moon, with 27.3% of the rdius of the Erth, hd chrge 27.3% s lrge, with the sme sign. Find the electric force the Erth would then exert on the Moon. (f) tte how the nswer to prt (e) compres with the grvittionl force the Erth exerts on the Moon. 55. Review. From lrge distnce wy, prticle of mss 2.00 g nd chrge 15.0 mc is fired t 21.0i^ m/s stright towrd second prticle, originlly sttionry ut free to move, with mss 5.00 g nd chrge 8.50 mc. Both prticles re constrined to move only long the x xis. () At the instnt of closest pproch, oth prticles will e moving t the sme velocity. Find this velocity. () Find the distnce of closest pproch. After the interction, the prticles will move fr prt gin. At this time, find the velocity of (c) the 2.00-g prticle nd (d) the 5.00-g prticle. 56. Review. From lrge distnce wy, prticle of mss m 1 nd positive chrge q 1 is fired t speed v in the positive x direction stright towrd second prticle, originlly sttionry ut free to move, with mss m 2 nd positive chrge q 2. Both prticles re constrined to move only long the x xis. () At the instnt of closest pproch, oth prticles will e moving t the sme velocity. Find this velocity. () Find the distnce of closest pproch. After the interction, the prticles will move fr prt gin. At this time, find the velocity of (c) the prticle of mss m 1 nd (d) the prticle of mss m The liquid-drop model of the tomic nucleus suggests M high-energy oscilltions of certin nuclei cn split the nucleus into two unequl frgments plus few

117 774 Chpter 25 Electric Potentil neutrons. The fission products cquire kinetic energy from their mutul Coulom repulsion. Assume the chrge is distriuted uniformly throughout the volume of ech sphericl frgment nd, immeditely efore seprting, ech frgment is t rest nd their surfces re in contct. The electrons surrounding the nucleus cn e ignored. Clculte the electric potentil energy (in electron volts) of two sphericl frgments from urnium nucleus hving the following chrges nd rdii: 38e nd m, nd 54e nd m. 58. On dry winter dy, you scuff your lether-soled shoes cross crpet nd get shock when you extend the tip of one finger towrd metl doorkno. In drk room, you see sprk perhps 5 mm long. Mke orderof-mgnitude estimtes of () your electric potentil nd () the chrge on your ody efore you touch the doorkno. Explin your resoning. 59. The electric potentil immeditely outside chrged conducting sphere is 200 V, nd 10.0 cm frther from the center of the sphere the potentil is 150 V. Determine () the rdius of the sphere nd () the chrge on it. The electric potentil immeditely outside nother chrged conducting sphere is 210 V, nd 10.0 cm frther from the center the mgnitude of the electric field is 400 V/m. Determine (c) the rdius of the sphere nd (d) its chrge on it. (e) Are the nswers to prts (c) nd (d) unique? 60. () Use the exct result from Exmple 25.4 to find the Q/C electric potentil creted y the dipole descried in the exmple t the point (3, 0). () Explin how this nswer compres with the result of the pproximte expression tht is vlid when x is much greter thn. 61. Clculte the work tht must e done on chrges rought from infinity to chrge sphericl shell of rdius R m to totl chrge Q mc. 62. Clculte the work tht must e done on chrges rought from infinity to chrge sphericl shell of rdius R to totl chrge Q. 63. The electric potentil everywhere on the xy plne is 36 V 5 "1x y "x y where V is in volts nd x nd y re in meters. Determine the position nd chrge on ech of the prticles tht crete this potentil. 64. Why is the following sitution impossile? You set up n pprtus in your lortory s follows. The x xis is the symmetry xis of sttionry, uniformly chrged ring of rdius R m nd chrge Q mc (Fig. P25.64). You plce prticle with chrge Q R x Figure P25.64 Q v x Q mc nd mss m kg t the center of the ring nd rrnge for it to e constrined to move only long the x xis. When it is displced slightly, the prticle is repelled y the ring nd ccelertes long the x xis. The prticle moves fster thn you expected nd strikes the opposite wll of your lortory t 40.0 m/s. 65. From Guss s lw, the electric field set up y uniform line of chrge is l E 5 2pP 0 r r^ where r^ is unit vector pointing rdilly wy from the line nd l is the liner chrge density long the line. Derive n expression for the potentil difference etween r 5 r 1 nd r 5 r A uniformly chrged filment lies long the x xis Q/C etween x m nd x 5 1, m s shown in Figure P The totl chrge on the filment is 1.60 nc. Clculte successive pproximtions for the electric potentil t the origin y modeling the filment s () single chrged prticle t x m, () two nC chrged prticles t x m nd x m, nd (c) four nC chrged prticles t x m, x m, x m, nd x m. (d) Explin how the results compre with the potentil given y the exct expression P y V 5 k eq ln, 1, Figure P The thin, uniformly chrged rod y shown in Figure P25.67 hs liner chrge density l. Find n P expression for the electric potentil t P. 68. A Geiger Mueller tue is rdition detector tht consists of x closed, hollow, metl cylinder L (the cthode) of inner rdius r nd coxil cylindricl wire (the Figure P25.67 node) of rdius r (Fig. P25.68). The chrge per unit length on the node is l, nd the chrge per unit length on the cthode is 2l. A gs fills the spce etween the electrodes. When the tue is in use (Fig. P25.68) nd high-energy elementry prticle psses through this spce, it cn ionize n tom of the gs. The strong electric field mkes the resulting ion nd electron ccelerte in opposite directions. They strike other molecules of the gs to ionize them, producing n vlnche of electricl dischrge. The x

118 Prolems 775 pulse of electric current etween the wire nd the cylinder is counted y n externl circuit. () how tht the mgnitude of the electric potentil difference etween the wire nd the cylinder is DV 5 2k e l ln r r () how tht the mgnitude of the electric field in the spce etween cthode nd node is E 5 DV ln 1r /r 2 1 r where r is the distnce from the xis of the node to the point where the field is to e clculted. r r Anode Cthode l Figure P Review. Two prllel pltes hving chrges of equl mgnitude ut opposite sign re seprted y 12.0 cm. Ech plte hs surfce chrge density of 36.0 nc/m 2. A proton is relesed from rest t the positive plte. Determine () the mgnitude of the electric field etween the pltes from the chrge density, () the potentil difference etween the pltes, (c) the kinetic energy of the proton when it reches the negtive plte, (d) the speed of the proton just efore it strikes the negtive plte, (e) the ccelertion of the proton, nd (f) the force on the proton. (g) From the force, find the mgnitude of the electric field. (h) How does your vlue of the electric field compre with tht found in prt ()? 70. When n unchrged conducting sphere of rdius is plced t the origin of n xyz coordinte system tht lies in n initilly uniform electric field E 5 E 0 k^, the resulting electric potentil is V(x, y, z) 5 V 0 for points inside the sphere nd E 0 3 z V 1x, y, z2 5 V 0 2 E 0 z 1 1x 2 1 y 2 1 z 2 2 3/2 l Hnk Morgn/Photo Reserchers, Inc. is fr from the dipole (r.. ), show tht the electric potentil is V 5 k e p cos u r 2 () Clculte the rdil component E r nd the perpendiculr component E u of the ssocited electric field. Note tht E u 5 2(1/r)('V/'u). Do these results seem resonle for (c) u nd 08? (d) For r 5 0? (e) For the dipole rrngement shown in Figure P25.71, express V in terms of Crtesin coordintes using r 5 (x 2 1 y 2 ) 1/2 nd y cos u 5 1x 2 1 y 2 2 1/2 q u (f) Using these results nd gin tking r.., clculte the field components E x nd E y. 72. A solid sphere of rdius R hs uniform chrge density r nd totl chrge Q. Derive n expression for its totl electric potentil energy. uggestion: Imgine the sphere is constructed y dding successive lyers of concentric shells of chrge dq 5 (4pr 2 dr)r nd use du 5 V dq. 73. A disk of rdius R (Fig. P25.73) hs nonuniform surfce chrge density s 5 R P Cr, where C is constnt x nd r is mesured from the center of the disk to point on the surfce of the disk. Find (y direct integrtion) the electric potentil t P. Figure P Four lls, ech with mss m, re 1 2 connected y four nonconducting strings to form squre with side s shown in Figure P The ssemly is plced on nonconducting, frictionless, horizontl surfce. Blls 1 nd 2 ech hve chrge q, nd lls 3 nd 4 re unchrged. 3 Figure P After the string connecting lls 1 nd 2 is cut, wht is the mximum speed of lls 3 nd 4? 75. () A uniformly chrged cylindricl shell with no end cps hs totl chrge Q, rdius R, nd length h. Determine the electric potentil t point distnce d from the right end of the cylinder s shown in Figure P y q for points outside the sphere, where V 0 is the (constnt) electric potentil on the conductor. Use this eqution to determine the x, y, nd z components of the resulting electric field () inside the sphere nd () outside the sphere. Chllenge Prolems 71. An electric dipole is locted long the y xis s shown in Figure P The mgnitude of its electric dipole moment is defined s p 5 2q. () At point P, which h R Figure P25.75 r 1 r r 2 P Figure P25.71 d E r E u x

119 776 Chpter 25 Electric Potentil uggestion: Use the result of Exmple 25.5 y treting the cylinder s collection of ring chrges. () Wht If? Use the result of Exmple 25.6 to solve the sme prolem for solid cylinder. 76. As shown in Figure P25.76, two lrge, prllel, verticl conducting pltes seprted y distnce d re chrged so tht their potentils re 1V 0 nd 2V 0. A smll conducting ll of mss m nd rdius R (where R,, d) hngs midwy etween the pltes. The thred of length L supporting the ll is conducting wire connected to ground, so the potentil of the ll is fixed t V 5 0. The ll hngs stright down in stle equilirium when V 0 is sufficiently smll. how tht the equilirium of the ll is unstle if V 0 exceeds the criticl vlue 3k e d 2 mg/14rl24 1/2. L uggestion: Consider the forces on the ll when it is displced distnce x,, L. 77. A prticle with chrge q is locted t x 5 2R, nd prticle with chrge 22q is locted V 0 V 0 d t the origin. Prove tht the Figure P25.76 equipotentil surfce tht hs zero potentil is sphere centered t (24R/3, 0, 0) nd hving rdius r 5 2 3R.

120 Cpcitnce nd Dielectrics c h p t e r 26 In this chpter, we introduce the first of three simple circuit elements tht cn e connected with wires to form n electric circuit. Electric circuits re the sis for the vst mjority of the devices used in our society. Here we shll discuss cpcitors, devices tht store electric chrge. This discussion is followed y the study of resistors in Chpter 27 nd inductors in Chpter 32. In lter chpters, we will study more sophisticted circuit elements such s diodes nd trnsistors. Cpcitors re commonly used in vriety of electric circuits. For instnce, they re used to tune the frequency of rdio receivers, s filters in power supplies, to eliminte sprking in utomoile ignition systems, nd s energy-storing devices in electronic flsh units Definition of Cpcitnce 26.2 Clculting Cpcitnce 26.3 Comintions of Cpcitors 26.4 Energy tored in Chrged Cpcitor 26.5 Cpcitors with Dielectrics 26.6 Electric Dipole in n Electric Field 26.7 An Atomic Description of Dielectrics When ptient receives shock from defirilltor, the energy delivered to the ptient is initilly stored in cpcitor. We will study cpcitors nd cpcitnce in this chpter. (Andrew Olney/Getty Imges) 26.1 Definition of Cpcitnce Consider two conductors s shown in Figure 26.1 (pge 778). uch comintion of two conductors is clled cpcitor. The conductors re clled pltes. If the conductors crry chrges of equl mgnitude nd opposite sign, potentil difference DV exists etween them. 777

121 778 Chpter 26 Cpcitnce nd Dielectrics Pitfll Prevention 26.1 Cpcitnce Is Cpcity To understnd cpcitnce, think of similr notions tht use similr word. The cpcity of milk crton is the volume of milk it cn store. The het cpcity of n oject is the mount of energy n oject cn store per unit of temperture difference. The cpcitnce of cpcitor is the mount of chrge the cpcitor cn store per unit of potentil difference. Pitfll Prevention 26.2 Potentil Difference Is DV, Not V We use the symol DV for the potentil difference cross circuit element or device ecuse this nottion is consistent with our definition of potentil difference nd with the mening of the delt sign. It is common ut confusing prctice to use the symol V without the delt sign for oth potentil nd potentil difference! Keep tht in mind if you consult other texts. Definition of cpcitnce When the cpcitor is connected to the terminls of ttery, electrons trnsfer etween the pltes nd the wires so tht the pltes ecome chrged. Q d Q Are A Figure 26.1 A cpcitor consists of two conductors. When the cpcitor is chrged, the conductors crry chrges of equl mgnitude nd opposite sign. Q Wht determines how much chrge is on the pltes of cpcitor for given voltge? Experiments show tht the quntity of chrge Q on cpcitor 1 is linerly proportionl to the potentil difference etween the conductors; tht is, Q ~ DV. The proportionlity constnt depends on the shpe nd seprtion of the conductors. 2 This reltionship cn e written s Q 5 C DV if we define cpcitnce s follows: The cpcitnce C of cpcitor is defined s the rtio of the mgnitude of the chrge on either conductor to the mgnitude of the potentil difference etween the conductors: C ; Q (26.1) DV By definition cpcitnce is lwys positive quntity. Furthermore, the chrge Q nd the potentil difference DV re lwys expressed in Eqution 26.1 s positive quntities. From Eqution 26.1, we see tht cpcitnce hs I units of couloms per volt. Nmed in honor of Michel Frdy, the I unit of cpcitnce is the frd (F): 1 F 5 1 C/V The frd is very lrge unit of cpcitnce. In prctice, typicl devices hve cpcitnces rnging from microfrds (10 26 F) to picofrds ( F). We shll use the symol mf to represent microfrds. In prctice, to void the use of Greek letters, physicl cpcitors re often leled mf for microfrds nd mmf for micromicrofrds or, equivlently, pf for picofrds. Let s consider cpcitor formed from pir of prllel pltes s shown in Figure Ech plte is connected to one terminl of ttery, which cts s source of potentil difference. If the cpcitor is initilly unchrged, the ttery estlishes n electric field in the connecting wires when the connections re mde. Let s focus on the plte connected to the negtive terminl of the ttery. The electric field in the wire pplies force on electrons in the wire immeditely outside this plte; this force cuses the electrons to move onto the plte. The movement continues until the plte, the wire, nd the terminl re ll t the sme electric potentil. Once this equilirium sitution is ttined, potentil difference no longer exists etween the terminl nd the plte; s result, no electric field is present in the wire nd Q Figure 26.2 A prllel-plte cpcitor consists of two prllel conducting pltes, ech of re A, seprted y distnce d. 1 Although the totl chrge on the cpcitor is zero (ecuse there is s much excess positive chrge on one conductor s there is excess negtive chrge on the other), it is common prctice to refer to the mgnitude of the chrge on either conductor s the chrge on the cpcitor. 2 The proportionlity etween Q nd DV cn e proven from Coulom s lw or y experiment.

122 26.2 Clculting Cpcitnce 779 the electrons stop moving. The plte now crries negtive chrge. A similr process occurs t the other cpcitor plte, where electrons move from the plte to the wire, leving the plte positively chrged. In this finl configurtion, the potentil difference cross the cpcitor pltes is the sme s tht etween the terminls of the ttery. Q uick Quiz 26.1 A cpcitor stores chrge Q t potentil difference DV. Wht hppens if the voltge pplied to the cpcitor y ttery is douled to 2 DV? () The cpcitnce flls to hlf its initil vlue, nd the chrge remins the sme. () The cpcitnce nd the chrge oth fll to hlf their initil vlues. (c) The cpcitnce nd the chrge oth doule. (d) The cpcitnce remins the sme, nd the chrge doules Clculting Cpcitnce We cn derive n expression for the cpcitnce of pir of oppositely chrged conductors hving chrge of mgnitude Q in the following mnner. First we clculte the potentil difference using the techniques descried in Chpter 25. We then use the expression C 5 Q /DV to evlute the cpcitnce. The clcultion is reltively esy if the geometry of the cpcitor is simple. Although the most common sitution is tht of two conductors, single conductor lso hs cpcitnce. For exmple, imgine single sphericl, chrged conductor. The electric field lines round this conductor re exctly the sme s if there were conducting, sphericl shell of infinite rdius, concentric with the sphere nd crrying chrge of the sme mgnitude ut opposite sign. Therefore, we cn identify the imginry shell s the second conductor of two-conductor cpcitor. The electric potentil of the sphere of rdius is simply k e Q / (see ection 25.6), nd setting V 5 0 for the infinitely lrge shell gives C 5 Q DV 5 Q k e Q / 5 5 4pP k 0 (26.2) e This expression shows tht the cpcitnce of n isolted, chrged sphere is proportionl to its rdius nd is independent of oth the chrge on the sphere nd its potentil, s is the cse with ll cpcitors. Eqution 26.1 is the generl definition of cpcitnce in terms of electricl prmeters, ut the cpcitnce of given cpcitor will depend only on the geometry of the pltes. The cpcitnce of pir of conductors is illustrted elow with three fmilir geometries, nmely, prllel pltes, concentric cylinders, nd concentric spheres. In these clcultions, we ssume the chrged conductors re seprted y vcuum. Prllel-Plte Cpcitors Two prllel, metllic pltes of equl re A re seprted y distnce d s shown in Figure One plte crries chrge 1Q, nd the other crries chrge 2Q. The surfce chrge density on ech plte is s 5 Q /A. If the pltes re very close together (in comprison with their length nd width), we cn ssume the electric field is uniform etween the pltes nd zero elsewhere. According to the Wht If? feture of Exmple 24.5, the vlue of the electric field etween the pltes is E 5 s 5 Q P 0 P 0 A Becuse the field etween the pltes is uniform, the mgnitude of the potentil difference etween the pltes equls Ed (see Eq. 25.6); therefore, DV 5 Ed 5 Qd P 0 A Pitfll Prevention 26.3 Too Mny Cs Do not confuse n itlic C for cpcitnce with nonitlic C for the unit coulom. W WCpcitnce of n isolted chrged sphere

123 780 Chpter 26 Cpcitnce nd Dielectrics Cpcitnce of prllel pltes Key Movle plte Insultor Fixed plte Exmple 26.1 The Cylindricl Cpcitor A solid cylindricl conductor of rdius nd chrge Q is coxil with cylindricl shell of negligile thickness, rdius., nd chrge 2Q (Fig. 26.4). Find the cpcitnce of this cylindricl cpcitor if its length is,. olution B Figure 26.3 (Quick Quiz 26.2) One type of computer keyord utton. ustituting this result into Eqution 26.1, we find tht the cpcitnce is C 5 Q DV 5 Q Qd/P 0 A C 5 P 0A (26.3) d Tht is, the cpcitnce of prllel-plte cpcitor is proportionl to the re of its pltes nd inversely proportionl to the plte seprtion. Let s consider how the geometry of these conductors influences the cpcity of the pir of pltes to store chrge. As cpcitor is eing chrged y ttery, electrons flow into the negtive plte nd out of the positive plte. If the cpcitor pltes re lrge, the ccumulted chrges re le to distriute themselves over sustntil re nd the mount of chrge tht cn e stored on plte for given potentil difference increses s the plte re is incresed. Therefore, it is resonle tht the cpcitnce is proportionl to the plte re A s in Eqution Now consider the region tht seprtes the pltes. Imgine moving the pltes closer together. Consider the sitution efore ny chrges hve hd chnce to move in response to this chnge. Becuse no chrges hve moved, the electric field etween the pltes hs the sme vlue ut extends over shorter distnce. Therefore, the mgnitude of the potentil difference etween the pltes DV 5 Ed (Eq. 25.6) is smller. The difference etween this new cpcitor voltge nd the terminl voltge of the ttery ppers s potentil difference cross the wires connecting the ttery to the cpcitor, resulting in n electric field in the wires tht drives more chrge onto the pltes nd increses the potentil difference etween the pltes. When the potentil difference etween the pltes gin mtches tht of the ttery, the flow of chrge stops. Therefore, moving the pltes closer together cuses the chrge on the cpcitor to increse. If d is incresed, the chrge decreses. As result, the inverse reltionship etween C nd d in Eqution 26.3 is resonle. Q uick Quiz 26.2 Mny computer keyord uttons re constructed of cpcitors s shown in Figure When key is pushed down, the soft insultor etween the movle plte nd the fixed plte is compressed. When the key is pressed, wht hppens to the cpcitnce? () It increses. () It decreses. (c) It chnges in wy you cnnot determine ecuse the electric circuit connected to the keyord utton my cuse chnge in DV. Q Q r Q Q Conceptulize Recll tht ny pir of conductors qulifies s cpcitor, so the system descried in this exmple therefore qulifies. Figure 26.4 helps visulize the electric field etween the conductors. We expect the cpcitnce to depend only on geometric fctors, which, in this cse, re,, nd,. Ctegorize Becuse of the cylindricl symmetry of the system, we cn use results from previous studies of cylindricl systems to find the cpcitnce. Gussin surfce Figure 26.4 (Exmple 26.1) () A cylindricl cpcitor consists of solid cylindricl conductor of rdius nd length, surrounded y coxil cylindricl shell of rdius. () End view. The electric field lines re rdil. The dshed line represents the end of cylindricl gussin surfce of rdius r nd length,.

124 26.2 Clculting Cpcitnce continued Anlyze Assuming, is much greter thn nd, we cn neglect end effects. In this cse, the electric field is perpendiculr to the long xis of the cylinders nd is confined to the region etween them (Fig. 26.4). Write n expression for the potentil difference etween the two cylinders from Eqution 25.3: V 2 V E? d s Apply Eqution 24.7 for the electric field outside cylindriclly symmetric chrge distriution nd notice from Figure 26.4 tht E is prllel to d s long rdil line: ustitute the solute vlue of DV into Eqution 26.1 nd use l 5 Q /,: V 2 V E r dr 5 22k e l 3 C 5 Q DV 5 Q 5 12k e Q /,2 ln 1/2 dr r 5 22k e l ln, 2k e ln 1/2 Finlize The cpcitnce depends on the rdii nd nd is proportionl to the length of the cylinders. Eqution 26.4 shows tht the cpcitnce per unit length of comintion of concentric cylindricl conductors is C, 5 1 2k e ln 1/2 (26.5) (26.4) An exmple of this type of geometric rrngement is coxil cle, which consists of two concentric cylindricl conductors seprted y n insultor. You proly hve coxil cle ttched to your television set if you re suscrier to cle television. The coxil cle is especilly useful for shielding electricl signls from ny possile externl influences. Wht If? uppose for the cylindricl cpcitor. You would like to increse the cpcitnce, nd you cn do so y choosing to increse either, y 10% or y 10%. Which choice is more effective t incresing the cpcitnce? Answer According to Eqution 26.4, C is proportionl to,, so incresing, y 10% results in 10% increse in C. For the result of the chnge in, let s use Eqution 26.4 to set up rtio of the cpcitnce C9 for the enlrged cylinder rdius 9 to the originl cpcitnce: C r C 5,/2k e ln 1/r 2 ln 1/2 5,/2k e ln 1/2 ln 1/r 2 We now sustitute nd , representing 10% increse in : C r C 5 ln 12.00/2 ln ln 12.00/1.102 ln which corresponds to 16% increse in cpcitnce. Therefore, it is more effective to increse thn to increse,. Note two more extensions of this prolem. First, it is dvntgeous to increse only for rnge of reltionships etween nd. If. 2.85, incresing, y 10% is more effective thn incresing (see Prolem 70). econd, if decreses, the cpcitnce increses. Incresing or decresing hs the effect of ringing the pltes closer together, which increses the cpcitnce. Exmple 26.2 The phericl Cpcitor A sphericl cpcitor consists of sphericl conducting shell of rdius nd chrge 2Q concentric with smller conducting sphere of rdius nd chrge Q (Fig. 26.5, pge 782). Find the cpcitnce of this device. olution Conceptulize As with Exmple 26.1, this system involves pir of conductors nd qulifies s cpcitor. We expect the cpcitnce to depend on the sphericl rdii nd. continued

125 782 Chpter 26 Cpcitnce nd Dielectrics 26.2 continued Ctegorize Becuse of the sphericl symmetry of the system, we cn use results from previous studies of sphericl systems to find the cpcitnce. Anlyze As shown in Chpter 24, the direction of the electric field outside sphericlly symmetric chrge distriution is rdil nd its mgnitude is given y the expression E 5 k e Q /r 2. In this cse, this result pplies to the field etween the spheres (, r, ). Write n expression for the potentil difference etween the two conductors from Eqution 25.3: Apply the result of Exmple 24.3 for the electric field outside sphericlly symmetric chrge distriution nd note tht E is prllel to d s long rdil line: V 2 V E? d s ustitute the solute vlue of DV into Eqution 26.1: C 5 Q DV 5 Q 0 V 2 V 0 dr V 2 V E r dr 5 2k e Q 3 r 5 k eq c 1 2 r d 26.3 Comintions of Cpcitors Two or more cpcitors often re comined in electric circuits. We cn clculte the equivlent cpcitnce of certin comintions using methods descried in this section. Throughout this section, we ssume the cpcitors to e comined re initilly unchrged. In studying electric circuits, we use simplified pictoril representtion clled circuit digrm. uch digrm uses circuit symols to represent vrious circuit elements. The circuit symols re connected y stright lines tht represent the wires etween the circuit elements. The circuit symols for cpcitors, tteries, nd switches s well s the color codes used for them in this text re given in Figure The symol for the cpcitor reflects the geometry of the most common model for cpcitor, pir of prllel pltes. The positive terminl of the ttery is t the higher potentil nd is represented in the circuit symol y the longer line. (1) V 2 V 5 k e Q k eq 2 5 k e (26.6) Finlize The cpcitnce depends on nd s expected. The potentil difference etween the spheres in Eqution (1) is negtive ecuse Q is positive nd.. Therefore, in Eqution 26.6, when we tke the solute vlue, we chnge 2 to 2. The result is positive numer. Wht If? If the rdius of the outer sphere pproches infinity, wht does the cpcitnce ecome? Answer In Eqution 26.6, we let `: C 5 lim ` k e k e pP k 0 e Notice tht this expression is the sme s Eqution 26.2, the cpcitnce of n isolted sphericl conductor. Cpcitor symol Bttery symol witch symol Open Closed Figure 26.5 (Exmple 26.2) A sphericl cpcitor consists of n inner sphere of rdius surrounded y concentric sphericl shell of rdius. The electric field etween the spheres is directed rdilly outwrd when the inner sphere is positively chrged. Q Q Figure 26.6 Circuit symols for cpcitors, tteries, nd switches. Notice tht cpcitors re in lue, tteries re in green, nd switches re in red. The closed switch cn crry current, wheres the open one cnnot. Prllel Comintion Two cpcitors connected s shown in Figure 26.7 re known s prllel comintion of cpcitors. Figure 26.7 shows circuit digrm for this comintion of cpcitors. The left pltes of the cpcitors re connected to the positive terminl of the ttery y conducting wire nd re therefore oth t the sme electric potentil

126 26.3 Comintions of Cpcitors 783 A pictoril representtion of two cpcitors connected in prllel to ttery A circuit digrm showing the two cpcitors connected in prllel to ttery A circuit digrm showing the equivlent cpcitnce of the cpcitors in prllel Figure 26.7 Two cpcitors connected in prllel. All three digrms re equivlent. C 1 C 1 Q 1 Q 1 V 1 C 2 Q 2 Q 2 V 2 V V c s the positive terminl. Likewise, the right pltes re connected to the negtive terminl nd so re oth t the sme potentil s the negtive terminl. Therefore, the individul potentil differences cross cpcitors connected in prllel re the sme nd re equl to the potentil difference pplied cross the comintion. Tht is, DV 1 5 DV 2 5 DV where DV is the ttery terminl voltge. After the ttery is ttched to the circuit, the cpcitors quickly rech their mximum chrge. Let s cll the mximum chrges on the two cpcitors Q 1 nd Q 2, where Q 1 5 C 1 DV 1 nd Q 2 5 C 2 DV 2. The totl chrge Q tot stored y the two cpcitors is the sum of the chrges on the individul cpcitors: Q tot 5 Q 1 1 Q 2 5 C 1 DV 1 1 C 2 DV 2 (26.7) uppose you wish to replce these two cpcitors y one equivlent cpcitor hving cpcitnce C eq s in Figure 26.7c. The effect this equivlent cpcitor hs on the circuit must e exctly the sme s the effect of the comintion of the two individul cpcitors. Tht is, the equivlent cpcitor must store chrge Q tot when connected to the ttery. Figure 26.7c shows tht the voltge cross the equivlent cpcitor is DV ecuse the equivlent cpcitor is connected directly cross the ttery terminls. Therefore, for the equivlent cpcitor, Q tot 5 C eq DV ustituting this result into Eqution 26.7 gives C eq DV 5 C 1 DV 1 1 C 2 DV 2 C eq 5 C 1 1 C 2 1prllel comintion2 where we hve cnceled the voltges ecuse they re ll the sme. If this tretment is extended to three or more cpcitors connected in prllel, the equivlent cpcitnce is found to e Q 1 Q 2 C 2 C eq C 1 C 2 V C eq 5 C 1 1 C 2 1 C 3 1 c 1prllel comintion2 (26.8) Therefore, the equivlent cpcitnce of prllel comintion of cpcitors is (1) the lgeric sum of the individul cpcitnces nd (2) greter thn ny of W WEquivlent cpcitnce for cpcitors in prllel

127 784 Chpter 26 Cpcitnce nd Dielectrics Figure 26.8 Two cpcitors connected in series. All three digrms re equivlent. A pictoril representtion of two cpcitors connected in series to ttery A circuit digrm showing the two cpcitors connected in series to ttery A circuit digrm showing the equivlent cpcitnce of the cpcitors in series C 1 V 1 V 2 C 2 C 1 C 2 Q Q Q Q V V 1 V 2 V c the individul cpcitnces. ttement (2) mkes sense ecuse we re essentilly comining the res of ll the cpcitor pltes when they re connected with conducting wire, nd cpcitnce of prllel pltes is proportionl to re (Eq. 26.3). eries Comintion Two cpcitors connected s shown in Figure 26.8 nd the equivlent circuit digrm in Figure 26.8 re known s series comintion of cpcitors. The left plte of cpcitor 1 nd the right plte of cpcitor 2 re connected to the terminls of ttery. The other two pltes re connected to ech other nd to nothing else; hence, they form n isolted system tht is initilly unchrged nd must continue to hve zero net chrge. To nlyze this comintion, let s first consider the unchrged cpcitors nd then follow wht hppens immeditely fter ttery is connected to the circuit. When the ttery is connected, electrons re trnsferred out of the left plte of C 1 nd into the right plte of C 2. As this negtive chrge ccumultes on the right plte of C 2, n equivlent mount of negtive chrge is forced off the left plte of C 2, nd this left plte therefore hs n excess positive chrge. The negtive chrge leving the left plte of C 2 cuses negtive chrges to ccumulte on the right plte of C 1. As result, oth right pltes end up with chrge 2Q nd oth left pltes end up with chrge 1Q. Therefore, the chrges on cpcitors connected in series re the sme: Q 1 5 Q 2 5 Q where Q is the chrge tht moved etween wire nd the connected outside plte of one of the cpcitors. Figure 26.8 shows the individul voltges DV 1 nd DV 2 cross the cpcitors. These voltges dd to give the totl voltge DV tot cross the comintion: DV tot 5 DV 1 1 DV 2 5 Q 1 C 1 1 Q 2 C 2 (26.9) In generl, the totl potentil difference cross ny numer of cpcitors connected in series is the sum of the potentil differences cross the individul cpcitors. uppose the equivlent single cpcitor in Figure 26.8c hs the sme effect on the circuit s the series comintion when it is connected to the ttery. After it is fully chrged, the equivlent cpcitor must hve chrge of 2Q on its right plte nd chrge of 1Q on its left plte. Applying the definition of cpcitnce to the circuit in Figure 26.8c gives DV tot 5 Q C eq C eq C 1 C 2 V

128 26.3 Comintions of Cpcitors 785 ustituting this result into Eqution 26.9, we hve Q C eq 5 Q 1 C 1 1 Q 2 C 2 Cnceling the chrges ecuse they re ll the sme gives C eq C 1 C 2 1series comintion2 When this nlysis is pplied to three or more cpcitors connected in series, the reltionship for the equivlent cpcitnce is C eq C 1 C 2 C c 1series comintion2 (26.10) 3 This expression shows tht (1) the inverse of the equivlent cpcitnce is the lgeric sum of the inverses of the individul cpcitnces nd (2) the equivlent cpcitnce of series comintion is lwys less thn ny individul cpcitnce in the comintion. Q uick Quiz 26.3 Two cpcitors re identicl. They cn e connected in series or in prllel. If you wnt the smllest equivlent cpcitnce for the comintion, how should you connect them? () in series () in prllel (c) either wy ecuse oth comintions hve the sme cpcitnce Exmple 26.3 Equivlent Cpcitnce Find the equivlent cpcitnce etween nd for the comintion of cpcitors shown in Figure All cpcitnces re in microfrds. olution Conceptulize tudy Figure 26.9 crefully nd mke sure you understnd how the cpcitors re connected. Verify tht there re only series nd prllel connections etween cpcitors. Ctegorize Figure 26.9 shows tht the circuit contins oth series nd prllel connections, so we use the rules for series nd prllel comintions discussed in this section. W WEquivlent cpcitnce for cpcitors in series Anlyze Using Equtions 26.8 nd 26.10, we reduce the comintion step y step s indicted in the figure. As you follow long elow, notice tht in ech step we replce the comintion of two cpcitors in the circuit digrm with single cpcitor hving the equivlent cpcitnce. The 1.0-mF nd 3.0-mF cpcitors (upper red-rown circle in Fig. 26.9) re in prllel. Find the equivlent cpcitnce from Eqution 26.8: c d Figure 26.9 (Exmple 26.3) To find the equivlent cpcitnce of the cpcitors in (), we reduce the vrious comintions in steps s indicted in (), (c), nd (d), using the series nd prllel rules descried in the text. All cpcitnces re in microfrds. The 2.0-mF nd 6.0-mF cpcitors (lower red-rown circle in Fig. 26.9) re lso in prllel: The circuit now looks like Figure The two 4.0-mF cpcitors (upper green circle in Fig. 26.9) re in series. Find the equivlent cpcitnce from Eqution 26.10: C eq 5 C 1 1 C mf C eq 5 C 1 1 C mf C eq C 1 C mf mf mf C eq mf continued

129 786 Chpter 26 Cpcitnce nd Dielectrics 26.3 continued The two 8.0-mF cpcitors (lower green circle in Fig. 26.9) re lso in series. Find the equivlent cpcitnce from Eqution 26.10: The circuit now looks like Figure 26.9c. The 2.0-mF nd 4.0-mF cpcitors re in prllel: C eq C 1 C mf mf mf C eq mf C eq 5 C 1 1 C mf Finlize This finl vlue is tht of the single equivlent cpcitor shown in Figure 26.9d. For further prctice in treting circuits with comintions of cpcitors, imgine ttery is connected etween points nd in Figure 26.9 so tht potentil difference DV is estlished cross the comintion. Cn you find the voltge cross nd the chrge on ech cpcitor? Figure () A circuit consisting of cpcitor, ttery, nd switch. () When the switch is closed, the ttery estlishes n electric field in the wire nd the cpcitor ecomes chrged. V 26.4 Energy tored in Chrged Cpcitor Becuse positive nd negtive chrges re seprted in the system of two conductors in cpcitor, electric potentil energy is stored in the system. Mny of those who work with electronic equipment hve t some time verified tht cpcitor cn store energy. If the pltes of chrged cpcitor re connected y conductor such s wire, chrge moves etween ech plte nd its connecting wire until the cpcitor is unchrged. The dischrge cn often e oserved s visile sprk. If you ccidentlly touch the opposite pltes of chrged cpcitor, your fingers ct s pthwy for dischrge nd the result is n electric shock. The degree of shock you receive depends on the cpcitnce nd the voltge pplied to the cpcitor. uch shock could e dngerous if high voltges re present s in the power supply of home theter system. Becuse the chrges cn e stored in cpcitor even when the system is turned off, unplugging the system does not mke it sfe to open the cse nd touch the components inside. Figure shows ttery connected to single prllel-plte cpcitor with switch in the circuit. Let us identify the circuit s system. When the switch is closed (Fig ), the ttery estlishes n electric field in the wires nd chrges With the switch open, the cpcitor remins unchrged. Electrons move from the plte to the wire, leving the plte positively chrged. Electric field in wire V eprtion of chrges represents potentil energy Electric field etween pltes E Electrons move from the wire to the plte. Electric field in wire Chemicl potentil energy in the ttery is reduced.

130 26.4 Energy tored in Chrged Cpcitor 787 flow etween the wires nd the cpcitor. As tht occurs, there is trnsformtion of energy within the system. Before the switch is closed, energy is stored s chemicl potentil energy in the ttery. This energy is trnsformed during the chemicl rection tht occurs within the ttery when it is operting in n electric circuit. When the switch is closed, some of the chemicl potentil energy in the ttery is trnsformed to electric potentil energy ssocited with the seprtion of positive nd negtive chrges on the pltes. To clculte the energy stored in the cpcitor, we shll ssume chrging process tht is different from the ctul process descried in ection 26.1 ut tht gives the sme finl result. This ssumption is justified ecuse the energy in the finl configurtion does not depend on the ctul chrge-trnsfer process. 3 Imgine the pltes re disconnected from the ttery nd you trnsfer the chrge mechniclly through the spce etween the pltes s follows. You gr smll mount of positive chrge on one plte nd pply force tht cuses this positive chrge to move over to the other plte. Therefore, you do work on the chrge s it is trnsferred from one plte to the other. At first, no work is required to trnsfer smll mount of chrge dq from one plte to the other, 4 ut once this chrge hs een trnsferred, smll potentil difference exists etween the pltes. Therefore, work must e done to move dditionl chrge through this potentil difference. As more nd more chrge is trnsferred from one plte to the other, the potentil difference increses in proportion nd more work is required. The overll process is descried y the nonisolted system model for energy. Eqution 8.2 reduces to W 5 DU E ; the work done on the system y the externl gent ppers s n increse in electric potentil energy in the system. uppose q is the chrge on the cpcitor t some instnt during the chrging process. At the sme instnt, the potentil difference cross the cpcitor is DV 5 q/c. This reltionship is grphed in Figure From ection 25.1, we know tht the work necessry to trnsfer n increment of chrge dq from the plte crrying chrge 2q to the plte crrying chrge q (which is t the higher electric potentil) is dw 5 DV dq 5 q C dq The work required to trnsfer the chrge dq is the re of the tn rectngle in Figure Becuse 1 V 5 1 J/C, the unit for the re is the joule. The totl work required to chrge the cpcitor from q 5 0 to some finl chrge q 5 Q is Q q W 5 3 C dq 5 1 Q C 3 q dq 5 Q 2 2C 0 The work done in chrging the cpcitor ppers s electric potentil energy U E stored in the cpcitor. Using Eqution 26.1, we cn express the potentil energy stored in chrged cpcitor s 0 U E 5 Q 2 2C 5 1 2Q DV 5 1 2C 1DV 2 2 (26.11) Becuse the curve in Figure is stright line, the totl re under the curve is tht of tringle of se Q nd height DV. Eqution pplies to ny cpcitor, regrdless of its geometry. For given cpcitnce, the stored energy increses s the chrge nd the potentil difference increse. In prctice, there is limit to the mximum energy (or chrge) tht cn e stored ecuse, t sufficiently lrge vlue of DV, dischrge ultimtely occurs The work required to move chrge dq through the potentil difference V cross the cpcitor pltes is given pproximtely y the re of the shded rectngle. V dq Figure A plot of potentil difference versus chrge for cpcitor is stright line hving slope 1/C. W WEnergy stored in chrged cpcitor Q q 3 This discussion is similr to tht of stte vriles in thermodynmics. The chnge in stte vrile such s temperture is independent of the pth followed etween the initil nd finl sttes. The potentil energy of cpcitor (or ny system) is lso stte vrile, so its chnge does not depend on the process followed to chrge the cpcitor. 4 We shll use lowercse q for the time-vrying chrge on the cpcitor while it is chrging to distinguish it from uppercse Q, which is the totl chrge on the cpcitor fter it is completely chrged.

131 788 Chpter 26 Cpcitnce nd Dielectrics Pitfll Prevention 26.4 Not New Kind of Energy The energy given y Eqution is not new kind of energy. The eqution descries fmilir electric potentil energy ssocited with system of seprted source chrges. Eqution provides new interprettion, or new wy of modeling the energy. Furthermore, Eqution correctly descries the energy density ssocited with ny electric field, regrdless of the source. Exmple 26.4 Energy density in n electric field Rewiring Two Chrged Cpcitors Two cpcitors C 1 nd C 2 (where C 1. C 2 ) re chrged to the sme initil potentil difference DV i. The chrged cpcitors re removed from the ttery, nd their pltes re connected with opposite polrity s in Figure The switches 1 nd 2 re then closed s in Figure (A) Find the finl potentil difference DV f etween nd fter the switches re closed. etween the pltes. For this reson, cpcitors re usully leled with mximum operting voltge. We cn consider the energy in cpcitor to e stored in the electric field creted etween the pltes s the cpcitor is chrged. This description is resonle ecuse the electric field is proportionl to the chrge on the cpcitor. For prllel-plte cpcitor, the potentil difference is relted to the electric field through the reltionship DV 5 Ed. Furthermore, its cpcitnce is C 5 P 0 A/d (Eq. 26.3). ustituting these expressions into Eqution gives U E P 0 A d 1 Ed P 0Ad2E 2 (26.12) Becuse the volume occupied y the electric field is Ad, the energy per unit volume u E 5 U E /Ad, known s the energy density, is u E P 0 E 2 (26.13) Although Eqution ws derived for prllel-plte cpcitor, the expression is generlly vlid regrdless of the source of the electric field. Tht is, the energy density in ny electric field is proportionl to the squre of the mgnitude of the electric field t given point. Q uick Quiz 26.4 You hve three cpcitors nd ttery. In which of the following comintions of the three cpcitors is the mximum possile energy stored when the comintion is ttched to the ttery? () series () prllel (c) no difference ecuse oth comintions store the sme mount of energy Q 2f C 2 olution Conceptulize Figure helps us understnd the initil nd finl configurtions of the system. When the switches re closed, the chrge on the system will redistriute etween the cpcitors until oth cpcitors hve the sme Figure (Exmple 26.4) () Two cpcitors re chrged to the sme initil potentil difference nd connected together with pltes of opposite sign to e in contct when the switches re closed. () When the switches re closed, the chrges redistriute. potentil difference. Becuse C 1. C 2, more chrge exists on C 1 thn on C 2, so the finl configurtion will hve positive chrge on the left pltes s shown in Figure Ctegorize In Figure 26.12, it might pper s if the cpcitors re connected in prllel, ut there is no ttery in this circuit to pply voltge cross the comintion. Therefore, we cnnot ctegorize this prolem s one in which cpcitors re connected in prllel. We cn ctegorize it s prolem involving n isolted system for electric chrge. The left-hnd pltes of the cpcitors form n isolted system ecuse they re not connected to the right-hnd pltes y conductors. Q 1i C Q 2i C 2 Q C 1 1f 1 2 Anlyze Write n expression for the totl chrge on the left-hnd pltes of the system efore the switches re closed, noting tht negtive sign for Q 2i is necessry ecuse the chrge on the left plte of cpcitor C 2 is negtive: (1) Q i 5 Q 1i 1 Q 2i 5 C 1 DV i 2 C 2 DV i 5 (C 1 2 C 2 )DV i

132 26.4 Energy tored in Chrged Cpcitor continued After the switches re closed, the chrges on the individul cpcitors chnge to new vlues Q 1f nd Q 2f such tht the potentil difference is gin the sme cross oth cpcitors, with vlue of DV f. Write n expression for the totl chrge on the left-hnd pltes of the system fter the switches re closed: Becuse the system is isolted, the initil nd finl chrges on the system must e the sme. Use this condition nd Equtions (1) nd (2) to solve for DV f : Q f 5 Q i 1C 1 1 C 2 2 DV f 5 1C 1 2 C 2 2 DV i (3) DV f 5 C 1 2 C 2 DV C 1 1 C i 2 (B) Find the totl energy stored in the cpcitors efore nd fter the switches re closed nd determine the rtio of the finl energy to the initil energy. olution Use Eqution to find n expression for the totl energy stored in the cpcitors efore the switches re closed: Write n expression for the totl energy stored in the cpcitors fter the switches re closed: Use the results of prt (A) to rewrite this expression in terms of DV i : Divide Eqution (5) y Eqution (4) to otin the rtio of the energies stored in the system: (2) Q f 5 Q 1f 1 Q 2f 5 C 1 DV f 1 C 2 DV f 5 (C 1 1 C 2 )DV f (4) U i 5 1 2C 1 1DV i C 2 1DV i C 1 1 C 2 2 1DV i 2 2 (5) U f C 1 1 C 2 2 c C 1 2 C 2 2 DV C 1 1 C i d 5 1C C DV i C 1 1 C 2 (6) U f 5 1 2C 1 1DV f C 2 1DV f C 1 1 C 2 2 1DV f 2 2 U f C 1 2 C DV i 2 2 / 1C 1 1 C U i 2 1C 1 1 C 2 2 1DV i 2 2 U f 5 C 1 2 C 2 U i C 1 1 C 2 Finlize The rtio of energies is less thn unity, indicting tht the finl energy is less thn the initil energy. At first, you might think the lw of energy conservtion hs een violted, ut tht is not the cse. The missing energy is trnsferred out of the system y the mechnism of electromgnetic wves (T ER in Eq. 8.2), s we shll see in Chpter 34. Therefore, this system is isolted for electric chrge, ut nonisolted for energy. Wht If? Wht if the two cpcitors hve the sme cpcitnce? Wht would you expect to hppen when the switches re closed? Answer Becuse oth cpcitors hve the sme initil potentil difference pplied to them, the chrges on the identicl cpcitors hve the sme mgnitude. When the cpcitors with opposite polrities re connected together, the equlmgnitude chrges should cncel ech other, leving the cpcitors unchrged. Let s test our results to see if tht is the cse mthemticlly. In Eqution (1), ecuse the cpcitnces re equl, the initil chrge Q i on the system of left-hnd pltes is zero. Eqution (3) shows tht DV f 5 0, which is consistent with unchrged cpcitors. Finlly, Eqution (5) shows tht U f 5 0, which is lso consistent with unchrged cpcitors. 2 One device in which cpcitors hve n importnt role is the portle defirilltor (see the chpter-opening photo on pge 777). When crdic firilltion (rndom contrctions) occurs, the hert produces rpid, irregulr pttern of ets. A fst dischrge of energy through the hert cn return the orgn to its norml et pttern. Emergency medicl tems use portle defirilltors tht contin tteries cple of chrging cpcitor to high voltge. (The circuitry ctully permits the cpcitor to e chrged to much higher voltge thn tht of the ttery.) Up to 360 J is stored

133 790 Chpter 26 Cpcitnce nd Dielectrics Pitfll Prevention 26.5 Is the Cpcitor Connected to Bttery? For prolems in which cpcitor is modified (y insertion of dielectric, for exmple), you must note whether modifictions to the cpcitor re eing mde while the cpcitor is connected to ttery or fter it is disconnected. If the cpcitor remins connected to the ttery, the voltge cross the cpcitor necessrily remins the sme. If you disconnect the cpcitor from the ttery efore mking ny modifictions to the cpcitor, the cpcitor is n isolted system for electric chrge nd its chrge remins the sme. Cpcitnce of cpcitor filled with mteril of dielectric constnt k in the electric field of lrge cpcitor in defirilltor when it is fully chrged. The stored energy is relesed through the hert y conducting electrodes, clled pddles, which re plced on oth sides of the victim s chest. The defirilltor cn deliver the energy to ptient in out 2 ms (roughly equivlent to times the power delivered to 60-W lightul!). The prmedics must wit etween pplictions of the energy ecuse of the time intervl necessry for the cpcitors to ecome fully chrged. In this ppliction nd others (e.g., cmer flsh units nd lsers used for fusion experiments), cpcitors serve s energy reservoirs tht cn e slowly chrged nd then quickly dischrged to provide lrge mounts of energy in short pulse Cpcitors with Dielectrics A dielectric is nonconducting mteril such s ruer, glss, or wxed pper. We cn perform the following experiment to illustrte the effect of dielectric in cpcitor. Consider prllel-plte cpcitor tht without dielectric hs chrge Q 0 nd cpcitnce C 0. The potentil difference cross the cpcitor is DV 0 5 Q 0 /C 0. Figure illustrtes this sitution. The potentil difference is mesured y device clled voltmeter. Notice tht no ttery is shown in the figure; lso, we must ssume no chrge cn flow through n idel voltmeter. Hence, there is no pth y which chrge cn flow nd lter the chrge on the cpcitor. If dielectric is now inserted etween the pltes s in Figure 26.13, the voltmeter indictes tht the voltge etween the pltes decreses to vlue DV. The voltges with nd without the dielectric re relted y fctor k s follows: DV 5 DV 0 k Becuse DV, DV 0, we see tht k. 1. The dimensionless fctor k is clled the dielectric constnt of the mteril. The dielectric constnt vries from one mteril to nother. In this section, we nlyze this chnge in cpcitnce in terms of electricl prmeters such s electric chrge, electric field, nd potentil difference; ection 26.7 descries the microscopic origin of these chnges. Becuse the chrge Q 0 on the cpcitor does not chnge, the cpcitnce must chnge to the vlue The potentil difference cross the chrged cpcitor is initilly V 0. C 0 Q 0 C 5 Q 0 DV 5 Q 0 DV 0 /k 5 k Q 0 DV 0 C 5 kc 0 (26.14) After the dielectric is inserted etween the pltes, the chrge remins the sme, ut the potentil difference decreses nd the cpcitnce increses. Dielectric C Q 0 Figure A chrged cpcitor () efore nd () fter insertion of dielectric etween the pltes. V 0 V

134 26.5 Cpcitors with Dielectrics 791 Tht is, the cpcitnce increses y the fctor k when the dielectric completely fills the region etween the pltes. 5 Becuse C 0 5 P 0 A/d (Eq. 26.3) for prllel-plte cpcitor, we cn express the cpcitnce of prllel-plte cpcitor filled with dielectric s C 5 k P 0A (26.15) d From Eqution 26.15, it would pper tht the cpcitnce could e mde very lrge y inserting dielectric etween the pltes nd decresing d. In prctice, the lowest vlue of d is limited y the electric dischrge tht could occur through the dielectric medium seprting the pltes. For ny given seprtion d, the mximum voltge tht cn e pplied to cpcitor without cusing dischrge depends on the dielectric strength (mximum electric field) of the dielectric. If the mgnitude of the electric field in the dielectric exceeds the dielectric strength, the insulting properties rek down nd the dielectric egins to conduct. Physicl cpcitors hve specifiction clled y vriety of nmes, including working voltge, rekdown voltge, nd rted voltge. This prmeter represents the lrgest voltge tht cn e pplied to the cpcitor without exceeding the dielectric strength of the dielectric mteril in the cpcitor. Consequently, when selecting cpcitor for given ppliction, you must consider its cpcitnce s well s the expected voltge cross the cpcitor in the circuit, mking sure the expected voltge is smller thn the rted voltge of the cpcitor. Insulting mterils hve vlues of k greter thn unity nd dielectric strengths greter thn tht of ir s Tle 26.1 indictes. Therefore, dielectric provides the following dvntges: An increse in cpcitnce An increse in mximum operting voltge Possile mechnicl support etween the pltes, which llows the pltes to e close together without touching, therey decresing d nd incresing C Tle 26.1 Approximte Dielectric Constnts nd Dielectric trengths of Vrious Mterils t Room Temperture Mteril Dielectric Constnt k Dielectric trength (10 6 V/m) Air (dry) Bkelite Fused qurtz Mylr Neoprene ruer Nylon Pper Prffin-impregnted pper Polystyrene Polyvinyl chloride Porcelin 6 12 Pyrex glss ilicone oil trontium titnte Teflon Vcuum Wter 80 The dielectric strength equls the mximum electric field tht cn exist in dielectric without electricl rekdown. These vlues depend strongly on the presence of impurities nd flws in the mterils. 5 If the dielectric is introduced while the potentil difference is held constnt y ttery, the chrge increses to vlue Q 5 kq 0. The dditionl chrge comes from the wires ttched to the cpcitor, nd the cpcitnce gin increses y the fctor k.

135 792 Chpter 26 Cpcitnce nd Dielectrics A tuulr cpcitor whose pltes re seprted y pper nd then rolled into cylinder Pper A high-voltge cpcitor consisting of mny prllel pltes seprted y insulting oil Pltes Cse Electrolyte An electrolytic cpcitor. Cengge Lerning/George emple Metl foil Oil c Figure Three commercil cpcitor designs. The mterils etween the pltes of the cpcitor re the wllord nd ir. Cpcitor pltes tud finder Wllord tud When the cpcitor moves cross stud in the wll, the mterils etween the pltes re the wllord nd the wood stud. The chnge in the dielectric constnt cuses signl light to illuminte. Figure (Quick Quiz 26.5) A stud finder. Contcts Metllic foil oxide lyer When one set of metl pltes is rotted so s to lie etween fixed set of pltes, the cpcitnce of the device chnges. Figure A vrile cpcitor. Types of Cpcitors Mny cpcitors re uilt into integrted circuit chips, ut some electricl devices still use stnd-lone cpcitors. Commercil cpcitors re often mde from metllic foil interlced with thin sheets of either prffin-impregnted pper or Mylr s the dielectric mteril. These lternte lyers of metllic foil nd dielectric re rolled into cylinder to form smll pckge (Fig ). High-voltge cpcitors commonly consist of numer of interwoven metllic pltes immersed in silicone oil (Fig ). mll cpcitors re often constructed from cermic mterils. Often, n electrolytic cpcitor is used to store lrge mounts of chrge t reltively low voltges. This device, shown in Figure 26.14c, consists of metllic foil in contct with n electrolyte, solution tht conducts electricity y virtue of the motion of ions contined in the solution. When voltge is pplied etween the foil nd the electrolyte, thin lyer of metl oxide (n insultor) is formed on the foil, nd this lyer serves s the dielectric. Very lrge vlues of cpcitnce cn e otined in n electrolytic cpcitor ecuse the dielectric lyer is very thin nd therefore the plte seprtion is very smll. Electrolytic cpcitors re not reversile s re mny other cpcitors. They hve polrity, which is indicted y positive nd negtive signs mrked on the device. When electrolytic cpcitors re used in circuits, the polrity must e correct. If the polrity of the pplied voltge is the opposite of wht is intended, the oxide lyer is removed nd the cpcitor conducts electricity insted of storing chrge. Vrile cpcitors (typiclly 10 to 500 pf) usully consist of two interwoven sets of metllic pltes, one fixed nd the other movle, nd contin ir s the dielectric (Fig ). These types of cpcitors re often used in rdio tuning circuits. Q uick Quiz 26.5 If you hve ever tried to hng picture or mirror, you know it cn e difficult to locte wooden stud in which to nchor your nil or screw. A crpenter s stud finder is cpcitor with its pltes rrnged side y side insted of fcing ech other s shown in Figure When the device is moved over stud, does the cpcitnce () increse or () decrese? Exmple 26.5 Energy tored Before nd After AM A prllel-plte cpcitor is chrged with ttery to chrge Q 0. The ttery is then removed, nd sl of mteril tht hs dielectric constnt k is inserted etween the pltes. Identify the system s the cpcitor nd the dielectric. Find the energy stored in the system efore nd fter the dielectric is inserted.

136 26.6 Electric Dipole in n Electric Field continued olution Conceptulize Think out wht hppens when the dielectric is inserted etween the pltes. Becuse the ttery hs een removed, the chrge on the cpcitor must remin the sme. We know from our erlier discussion, however, tht the cpcitnce must chnge. Therefore, we expect chnge in the energy of the system. Ctegorize Becuse we expect the energy of the system to chnge, we model it s nonisolted system for energy involving cpcitor nd dielectric. Anlyze From Eqution 26.11, find the energy stored in the sence of the dielectric: Find the energy stored in the cpcitor fter the dielectric is inserted etween the pltes: U 0 5 Q 0 2 2C 0 U 5 Q 0 2 2C Use Eqution to replce the cpcitnce C : U 5 Q U 0 2kC 0 k Finlize Becuse k. 1, the finl energy is less thn the initil energy. We cn ccount for the decrese in energy of the system y performing n experiment nd noting tht the dielectric, when inserted, is pulled into the device. To keep the dielectric from ccelerting, n externl gent must do negtive work on the dielectric. Eqution 8.2 ecomes DU 5 W, where oth sides of the eqution re negtive Electric Dipole in n Electric Field We hve discussed the effect on the cpcitnce of plcing dielectric etween the pltes of cpcitor. In ection 26.7, we shll descrie the microscopic origin of this effect. Before we cn do so, however, let s expnd the discussion of the electric dipole introduced in ection 23.4 (see Exmple 23.6). The electric dipole consists of two chrges of equl mgnitude nd opposite sign seprted y distnce 2 s shown in Figure The electric dipole moment of this configurtion is defined s the vector p directed from 2q towrd 1q long the line joining the chrges nd hving mgnitude p ; 2q (26.16) Now suppose n electric dipole is plced in uniform electric field E nd mkes n ngle u with the field s shown in Figure We identify E s the field externl to the dipole, estlished y some other chrge distriution, to distinguish it from the field due to the dipole, which we discussed in ection Ech of the chrges is modeled s prticle in n electric field. The electric forces cting on the two chrges re equl in mgnitude (F 5 qe) nd opposite in direction s shown in Figure Therefore, the net force on the dipole is zero. The two forces produce net torque on the dipole, however; the dipole is therefore descried y the rigid oject under net torque model. As result, the dipole rottes in the direction tht rings the dipole moment vector into greter lignment with the field. The torque due to the force on the positive chrge out n xis through O in Figure hs mgnitude F sin u, where sin u is the moment rm of F out O. This force tends to produce clockwise rottion. The torque out O on the negtive chrge is lso of mgnitude F sin u; here gin, the force tends to produce clockwise rottion. Therefore, the mgnitude of the net torque out O is t 5 2F sin u Becuse F 5 qe nd p 5 2q, we cn express t s t 5 2qE sin u 5 pe sin u (26.17) The electric dipole moment p is directed from q towrd q. q 2 p q Figure An electric dipole consists of two chrges of equl mgnitude nd opposite sign seprted y distnce of 2. F The dipole moment p is t n ngle u to the field, cusing the dipole to experience torque. p q O q u E F Figure An electric dipole in uniform externl electric field.

137 794 Chpter 26 Cpcitnce nd Dielectrics Torque on n electric dipole in n externl electric field Potentil energy of the system of n electric dipole in n externl electric field H O 105 H The center of the positive chrge distriution is t the point. Figure The wter molecule, H 2 O, hs permnent polriztion resulting from its nonliner geometry. Bsed on this expression, it is convenient to express the torque in vector form s the cross product of the vectors p nd E : t 5 p 3 E (26.18) We cn lso model the system of the dipole nd the externl electric field s n isolted system for energy. Let s determine the potentil energy of the system s function of the dipole s orienttion with respect to the field. To do so, recognize tht work must e done y n externl gent to rotte the dipole through n ngle so s to cuse the dipole moment vector to ecome less ligned with the field. The work done is then stored s electric potentil energy in the system. Notice tht this potentil energy is ssocited with rottionl configurtion of the system. Previously, we hve seen potentil energies ssocited with trnsltionl configurtions: n oject with mss ws moved in grvittionl field, chrge ws moved in n electric field, or spring ws extended. The work dw required to rotte the dipole through n ngle du is dw 5 t du (see Eq ). Becuse t 5 pe sin u nd the work results in n increse in the electric potentil energy U, we find tht for rottion from u i to u f, the chnge in potentil energy of the system is U f 2 U i 5 3 u f u f t du 5 3 pe sin u du 5 pe 3 sin u du u i u i u i 5 pe 32cos u 4 uf u i 5 pe 1cos u i 2 cos u f 2 The term tht contins cos u i is constnt tht depends on the initil orienttion of the dipole. It is convenient to choose reference ngle of u i so tht cos u i 5 cos Furthermore, let s choose U i 5 0 t u i s our reference vlue of potentil energy. Hence, we cn express generl vlue of U E 5 U f s u f U E 5 2pE cos u (26.19) We cn write this expression for the potentil energy of dipole in n electric field s the dot product of the vectors p nd E : U E 5 2p? E (26.20) To develop conceptul understnding of Eqution 26.19, compre it with the expression for the potentil energy of the system of n oject in the Erth s grvittionl field, U g 5 mgy (Eq. 7.19). First, oth expressions contin prmeter of the entity plced in the field: mss for the oject, dipole moment for the dipole. econd, oth expressions contin the field, g for the oject, E for the dipole. Finlly, oth expressions contin configurtion description: trnsltionl position y for the oject, rottionl position u for the dipole. In oth cses, once the configurtion is chnged, the system tends to return to the originl configurtion when the oject is relesed: the oject of mss m flls towrd the ground, nd the dipole egins to rotte ck towrd the configurtion in which it is ligned with the field. Molecules re sid to e polrized when seprtion exists etween the verge position of the negtive chrges nd the verge position of the positive chrges in the molecule. In some molecules such s wter, this condition is lwys present; such molecules re clled polr molecules. Molecules tht do not possess permnent polriztion re clled nonpolr molecules. We cn understnd the permnent polriztion of wter y inspecting the geometry of the wter molecule. The oxygen tom in the wter molecule is onded to the hydrogen toms such tht n ngle of 1058 is formed etween the two onds (Fig ). The center of the negtive chrge distriution is ner the oxygen tom, nd the center of the positive chrge distriution lies t point midwy long the line joining the hydrogen toms (the point leled 3 in Fig ). We cn model the wter molecule nd other polr molecules s dipoles ecuse the verge positions of the positive nd negtive chrges ct s point chrges. As result, we cn pply our discussion of dipoles to the ehvior of polr molecules.

138 26.7 An Atomic Description of Dielectrics 795 Wshing with sop nd wter is household scenrio in which the dipole structure of wter is exploited. Grese nd oil re mde up of nonpolr molecules, which re generlly not ttrcted to wter. Plin wter is not very useful for removing this type of grime. op contins long molecules clled surfctnts. In long molecule, the polrity chrcteristics of one end of the molecule cn e different from those t the other end. In surfctnt molecule, one end cts like nonpolr molecule nd the other cts like polr molecule. The nonpolr end cn ttch to grese or oil molecule, nd the polr end cn ttch to wter molecule. Therefore, the sop serves s chin, linking the dirt nd wter molecules together. When the wter is rinsed wy, the grese nd oil go with it. A symmetric molecule (Fig ) hs no permnent polriztion, ut polriztion cn e induced y plcing the molecule in n electric field. A field directed to the left s in Figure cuses the center of the negtive chrge distriution to shift to the right reltive to the positive chrges. This induced polriztion is the effect tht predomintes in most mterils used s dielectrics in cpcitors. Exmple 26.6 The H 2 O Molecule AM E Figure () A liner symmetric molecule hs no permnent polriztion. () An externl electric field induces polriztion in the molecule. The wter (H 2 O) molecule hs n electric dipole moment of C? m. A smple contins wter molecules, with the dipole moments ll oriented in the direction of n electric field of mgnitude N/C. How much work is required to rotte the dipoles from this orienttion (u 5 08) to one in which ll the moments re perpendiculr to the field (u 5 908)? olution Conceptulize When ll the dipoles re ligned with the electric field, the dipoles electric field system hs the minimum potentil energy. This energy hs negtive vlue given y the product of the right side of Eqution 26.19, evluted t 08, nd the numer N of dipoles. Ctegorize The comintion of the dipoles nd the electric field is identified s system. We use the nonisolted system model ecuse n externl gent performs work on the system to chnge its potentil energy. Anlyze Write the pproprite reduction of the conservtion of energy eqution, Eqution 8.2, for this sitution: Use Eqution to evlute the initil nd finl potentil energies of the system nd Eqution (1) to clculte the work required to rotte the dipoles: 26.7 An Atomic Description of Dielectrics (1) DU E 5 W W 5 U U 08 5 (2NpE cos 908) 2 (2NpE cos 08) 5 NpE 5 (10 21 )( C? m)( N/C) J Finlize Notice tht the work done on the system is positive ecuse the potentil energy of the system hs een rised from negtive vlue to vlue of zero. In ection 26.5, we found tht the potentil difference DV 0 etween the pltes of cpcitor is reduced to DV 0 /k when dielectric is introduced. The potentil difference is reduced ecuse the mgnitude of the electric field decreses etween the pltes. In prticulr, if E 0 is the electric field without the dielectric, the field in the presence of dielectric is E 0 E 5 (26.21) k First consider dielectric mde up of polr molecules plced in the electric field etween the pltes of cpcitor. The dipoles (tht is, the polr molecules mking

139 796 Chpter 26 Cpcitnce nd Dielectrics Figure () Polr molecules in dielectric. () An electric field is pplied to the dielectric. (c) Detils of the electric field inside the dielectric. Polr molecules re rndomly oriented in the sence of n externl electric field. When n externl electric field is pplied, the molecules prtilly lign with the field. The chrged edges of the dielectric cn e modeled s n dditionl pir of prllel pltes estlishing n electric field E ind in the direction opposite tht of E 0. The induced chrge density s ind on the dielectric is less thn the chrge density s on the pltes. s s ind s ind s Figure Induced chrge on dielectric plced etween the pltes of chrged cpcitor. E 0 c up the dielectric) re rndomly oriented in the sence of n electric field s shown in Figure When n externl field E 0 due to chrges on the cpcitor pltes is pplied, torque is exerted on the dipoles, cusing them to prtilly lign with the field s shown in Figure The dielectric is now polrized. The degree of lignment of the molecules with the electric field depends on temperture nd the mgnitude of the field. In generl, the lignment increses with decresing temperture nd with incresing electric field. If the molecules of the dielectric re nonpolr, the electric field due to the pltes produces n induced polriztion in the molecule. These induced dipole moments tend to lign with the externl field, nd the dielectric is polrized. Therefore, dielectric cn e polrized y n externl field regrdless of whether the molecules in the dielectric re polr or nonpolr. With these ides in mind, consider sl of dielectric mteril plced etween the pltes of cpcitor so tht it is in uniform electric field E 0 s shown in Figure The electric field due to the pltes is directed to the right nd polrizes the dielectric. The net effect on the dielectric is the formtion of n induced positive surfce chrge density s ind on the right fce nd n equl-mgnitude negtive surfce chrge density 2s ind on the left fce s shown in Figure 26.21c. Becuse we cn model these surfce chrge distriutions s eing due to chrged prllel pltes, the induced surfce chrges on the dielectric give rise to n induced electric field E ind in the direction opposite the externl field E 0. Therefore, the net electric field E in the dielectric hs mgnitude E 5 E 0 2 E ind (26.22) In the prllel-plte cpcitor shown in Figure 26.22, the externl field E 0 is relted to the chrge density s on the pltes through the reltionship E 0 5 s/p 0. The induced electric field in the dielectric is relted to the induced chrge density s ind through the reltionship E ind 5 s ind /P 0. Becuse E 5 E 0 /k 5 s/kp 0, sustitution into Eqution gives s kp 0 5 s P 0 2 s ind P 0 s ind 5 k 2 1 s (26.23) k Becuse k. 1, this expression shows tht the chrge density s ind induced on the dielectric is less thn the chrge density s on the pltes. For instnce, if k 5 3, the induced chrge density is two-thirds the chrge density on the pltes. If no dielectric is present, then k 5 1 nd s ind 5 0 s expected. If the dielectric is replced y n electricl conductor for which E 5 0, however, Eqution indictes tht E 0 5 E ind, which corresponds to s ind 5 s. Tht is, the surfce chrge induced on s ind E 0 E ind s ind

140 26.7 An Atomic Description of Dielectrics 797 the conductor is equl in mgnitude ut opposite in sign to tht on the pltes, resulting in net electric field of zero in the conductor (see Fig ). Exmple 26.7 Effect of Metllic l A prllel-plte cpcitor hs plte seprtion d nd plte re A. An unchrged metllic sl of thickness is inserted midwy etween the pltes. (A) Find the cpcitnce of the device. olution Conceptulize Figure shows the metllic sl etween the pltes of the cpcitor. Any chrge tht ppers on one plte of the cpcitor must induce chrge of equl mgnitude nd opposite sign on the ner side of the sl s shown in Figure Consequently, the net chrge on the sl remins zero nd the electric field inside the sl is zero. Ctegorize The plnes of chrge on the metllic sl s upper nd lower edges re identicl to the distriution of chrges on the pltes of cpcitor. The metl etween the sl s edges serves only to mke n electricl connection etween the edges. Therefore, we cn model the edges of the sl s conducting plnes nd the ulk of the sl s wire. As result, the cpcitor in Figure is equivlent to two cpcitors in series, ech hving plte seprtion (d 2 )/2 s shown in Figure Anlyze Use Eqution 26.3 nd the rule for dding two cpcitors in series (Eq ) to find the equivlent cpcitnce in Figure 26.23: d (d )/2 s s s (d )/2 s (d )/2 (d )/2 Figure (Exmple 26.7) () A prllel-plte cpcitor of plte seprtion d prtilly filled with metllic sl of thickness. () The equivlent circuit of the device in () consists of two cpcitors in series, ech hving plte seprtion (d 2 )/2. 1 C C 1 C 2 C 5 P 0 A d P 0 A 1d 2 2 /2 1 P 0 A 1d 2 2 /2 (B) how tht the cpcitnce of the originl cpcitor is unffected y the insertion of the metllic sl if the sl is infinitesimlly thin. olution In the result for prt (A), let 0: C 5 lim P 0A 0 d 2 5 P 0A d Finlize The result of prt (B) is the originl cpcitnce efore the sl is inserted, which tells us tht we cn insert n infinitesimlly thin metllic sheet etween the pltes of cpcitor without ffecting the cpcitnce. We use this fct in the next exmple. Wht If? Wht if the metllic sl in prt (A) is not midwy etween the pltes? How would tht ffect the cpcitnce? Answer Let s imgine moving the sl in Figure upwrd so tht the distnce etween the upper edge of the sl nd the upper plte is. Then, the distnce etween the lower edge of the sl nd the lower plte is d 2 2. As in prt (A), we find the totl cpcitnce of the series comintion: 1 C C 1 C 2 P 0 A/ 1 1 P 0 A/ 1d P 0 A 1 d d 2 P 0 A P 0 A C 5 P 0A d 2 which is the sme result s found in prt (A). The cpcitnce is independent of the vlue of, so it does not mtter where the sl is locted. In Figure 26.23, when the centrl structure is moved up or down, the decrese in plte seprtion of one cpcitor is compensted y the increse in plte seprtion for the other.

141 798 Chpter 26 Cpcitnce nd Dielectrics Exmple 26.8 A Prtilly Filled Cpcitor A prllel-plte cpcitor with plte seprtion d hs cpcitnce C 0 in the sence of dielectric. Wht is the cpcitnce when sl of dielectric mteril of dielectric constnt k nd thickness fd is inserted etween the pltes (Fig ), where f is frction etween 0 nd 1? fd k fd k C 1 olution Conceptulize In our previous discussions of dielectrics etween the pltes of cpcitor, the dielectric filled the volume etween the pltes. In this exmple, only prt of the volume etween the pltes contins the dielectric mteril. Ctegorize In Exmple 26.7, we found tht n infinitesimlly thin metllic sheet inserted etween the pltes of cpcitor does not ffect the cpcitnce. Imgine sliding n infinitesimlly thin metllic sl long the ottom fce of the dielectric shown in Figure We cn model this system s series comintion of two cpcitors s shown (1 f )d (1 f )d in Figure One cpcitor hs plte seprtion fd nd is filled with dielectric; the other hs plte seprtion (1 2 f )d nd hs ir etween its pltes. Anlyze Evlute the two cpcitnces in Figure from Eqution 26.15: Find the equivlent cpcitnce C from Eqution for two cpcitors comined in series: Invert nd sustitute for the cpcitnce without the dielectric, C 0 5 P 0 A/d: d Figure (Exmple 26.8) () A prllel-plte cpcitor of plte seprtion d prtilly filled with dielectric of thickness fd. () The equivlent circuit of the cpcitor consists of two cpcitors connected in series. C 1 5 kp 0A fd nd C 2 5 P 0A 11 2 f 2d 1 C fd 11 2 f 2d 1 C 1 C 2 kp 0 A P 0 A 1 C 5 fd k11 2 f 2d 1 5 f 1 k11 2 f 2 d kp 0 A kp 0 A k P 0 A k P 0 A C 5 f 1 k 11 2 f 2 d 5 k f 1 k 11 2 f 2 C 0 Finlize Let s test this result for some known limits. If f 0, the dielectric should dispper. In this limit, C C 0, which is consistent with cpcitor with ir etween the pltes. If f 1, the dielectric fills the volume etween the pltes. In this limit, C kc 0, which is consistent with Eqution ummry Definitions C 2 A cpcitor consists of two conductors crrying chrges of equl mgnitude nd opposite sign. The cpcitnce C of ny cpcitor is the rtio of the chrge Q on either conductor to the potentil difference DV etween them: C ; Q (26.1) DV The cpcitnce depends only on the geometry of the conductors nd not on n externl source of chrge or potentil difference. The I unit of cpcitnce is couloms per volt, or the frd (F): 1 F 5 1 C/V. The electric dipole moment p of n electric dipole hs mgnitude p ; 2q (26.16) where 2 is the distnce etween the chrges q nd 2q. The direction of the electric dipole moment vector is from the negtive chrge towrd the positive chrge.

142 Ojective Questions 799 Concepts nd Principles If two or more cpcitors re connected in prllel, the potentil difference is the sme cross ll cpcitors. The equivlent cpcitnce of prllel comintion of cpcitors is C eq 5 C 1 1 C 2 1 C (26.8) If two or more cpcitors re connected in series, the chrge is the sme on ll cpcitors, nd the equivlent cpcitnce of the series comintion is given y C eq C 1 C 2 C c (26.10) 3 These two equtions enle you to simplify mny electric circuits y replcing multiple cpcitors with single equivlent cpcitnce. When dielectric mteril is inserted etween the pltes of cpcitor, the cpcitnce increses y dimensionless fctor k, clled the dielectric constnt: C 5 kc 0 (26.14) where C 0 is the cpcitnce in the sence of the dielectric. Ojective Questions 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. A fully chrged prllel-plte cpcitor remins connected to ttery while you slide dielectric etween the pltes. Do the following quntities () increse, () decrese, or (c) sty the sme? (i) C (ii) Q (iii) DV (iv) the energy stored in the cpcitor 2. By wht fctor is the cpcitnce of metl sphere multiplied if its volume is tripled? () 3 () 3 1/3 (c) 1 (d) 3 21/3 (e) An electronics technicin wishes to construct prllel-plte cpcitor using rutile (k 5 100) s the dielectric. The re of the pltes is 1.00 cm 2. Wht is the cpcitnce if the rutile thickness is 1.00 mm? () 88.5 pf () 177 pf (c) 8.85 mf (d) 100 mf (e) 35.4 mf 4. A prllel-plte cpcitor is connected to ttery. Wht hppens to the stored energy if the plte seprtion is douled while the cpcitor remins connected to the ttery? () It remins the sme. () It is douled. (c) It decreses y fctor of 2. (d) It decreses y fctor of 4. (e) It increses y fctor of If three unequl cpcitors, initilly unchrged, re connected in series cross ttery, which of the following sttements is true? () The equivlent cpcitnce is greter thn ny of the individul cpcitnces. () The lrgest voltge ppers cross the smllest cpcitnce. (c) The lrgest voltge ppers cross the lrgest cpcitnce. (d) The cpcitor with the lrgest cpcitnce hs the gretest chrge. (e) The cpcitor with the smllest cpcitnce hs the smllest chrge. Energy is stored in chrged cpcitor ecuse the chrging process is equivlent to the trnsfer of chrges from one conductor t lower electric potentil to nother conductor t higher potentil. The energy stored in cpcitor of cpcitnce C with chrge Q nd potentil difference DV is U E 5 Q 2 2C 5 1 2Q DV 5 1 2C 1DV 2 2 (26.11) The torque cting on n electric dipole in uniform electric field E is t 5 p 3 E (26.18) The potentil energy of the system of n electric dipole in uniform externl electric field E is U E 5 2p? E (26.20) 6. Assume device is designed to otin lrge potentil difference y first chrging nk of cpcitors connected in prllel nd then ctivting switch rrngement tht in effect disconnects the cpcitors from the chrging source nd from ech other nd reconnects them ll in series rrngement. The group of chrged cpcitors is then dischrged in series. Wht is the mximum potentil difference tht cn e otined in this mnner y using ten 500-mF cpcitors nd n 800-V chrging source? () 500 V () 8.00 kv (c) 400 kv (d) 800 V (e) 0 7. (i) Wht hppens to the mgnitude of the chrge on ech plte of cpcitor if the potentil difference etween the conductors is douled? () It ecomes four times lrger. () It ecomes two times lrger. (c) It is unchnged. (d) It ecomes one-hlf s lrge. (e) It ecomes one-fourth s lrge. (ii) If the potentil difference cross cpcitor is douled, wht hppens to the energy stored? Choose from the sme possiilities s in prt (i). 8. A cpcitor with very lrge cpcitnce is in series with nother cpcitor with very smll cpcitnce. Wht is the equivlent cpcitnce of the comintion? () slightly greter thn the cpcitnce of the lrge cpcitor () slightly less thn the cpcitnce of the lrge cpcitor (c) slightly greter thn the cpcitnce of the smll cpcitor (d) slightly less thn the cpcitnce of the smll cpcitor

143 800 Chpter 26 Cpcitnce nd Dielectrics 9. A prllel-plte cpcitor filled with ir crries chrge Q. The ttery is disconnected, nd sl of mteril with dielectric constnt k 5 2 is inserted etween the pltes. Which of the following sttements is true? () The voltge cross the cpcitor decreses y fctor of 2. () The voltge cross the cpcitor is douled. (c) The chrge on the pltes is douled. (d) The chrge on the pltes decreses y fctor of 2. (e) The electric field is douled. 10. (i) A ttery is ttched to severl different cpcitors connected in prllel. Which of the following sttements is true? () All cpcitors hve the sme chrge, nd the equivlent cpcitnce is greter thn the cpcitnce of ny of the cpcitors in the group. () The cpcitor with the lrgest cpcitnce crries the smllest chrge. (c) The potentil difference cross ech cpcitor is the sme, nd the equivlent cpcitnce is greter thn ny of the cpcitors in the group. (d) The cpcitor with the smllest cpcitnce crries the lrgest chrge. (e) The potentil differences cross the cpcitors re the sme only if the cpcitnces re the sme. (ii) The cpcitors re reconnected in series, nd the comintion is gin connected to the ttery. From the sme choices, choose the one tht is true. 11. A prllel-plte cpcitor is chrged nd then is disconnected from the ttery. By wht fctor does the stored energy chnge when the plte seprtion is then douled? () It ecomes four times lrger. () It Conceptul Questions 1. () Why is it dngerous to touch the terminls of high-voltge cpcitor even fter the voltge source tht chrged the cpcitor is disconnected from the cpcitor? () Wht cn e done to mke the cpcitor sfe to hndle fter the voltge source hs een removed? 2. Assume you wnt to increse the mximum operting voltge of prllel-plte cpcitor. Descrie how you cn do tht with fixed plte seprtion. 3. If you were sked to design cpcitor in which smll size nd lrge cpcitnce were required, wht would e the two most importnt fctors in your design? 4. Explin why dielectric increses the mximum operting voltge of cpcitor even though the physicl size of the cpcitor doesn t chnge. Prolems 1. denotes nswer ville in tudent olutions Mnul/tudy Guide ecomes two times lrger. (c) It stys the sme. (d) It ecomes one-hlf s lrge. (e) It ecomes one-fourth s lrge. 12. (i) Rnk the following five cpcitors from gretest to smllest cpcitnce, noting ny cses of equlity. () 20-mF cpcitor with 4-V potentil difference etween its pltes () 30-mF cpcitor with chrges of mgnitude 90 mc on ech plte (c) cpcitor with chrges of mgnitude 80 mc on its pltes, differing y 2 V in potentil, (d) 10-mF cpcitor storing energy 125 mj (e) cpcitor storing energy 250 mj with 10-V potentil difference (ii) Rnk the sme cpcitors in prt (i) from lrgest to smllest ccording to the potentil difference etween the pltes. (iii) Rnk the cpcitors in prt (i) in the order of the mgnitudes of the chrges on their pltes. (iv) Rnk the cpcitors in prt (i) in the order of the energy they store. 13. True or Flse? () From the definition of cpcitnce C 5 Q /DV, it follows tht n unchrged cpcitor hs cpcitnce of zero. () As descried y the definition of cpcitnce, the potentil difference cross n unchrged cpcitor is zero. 14. You chrge prllel-plte cpcitor, remove it from the ttery, nd prevent the wires connected to the pltes from touching ech other. When you increse the plte seprtion, do the following quntities () increse, () decrese, or (c) sty the sme? (i) C (ii) Q (iii) E etween the pltes (iv) DV 5. Explin why the work needed to move prticle with chrge Q through potentil difference DV is W 5 Q DV, wheres the energy stored in chrged cpcitor is U E 5 1 2Q DV. Where does the fctor 1 2 come from? 6. An ir-filled cpcitor is chrged, then disconnected from the power supply, nd finlly connected to voltmeter. Explin how nd why the potentil difference chnges when dielectric is inserted etween the pltes of the cpcitor. 7. The sum of the chrges on oth pltes of cpcitor is zero. Wht does cpcitor store? 8. Becuse the chrges on the pltes of prllel-plte cpcitor re opposite in sign, they ttrct ech other. Hence, it would tke positive work to increse the plte seprtion. Wht type of energy in the system chnges due to the externl work done in this process? The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign BIO Q/C

144 Prolems 801 ection 26.1 Definition of Cpcitnce 1. () When ttery is connected to the pltes of 3.00-mF cpcitor, it stores chrge of 27.0 mc. Wht is the voltge of the ttery? () If the sme cpcitor is connected to nother ttery nd 36.0 mc of chrge is stored on the cpcitor, wht is the voltge of the ttery? 2. Two conductors hving net chrges of mc nd W mc hve potentil difference of 10.0 V etween them. () Determine the cpcitnce of the system. () Wht is the potentil difference etween the two conductors if the chrges on ech re incresed to 1100 mc nd 2100 mc? 3. () How much chrge is on ech plte of 4.00-mF W cpcitor when it is connected to 12.0-V ttery? () If this sme cpcitor is connected to 1.50-V ttery, wht chrge is stored? ection 26.2 Clculting Cpcitnce 4. An ir-filled sphericl cpcitor is constructed with M inner- nd outer-shell rdii of 7.00 cm nd 14.0 cm, respectively. () Clculte the cpcitnce of the device. () Wht potentil difference etween the spheres results in 4.00-mC chrge on the cpcitor? 5. A 50.0-m length of coxil cle hs n inner conductor tht hs dimeter of 2.58 mm nd crries M chrge of 8.10 mc. The surrounding conductor hs n inner dimeter of 7.27 mm nd chrge of mc. Assume the region etween the conductors is ir. () Wht is the cpcitnce of this cle? () Wht is the potentil difference etween the two conductors? 6. () Regrding the Erth nd cloud lyer 800 m W ove the Erth s the pltes of cpcitor, clculte the cpcitnce of the Erth cloud lyer system. Assume the cloud lyer hs n re of 1.00 km 2 nd the ir etween the cloud nd the ground is pure nd dry. Assume chrge uilds up on the cloud nd on the ground until uniform electric field of N/C throughout the spce etween them mkes the ir rek down nd conduct electricity s lightning olt. () Wht is the mximum chrge the cloud cn hold? 7. When potentil difference of 150 V is pplied to the pltes of prllel-plte cpcitor, the pltes crry surfce chrge density of 30.0 nc/cm 2. Wht is the spcing etween the pltes? 8. An ir-filled prllel-plte cpcitor hs pltes of re 2.30 cm 2 seprted y 1.50 mm. () Find the vlue of its cpcitnce. The cpcitor is connected to 12.0-V ttery. () Wht is the chrge on the cpcitor? (c) Wht is the mgnitude of the uniform electric field etween the pltes? 9. An ir-filled cpcitor consists of two prllel pltes, M ech with n re of 7.60 cm 2, seprted y distnce of 1.80 mm. A 20.0-V potentil difference is pplied to these pltes. Clculte () the electric field etween the pltes, () the surfce chrge density, (c) the cpcitnce, nd (d) the chrge on ech plte. 10. A vrile ir cpcitor used in rdio tuning circuit is mde of N semicirculr pltes, ech of rdius R nd positioned distnce d d from its neighors, to u which it is electriclly connected. As shown in Figure P26.10, second identicl R set of pltes is enmeshed with the first set. Ech plte in the second set is hlfwy Figure P26.10 etween two pltes of the first set. The second set cn rotte s unit. Determine the cpcitnce s function of the ngle of rottion u, where u 5 0 corresponds to the mximum cpcitnce. 11. An isolted, chrged conducting sphere of rdius 12.0 cm cretes n electric field of N/C t distnce 21.0 cm from its center. () Wht is its surfce chrge density? () Wht is its cpcitnce? 12. Review. A smll oject of mss m crries chrge q nd is suspended y thred etween the verticl pltes of prllel-plte cpcitor. The plte seprtion is d. If the thred mkes n ngle u with the verticl, wht is the potentil difference etween the pltes? ection 26.3 Comintions of Cpcitors 13. Two cpcitors, C mf nd C mf, re W connected in prllel, nd the resulting comintion is connected to 9.00-V ttery. Find () the equivlent cpcitnce of the comintion, () the potentil difference cross ech cpcitor, nd (c) the chrge stored on ech cpcitor. 14. Wht If? The two cpcitors of Prolem 13 (C mf W nd C mf) re now connected in series nd to 9.00-V ttery. Find () the equivlent cpcitnce of the comintion, () the potentil difference cross ech cpcitor, nd (c) the chrge on ech cpcitor. 15. Find the equivlent cpcitnce of 4.20-mF cpcitor nd n 8.50-mF cpcitor when they re connected () in series nd () in prllel. 16. Given 2.50-mF cpcitor, 6.25-mF cpcitor, nd 6.00-V ttery, find the chrge on ech cpcitor if you connect them () in series cross the ttery nd () in prllel cross the ttery. 17. According to its design specifiction, the timer circuit delying the closing of n elevtor door is to hve cpcitnce of 32.0 mf etween two points A nd B. When one circuit is eing constructed, the inexpensive ut durle cpcitor instlled etween these two points is found to hve cpcitnce 34.8 mf. To meet the specifiction, one dditionl cpcitor cn e plced etween the two points. () hould it e in series or in prllel with the 34.8-mF cpcitor? () Wht should e its cpcitnce? (c) Wht If? The next circuit comes down the ssemly line with cpcitnce 29.8 mf etween A nd B. To meet the specifiction, wht dditionl cpcitor should e instlled in series or in prllel in tht circuit?

145 802 Chpter 26 Cpcitnce nd Dielectrics 18. Why is the following sitution impossile? A technicin is testing circuit tht contins cpcitnce C. He relizes tht etter design for the circuit would include cpcitnce 7 3C rther thn C. He hs three dditionl cpcitors, ech with cpcitnce C. By comining these dditionl cpcitors in certin comintion tht is then plced in prllel with the originl cpcitor, he chieves the desired cpcitnce. 19. For the system of four cpcitors shown in Figure P26.19, find () the equivlent cpcitnce of the system, () the chrge on ech cpcitor, nd (c) the potentil difference cross ech cpcitor. 20. Three cpcitors re connected 90.0 V to ttery s shown Q/C in Figure P Their Figure P26.19 Prolems 19 nd 56. cpcitnces re C 1 5 3C, C 2 5 C, nd C 3 5 5C. () Wht is the equivlent cpcitnce C 1 of this set of cpcitors? () tte the rnking of the cpcitors ccording to the chrge they store from lrgest C 2 C 3 to smllest. (c) Rnk the cpcitors ccording to the potentil differences cross Figure P26.20 them from lrgest to smllest. (d) Wht If? Assume C 3 is incresed. Explin wht hppens to the chrge stored y ech cpcitor. 21. A group of identicl cpcitors is connected first in M series nd then in prllel. The comined cpcitnce in prllel is 100 times lrger thn for the series connection. How mny cpcitors re in the group? 22. () Find the equivlent cpcitnce C etween points nd for the 1 W C 1 group of cpcitors connected s shown in Figure P Tke C 1 5 C 2 C C mf, C mf, nd C mf. () Wht chrge is stored on C 3 if the potentil difference etween points nd is 60.0 V? C 2 C Four cpcitors re connected s M shown in Figure P () Find the equivlent cpcitnce etween Figure P26.22 points nd. () Clculte the chrge on ech cpcitor, tking DV V mf 3.00 mf 3.00 µ F 2.00 µ F 6.00 µ F 4.00 µ F is first chrged y closing switch 1. witch 1 is then opened, nd the chrged cpcitor is connected to the unchrged cpcitor y closing 2. Clculte () the initil chrge cquired y C 1 nd () the finl chrge on ech cpcitor. 25. Find the equivlent cpcitnce 7.0 mf etween points nd in the comintion of cpcitors shown in Figure P mf 6.0 mf 26. Find () the equivlent cpcitnce of the cpcitors in Figure P26.25 Figure P26.26, () the chrge on ech cpcitor, nd (c) the potentil difference cross ech cpcitor µ F 8.00 µ F 2.00 µ F 9.00 V Figure P µ F C 1 C Two cpcitors give n equivlent cpcitnce of 9.00 pf when connected in prllel nd n equivlent cpcitnce of 2.00 pf when connected in series. Wht is the cpcitnce of ech cpcitor? 28. Two cpcitors give n equivlent cpcitnce of C p when connected in prllel nd n equivlent cpcitnce of C s when connected in series. Wht is the cpcitnce of ech cpcitor? V 29. Consider three cpcitors C 1, C 2, nd C 3 nd ttery. If only C 1 is connected to the ttery, the chrge on C 1 is 30.8 mc. Now C 1 is disconnected, dischrged, nd connected in series with C 2. When the series comintion of C 2 nd C 1 is connected cross the ttery, the chrge on C 1 is 23.1 mc. The circuit is disconnected, nd oth cpcitors re dischrged. Next, C 3, C 1, nd the ttery re connected in series, resulting in chrge on C 1 of 25.2 mc. If, fter eing disconnected nd dischrged, C 1, C 2, nd C 3 re connected in series with one nother nd with the ttery, wht is the chrge on C 1? 1 Figure P mf mf 6.00 mf Figure P Consider the circuit shown in Figure P26.24, where C 1 5 M 6.00 mf, C mf, nd DV V. Cpcitor C 1 ection 26.4 Energy tored in Chrged Cpcitor 30. The immedite cuse of mny deths is ventriculr BIO firilltion, which is n uncoordinted quivering of the hert. An electric shock to the chest cn cuse momentry prlysis of the hert muscle, fter which the hert sometimes resumes its proper eting. One type of defirilltor (chpter- opening photo, pge 777) pplies strong electric shock to the chest over time intervl of few milliseconds. This device contins

146 Prolems 803 cpcitor of severl microfrds, chrged to severl thousnd volts. Electrodes clled pddles re held ginst the chest on oth sides of the hert, nd the cpcitor is dischrged through the ptient s chest. Assume n energy of 300 J is to e delivered from 30.0-mF cpcitor. To wht potentil difference must it e chrged? 31. A 12.0-V ttery is connected to cpcitor, resulting in 54.0 mc of chrge stored on the cpcitor. How much energy is stored in the cpcitor? 32. () A 3.00-mF cpcitor is connected to 12.0-V ttery. W How much energy is stored in the cpcitor? () Hd the cpcitor een connected to 6.00-V ttery, how much energy would hve een stored? 33. As person moves out in dry environment, electric chrge ccumultes on the person s ody. Once BIO it is t high voltge, either positive or negtive, the ody cn dischrge vi sprks nd shocks. Consider humn ody isolted from ground, with the typicl cpcitnce 150 pf. () Wht chrge on the ody will produce potentil of 10.0 kv? () ensitive electronic devices cn e destroyed y electrosttic dischrge from person. A prticulr device cn e destroyed y dischrge relesing n energy of 250 mj. To wht voltge on the ody does this sitution correspond? 34. Two cpcitors, C mf nd C mf, re connected in series, nd 12.0-V ttery is connected cross Q/C the two cpcitors. Find () the equivlent cpcitnce nd () the energy stored in this equivlent cpcitnce. (c) Find the energy stored in ech individul cpcitor. (d) how tht the sum of these two energies is the sme s the energy found in prt (). (e) Will this equlity lwys e true, or does it depend on the numer of cpcitors nd their cpcitnces? (f) If the sme cpcitors were connected in prllel, wht potentil difference would e required cross them so tht the comintion stores the sme energy s in prt ()? (g) Which cpcitor stores more energy in this sitution, C 1 or C 2? 35. Two identicl prllel-plte cpcitors, ech with Q/C cpcitnce 10.0 mf, re chrged to potentil difference 50.0 V nd then disconnected from the ttery. They re then connected to ech other in prllel with pltes of like sign connected. Finlly, the plte seprtion in one of the cpcitors is douled. () Find the totl energy of the system of two cpcitors efore the plte seprtion is douled. () Find the potentil difference cross ech cpcitor fter the plte seprtion is douled. (c) Find the totl energy of the system fter the plte seprtion is douled. (d) Reconcile the difference in the nswers to prts () nd (c) with the lw of conservtion of energy. 36. Two identicl prllel-plte cpcitors, ech with cpcitnce C, re chrged to potentil difference DV nd Q/C then disconnected from the ttery. They re then connected to ech other in prllel with pltes of like sign connected. Finlly, the plte seprtion in one of the cpcitors is douled. () Find the totl energy of the system of two cpcitors efore the plte seprtion is douled. () Find the potentil difference cross ech cpcitor fter the plte seprtion is douled. (c) Find the totl energy of the system fter the plte seprtion is douled. (d) Reconcile the difference in the nswers to prts () nd (c) with the lw of conservtion of energy. 37. Two cpcitors, C mf nd C mf, re connected in prllel nd chrged with 100-V power supply. () Drw circuit digrm nd () clculte the totl energy stored in the two cpcitors. (c) Wht If? Wht potentil difference would e required cross the sme two cpcitors connected in series for the comintion to store the sme mount of energy s in prt ()? (d) Drw circuit digrm of the circuit descried in prt (c). 38. A prllel-plte cpcitor hs chrge Q nd pltes of re A. Wht force cts on one plte to ttrct it towrd the other plte? Becuse the electric field etween the pltes is E 5 Q /AP 0, you might think the force is F 5 QE 5 Q 2 /AP 0. This conclusion is wrong ecuse the field E includes contriutions from oth pltes, nd the field creted y the positive plte cnnot exert ny force on the positive plte. how tht the force exerted on ech plte is ctully F 5 Q 2 /2AP 0. uggestion: Let C 5 P 0 A/x for n ritrry plte seprtion x nd note tht the work done in seprting the two chrged pltes is W 5 e F dx. 39. Review. A storm cloud nd the ground represent the AMT pltes of cpcitor. During storm, the cpcitor hs potentil difference of V etween its pltes nd chrge of 50.0 C. A lightning strike delivers 1.00% of the energy of the cpcitor to tree on the ground. How much sp in the tree cn e oiled wy? Model the sp s wter initilly t 30.08C. Wter hs specific het of J/kg? 8C, oiling point of 1008C, nd ltent het of vporiztion of J/kg. 40. Consider two conducting spheres with rdii R 1 nd GP R 2 seprted y distnce much greter thn either rdius. A totl chrge Q is shred etween the spheres. We wish to show tht when the electric potentil energy of the system hs minimum vlue, the potentil difference etween the spheres is zero. The totl chrge Q is equl to q 1 1 q 2, where q 1 represents the chrge on the first sphere nd q 2 the chrge on the second. Becuse the spheres re very fr prt, you cn ssume the chrge of ech is uniformly distriuted over its surfce. () how tht the energy ssocited with single conducting sphere of rdius R nd chrge q surrounded y vcuum is U 5 k e q 2 /2R. () Find the totl energy of the system of two spheres in terms of q 1, the totl chrge Q, nd the rdii R 1 nd R 2. (c) To minimize the energy, differentite the result to prt () with respect to q 1 nd set the derivtive equl to zero. olve for q 1 in terms of Q nd the rdii. (d) From the result to prt (c), find the chrge q 2. (e) Find the potentil of ech sphere. (f) Wht is the potentil difference etween the spheres? 41. Review. The circuit in Figure P26.41 (pge 804) consists of two identicl, prllel metl pltes connected to AMT identicl metl springs, switch, nd 100-V ttery.

147 804 Chpter 26 Cpcitnce nd Dielectrics With the switch open, the pltes re unchrged, re seprted y distnce d mm, nd hve cpcitnce C mf. When the switch is closed, the distnce etween the pltes decreses y fctor of () How much chrge collects on ech plte? () Wht is the spring constnt for ech spring? ection 26.5 Cpcitors with Dielectrics 42. A supermrket sells rolls of luminum foil, plstic wrp, nd wxed pper. () Descrie cpcitor mde from such mterils. Compute order-of-mgnitude estimtes for () its cpcitnce nd (c) its rekdown voltge. 43. () How much chrge cn e plced on cpcitor with W ir etween the pltes efore it reks down if the re of ech plte is 5.00 cm 2? () Wht If? Find the mximum chrge if polystyrene is used etween the pltes insted of ir. 44. The voltge cross n ir-filled prllel-plte cpcitor Q/C is mesured to e 85.0 V. When dielectric is inserted nd completely fills the spce etween the pltes s in Figure P26.44, the voltge drops to 25.0 V. () Wht is the dielectric constnt of the inserted mteril? () Cn you identify the dielectric? If so, wht is it? (c) If the dielectric does not completely fill the spce etween the pltes, wht could you conclude out the voltge cross the pltes? C 0 V 0 C Figure P26.44 V Dielectric 45. Determine () the cpcitnce nd () the mximum W potentil difference tht cn e pplied to Teflonfilled prllel-plte cpcitor hving plte re of 1.75 cm 2 nd plte seprtion of mm. 46. A commercil cpcitor is to e constructed s shown in Figure P This prticulr cpcitor is mde from two strips of luminum foil seprted y strip of prffin-coted pper. Ech strip of foil nd pper is 7.00 cm wide. The foil is mm thick, nd the pper is mm thick nd hs dielectric constnt of Wht length should the strips hve if cpci- k d V k Figure P26.41 tnce of F is desired efore the cpcitor is rolled up? (Adding second strip of pper nd rolling the cpcitor would effectively doule its cpcitnce y llowing chrge storge on oth sides of ech strip of foil.) Pper Aluminum Figure P cm 47. A prllel-plte cpcitor in ir hs plte seprtion of 1.50 cm nd plte re of 25.0 cm 2. The pltes re chrged to potentil difference of 250 V nd disconnected from the source. The cpcitor is then immersed in distilled wter. Assume the liquid is n insultor. Determine () the chrge on the pltes efore nd fter immersion, () the cpcitnce nd potentil difference fter immersion, nd (c) the chnge in energy of the cpcitor. 48. Ech cpcitor in the comintion shown in Figure P26.48 hs rekdown voltge of 15.0 V. Wht is the rekdown voltge of the comintion? 20.0 mf 20.0 mf 10.0 mf Figure P mf 20.0 mf 49. A 2.00-nF prllel-plte cpcitor is chrged to n initil potentil difference DV i V nd is then iso- AMT lted. The dielectric mteril etween the pltes is mic, with dielectric constnt of () How much work is required to withdrw the mic sheet? () Wht is the potentil difference cross the cpcitor fter the mic is withdrwn? ection 26.6 Electric Dipole in n Electric Field 50. A smll, rigid oject crries positive nd negtive M 3.50-nC chrges. It is oriented so tht the positive chrge hs coordintes (21.20 mm, 1.10 mm) nd the negtive chrge is t the point (1.40 mm, mm). () Find the electric dipole moment of the oject. The oject is plced in n electric field E i^ j^2 N/C. () Find the torque cting on the oject. (c) Find the potentil energy of the oject field system when the oject is in this orienttion. (d) Assuming the orienttion of the oject cn chnge, find the difference etween the mximum nd minimum potentil energies of the system. 51. An infinite line of positive chrge lies long the y xis, AMT with chrge density l mc/m. A dipole is plced

148 Prolems 805 with its center long the x xis t x cm. The dipole consists of two chrges mc seprted y 2.00 cm. The xis of the dipole mkes n ngle of with the x xis, nd the positive chrge is frther from the line of chrge thn the negtive chrge. Find the net force exerted on the dipole. 52. A smll oject with electric dipole moment p is plced in nonuniform electric field E 5 E1x2i^. Tht is, the field is in the x direction, nd its mgnitude depends only on the coordinte x. Let u represent the ngle etween the dipole moment nd the x direction. Prove tht the net force on the dipole is F 5 p de dx cos u cting in the direction of incresing field. ection 26.7 An Atomic Description of Dielectrics 53. The generl form of Guss s lw descries how chrge cretes n electric field in mteril, s well s in vcuum: 3 E? d A 5 q in P where P 5 kp 0 is the permittivity of the mteril. () A sheet with chrge Q uniformly distriuted over its re A is surrounded y dielectric. how tht the sheet cretes uniform electric field t nery points with mgnitude E 5 Q /2AP. () Two lrge sheets of re A, crrying opposite chrges of equl mgnitude Q, re smll distnce d prt. how tht they crete uniform electric field in the spce etween them with mgnitude E 5 Q /AP. (c) Assume the negtive plte is t zero potentil. how tht the positive plte is t potentil Qd/AP. (d) how tht the cpcitnce of the pir of pltes is given y C 5 AP/d 5 kap 0 /d. Additionl Prolems 54. Find the equivlent cpcitnce of the group of cpcitors shown in Figure P µ F 6.00 µ F 5.00 µ F 3.00 µ F 2.00 µ F 3.00 µ F 7.00 µ F Figure P Four prllel metl pltes P 1, P 2, P 3, nd P 4, ech of re 7.50 cm 2, re seprted successively y distnce d mm s shown in Figure P Plte P 1 is connected to the negtive terminl of ttery, nd P 2 is connected to the positive terminl. The ttery mintins potentil difference of 12.0 V. () If P 3 is connected to the negtive terminl, wht is the cpcitnce of the three-plte system P 1 P 2 P 3? () Wht is the chrge on P 2? (c) If P 4 is now connected to the positive terminl, wht is the cpcitnce of the four-plte system P 1 P 2 P 3 P 4? (d) Wht is the chrge on P 4? 12.0 V P 1 d d d Figure P26.55 P 2 P 3 P For the system of four cpcitors shown in Figure Q/C P26.19, find () the totl energy stored in the system nd () the energy stored y ech cpcitor. (c) Compre the sum of the nswers in prt () with your result to prt () nd explin your oservtion. 57. A uniform electric field E V/m exists within certin region. Wht volume of spce contins n energy equl to J? Express your nswer in cuic meters nd in liters. 58. Two lrge, prllel metl pltes, ech of re A, re oriented horizontlly nd seprted y distnce 3d. A grounded conducting wire joins them, nd initilly ech plte crries no chrge. Now third identicl plte crrying chrge Q is inserted etween the two pltes, prllel to them nd locted distnce d from the upper plte s shown in Figure P () Wht induced chrge ppers on ech of the two originl pltes? () Wht potentil difference ppers etween the middle plte nd ech of the other pltes? Figure P A prllel-plte cpcitor is constructed using M dielectric mteril whose dielectric constnt is 3.00 nd whose dielectric strength is V/m. The desired cpcitnce is mf, nd the cpcitor must withstnd mximum potentil difference of 4.00 kv. Find the minimum re of the cpcitor pltes. 60. Why is the following sitution impossile? A 10.0-mF cpcitor hs pltes with vcuum etween them. The cpcitor is chrged so tht it stores J of energy. A prticle with chrge mc is fired from the positive plte towrd the negtive plte with n initil kinetic energy equl to J. The prticle rrives t the negtive plte with reduced kinetic energy. d 2d

149 806 Chpter 26 Cpcitnce nd Dielectrics 61. A model of red lood cell portrys the cell s cpcitor with two sphericl pltes. It is positively chrged BIO conducting liquid sphere of re A, seprted y n insulting memrne of thickness t from the surrounding negtively chrged conducting fluid. Tiny electrodes introduced into the cell show potentil difference of 100 mv cross the memrne. Tke the memrne s thickness s 100 nm nd its dielectric constnt s () Assume tht typicl red lood cell hs mss of kg nd density kg/m 3. Clculte its volume nd its surfce re. () Find the cpcitnce of the cell. (c) Clculte the chrge on the surfces of the memrne. How mny electronic chrges does this chrge represent? 62. A prllel-plte cpcitor with vcuum etween its Q/C horizontl pltes hs cpcitnce of 25.0 mf. A nonconducting liquid with dielectric constnt 6.50 is poured into the spce etween the pltes, filling up frction f of its volume. () Find the new cpcitnce s function of f. () Wht should you expect the cpcitnce to e when f 5 0? Does your expression from prt () gree with your nswer? (c) Wht cpcitnce should you expect when f 5 1? Does the expression from prt () gree with your nswer? 63. A 10.0-mF cpcitor is chrged 5.00 mf 10.0 mf to 15.0 V. It is next connected in series with n unchrged 5.00-mF V i 15.0 V cpcitor. The series comintion is finlly connected cross 50.0-V ttery s digrmmed in Figure P Find the new 50.0 V potentil differences cross the 5.00-mF nd 10.0-mF cpcitors Figure P26.63 fter the switch is thrown closed. 64. Assume tht the internl dimeter of the Geiger Mueller tue descried in Prolem 68 in Chpter 25 is 2.50 cm nd tht the wire long the xis hs dimeter of mm. The dielectric strength of the gs etween the centrl wire nd the cylinder is V/m. Use the result of tht prolem to clculte the mximum potentil difference tht cn e pplied etween the wire nd the cylinder efore rekdown occurs in the gs. 65. Two squre pltes of sides, re plced prllel to Q/C ech other with seprtion d s suggested in Figure P You my ssume d is much less thn,. The pltes crry uniformly distriuted sttic chrges 1Q 0 nd 2Q 0. A lock of metl hs width,, length,, nd thickness slightly less thn d. It is inserted distnce x into the spce etween the pltes. The chrges on the pltes remin uniformly distriuted s the lock slides in. In sttic sitution, metl prevents n electric field from penetrting inside it. The metl cn e thought of s perfect dielectric, with k `. () Clculte the stored energy in the system s function of x. () Find the direction nd mgnitude of the force tht cts on the metllic lock. (c) The re of the dvncing front fce of the lock is essentilly equl to,d. Considering the force on the lock s cting on this fce, find the stress (force per re) on it. (d) Express the energy density in the electric field etween the chrged pltes in terms of Q 0,,, d, nd P 0. (e) Explin how the nswers to prts (c) nd (d) compre with ech other. Q 0 x d Figure P26.65 Q () Two spheres hve rdii nd, nd their centers re distnce d prt. how tht the cpcitnce of this system is 4pP 0 C d provided d is lrge compred with nd. uggestion: Becuse the spheres re fr prt, ssume the potentil of ech equls the sum of the potentils due to ech sphere. () how tht s d pproches infinity, the ove result reduces to tht of two sphericl cpcitors in series. 67. A cpcitor of unknown cpcitnce hs een chrged to potentil difference of 100 V nd then disconnected from the ttery. When the chrged cpcitor is then connected in prllel to n unchrged 10.0-mF cpcitor, the potentil difference cross the comintion is 30.0 V. Clculte the unknown cpcitnce. 68. A prllel-plte cpcitor of plte seprtion d is Q/C chrged to potentil difference DV 0. A dielectric sl of thickness d nd dielectric constnt k is introduced etween the pltes while the ttery remins connected to the pltes. () how tht the rtio of energy stored fter the dielectric is introduced to the energy stored in the empty cpcitor is U/U 0 5 k. () Give physicl explntion for this increse in stored energy. (c) Wht hppens to the chrge on the cpcitor? Note: This sitution is not the sme s in Exmple 26.5, in which the ttery ws removed from the circuit efore the dielectric ws introduced. 69. Cpcitors C mf nd C mf re chrged s prllel comintion cross 250-V ttery. The cpcitors re disconnected from the ttery nd from ech other. They re then connected positive plte to negtive plte nd negtive plte to positive plte. Clculte the resulting chrge on ech cpcitor. 70. Exmple 26.1 explored cylindricl cpcitor of length, with rdii nd for the two conductors. In the Wht If? section of tht exmple, it ws climed tht incresing, y 10% is more effective in terms of incresing the cpcitnce thn incresing y 10% if Verify this clim mthemticlly. 71. To repir power supply for stereo mplifier, n electronics technicin needs 100-mF cpcitor cple of Q/C withstnding potentil difference of 90 V etween the

150 Prolems 807 pltes. The immeditely ville supply is ox of five 100-mF cpcitors, ech hving mximum voltge cpility of 50 V. () Wht comintion of these cpcitors hs the proper electricl chrcteristics? Will the technicin use ll the cpcitors in the ox? Explin your nswers. () In the comintion of cpcitors otined in prt (), wht will e the mximum voltge cross ech of the cpcitors used? Chllenge Prolems 72. The inner conductor of coxil cle hs rdius of mm, nd the outer conductor s inside rdius is 3.00 mm. The spce etween the conductors is filled with polyethylene, which hs dielectric constnt of 2.30 nd dielectric strength of V/m. Wht is the mximum potentil difference this cle cn withstnd? 73. ome physicl systems possessing cpcitnce continuously distriuted over spce cn e modeled s n infi- nite rry of discrete circuit elements. Exmples re microwve wveguide nd the xon of nerve cell. To prctice nlysis of n infinite rry, determine the equivlent cpcitnce C etween terminls X nd Y of the infinite set of cpcitors represented in Figure P Ech cpcitor hs cpcitnce C 0. uggestions: Imgine tht the ldder is cut t the line AB nd note tht the equivlent cpcitnce of the infinite section to the right of AB is lso C. X Y C 0 C 0 C 0 A B Figure P Consider two long, prllel, nd oppositely chrged wires of rdius r with their centers seprted y distnce D tht is much lrger thn r. Assuming the chrge is distriuted uniformly on the surfce of ech wire, show tht the cpcitnce per unit length of this pir of wires is C, 5 pp 0 ln 1D/r2 75. Determine the equivlent cpcitnce of the comintion shown in Figure P uggestion: Consider the symmetry involved. 76. A prllel-plte cpcitor with pltes of re LW nd Q/C plte seprtion t hs the region etween its pltes filled with wedges of two dielectric mterils s shown in Figure P Assume t is much less thn oth L nd W. () Determine its cpcitnce. () hould the cpcitnce e the sme if the lels k 1 nd k 2 re interchnged? Demonstrte tht your expression does or does not hve this property. (c) how tht if k 1 nd k 2 pproch equlity to common vlue k, your result ecomes the sme s the cpcitnce of cpcitor contining single dielectric: C 5 kp 0 LW/t. t k 1 L k 2 W Figure P Clculte the equivlent cpcitnce etween points nd in Figure P Notice tht this system is not simple series or prllel comintion. uggestion: Assume potentil difference DV etween points nd. Write expressions for DV in terms of the chrges nd cpcitnces for the vrious possile pthwys from to nd require conservtion of chrge for those cpcitor pltes tht re connected to ech other mf 2.00 mf 8.00 mf 4.00 mf 2.00 mf Figure P A cpcitor is constructed from two squre, metllic pltes of sides, nd seprtion d. Chrges 1Q nd 2Q re plced on the pltes, nd the power supply is then removed. A mteril of dielectric constnt k is inserted distnce x into the cpcitor s shown in Figure P Assume d is much smller thn x. () Find the equivlent cpcitnce of the device. () Clculte the energy stored in the cpcitor. (c) Find the direction nd mgnitude of the force exerted y the pltes on the dielectric. (d) Otin numericl vlue for the force when x 5,/2, ssuming, cm, d mm, the dielectric is glss (k ), nd the cpcitor ws chrged to V efore the dielectric ws inserted. uggestion: The system cn e considered s two cpcitors connected in prllel. C 3C 2C k x d Q C 2C Figure P26.75 Figure P26.78 Q

151 chpter 27 Current nd Resistnce 27.1 Electric Current 27.2 Resistnce 27.3 A Model for Electricl Conduction 27.4 Resistnce nd Temperture 27.5 uperconductors 27.6 Electricl Power These two lightuls provide similr power output y visile light (electromgnetic rdition). The compct fluorescent ul on the left, however, produces this light output with fr less input y electricl trnsmission thn the incndescent ul on the right. The fluorescent ul, therefore, is less costly to operte nd sves vlule resources needed to generte electricity. (Christin Richrds/ hutterstock.com) We now consider situtions involving electric chrges tht re in motion through some region of spce. We use the term electric current, or simply current, to descrie the rte of flow of chrge. Most prcticl pplictions of electricity del with electric currents, including vriety of home pplinces. For exmple, the voltge from wll plug produces current in the coils of toster when it is turned on. In these common situtions, current exists in conductor such s copper wire. Currents cn lso exist outside conductor. For instnce, em of electrons in prticle ccelertor constitutes current. This chpter egins with the definition of current. A microscopic description of current is given, nd some fctors tht contriute to the opposition to the flow of chrge in conductors re discussed. A clssicl model is used to descrie electricl conduction in metls, nd some limittions of this model re cited. We lso define electricl resistnce nd introduce new circuit element, the resistor. We conclude y discussing the rte t which energy is trnsferred to device in n electric circuit. The energy trnsfer mechnism in Eqution 8.2 tht corresponds to this process is electricl trnsmission T ET Electric Current 808 In this section, we study the flow of electric chrges through piece of mteril. The mount of flow depends on oth the mteril through which the chrges re

152 27.1 Electric Current 809 pssing nd the potentil difference cross the mteril. Whenever there is net flow of chrge through some region, n electric current is sid to exist. It is instructive to drw n nlogy etween wter flow nd current. The flow of wter in pluming pipe cn e quntified y specifying the mount of wter tht emerges from fucet during given time intervl, often mesured in liters per minute. A river current cn e chrcterized y descriing the rte t which the wter flows pst prticulr loction. For exmple, the flow over the rink t Nigr Flls is mintined t rtes etween m 3 /s nd m 3 /s. There is lso n nlogy etween therml conduction nd current. In ection 20.7, we discussed the flow of energy y het through smple of mteril. The rte of energy flow is determined y the mteril s well s the temperture difference cross the mteril s descried y Eqution To define current quntittively, suppose chrges re moving perpendiculr to surfce of re A s shown in Figure (This re could e the cross-sectionl re of wire, for exmple.) The current is defined s the rte t which chrge flows through this surfce. If DQ is the mount of chrge tht psses through this surfce in time intervl Dt, the verge current I vg is equl to the chrge tht psses through A per unit time: I vg 5 DQ Dt (27.1) If the rte t which chrge flows vries in time, the current vries in time; we define the instntneous current I s the limit of the verge current s Dt 0: The I unit of current is the mpere (A): I ; dq dt (27.2) 1 A 5 1 C/s (27.3) Tht is, 1 A of current is equivlent to 1 C of chrge pssing through surfce in 1 s. The chrged prticles pssing through the surfce in Figure 27.1 cn e positive, negtive, or oth. It is conventionl to ssign to the current the sme direction s the flow of positive chrge. In electricl conductors such s copper or luminum, the current results from the motion of negtively chrged electrons. Therefore, in n ordinry conductor, the direction of the current is opposite the direction of flow of electrons. For em of positively chrged protons in n ccelertor, however, the current is in the direction of motion of the protons. In some cses such s those involving gses nd electrolytes, for instnce the current is the result of the flow of oth positive nd negtive chrges. It is common to refer to moving chrge (positive or negtive) s moile chrge crrier. If the ends of conducting wire re connected to form loop, ll points on the loop re t the sme electric potentil; hence, the electric field is zero within nd t the surfce of the conductor. Becuse the electric field is zero, there is no net trnsport of chrge through the wire; therefore, there is no current. If the ends of the conducting wire re connected to ttery, however, ll points on the loop re not t the sme potentil. The ttery sets up potentil difference etween the ends of the loop, creting n electric field within the wire. The electric field exerts forces on the electrons in the wire, cusing them to move in the wire nd therefore creting current. Microscopic Model of Current We cn relte current to the motion of the chrge crriers y descriing microscopic model of conduction in metl. Consider the current in cylindricl A WWElectric current The direction of the current is the direction in which positive chrges flow when free to do so. Figure 27.1 Chrges in motion through n re A. The time rte t which chrge flows through the re is defined s the current I. Pitfll Prevention 27.1 Current Flow Is Redundnt The phrse current flow is commonly used, lthough it is techniclly incorrect ecuse current is flow (of chrge). This wording is similr to the phrse het trnsfer, which is lso redundnt ecuse het is trnsfer (of energy). We will void this phrse nd spek of flow of chrge or chrge flow. Pitfll Prevention 27.2 Btteries Do Not upply Electrons A ttery does not supply electrons to the circuit. It estlishes the electric field tht exerts force on electrons lredy in the wires nd elements of the circuit. I

153 810 Chpter 27 Current nd Resistnce q v d x t v d Figure 27.2 A segment of uniform conductor of cross-sectionl re A. E The rndom motion of the chrge crriers is modified y the field, nd they hve drift velocity opposite the direction of the electric field. A v d Figure 27.3 () A schemtic digrm of the rndom motion of two chrge crriers in conductor in the sence of n electric field. The drift velocity is zero. () The motion of the chrge crriers in conductor in the presence of n electric field. Becuse of the ccelertion of the chrge crriers due to the electric force, the pths re ctully prolic. The drift speed, however, is much smller thn the verge speed, so the prolic shpe is not visile on this scle. conductor of cross-sectionl re A (Fig. 27.2). The volume of segment of the conductor of length Dx (etween the two circulr cross sections shown in Fig. 27.2) is A Dx. If n represents the numer of moile chrge crriers per unit volume (in other words, the chrge crrier density), the numer of crriers in the segment is na Dx. Therefore, the totl chrge DQ in this segment is DQ 5 (na Dx)q where q is the chrge on ech crrier. If the crriers move with velocity v d prllel to the xis of the cylinder, the mgnitude of the displcement they experience in the x direction in time intervl Dt is Dx 5 v d Dt. Let Dt e the time intervl required for the chrge crriers in the segment to move through displcement whose mgnitude is equl to the length of the segment. This time intervl is lso the sme s tht required for ll the chrge crriers in the segment to pss through the circulr re t one end. With this choice, we cn write DQ s DQ 5 (nav d Dt)q Dividing oth sides of this eqution y Dt, we find tht the verge current in the conductor is I vg 5 DQ Dt 5 nqv da (27.4) In relity, the speed of the chrge crriers v d is n verge speed clled the drift speed. To understnd the mening of drift speed, consider conductor in which the chrge crriers re free electrons. If the conductor is isolted tht is, the potentil difference cross it is zero these electrons undergo rndom motion tht is nlogous to the motion of gs molecules. The electrons collide repetedly with the metl toms, nd their resultnt motion is complicted nd zigzgged s in Figure As discussed erlier, when potentil difference is pplied cross the conductor (for exmple, y mens of ttery), n electric field is set up in the conductor; this field exerts n electric force on the electrons, producing current. In ddition to the zigzg motion due to the collisions with the metl toms, the electrons move slowly long the conductor (in direction opposite tht of E ) t the drift velocity v d s shown in Figure You cn think of the tom electron collisions in conductor s n effective internl friction (or drg force) similr to tht experienced y liquid s molecules flowing through pipe stuffed with steel wool. The energy trnsferred from the electrons to the metl toms during collisions cuses n increse in the tom s virtionl energy nd corresponding increse in the conductor s temperture. Q uick Quiz 27.1 Consider positive nd negtive chrges moving horizontlly through the four regions shown in Figure Rnk the current in these four regions from highest to lowest. c d Figure 27.4 (Quick Quiz 27.1) Chrges move through four regions.

154 27.2 Resistnce 811 Exmple 27.1 Drift peed in Copper Wire The 12-guge copper wire in typicl residentil uilding hs cross-sectionl re of m 2. It crries constnt current of 10.0 A. Wht is the drift speed of the electrons in the wire? Assume ech copper tom contriutes one free electron to the current. The density of copper is 8.92 g/cm 3. olution Conceptulize Imgine electrons following zigzg motion such s tht in Figure 27.3, with drift velocity prllel to the wire superimposed on the motion s in Figure As mentioned erlier, the drift speed is smll, nd this exmple helps us quntify the speed. Ctegorize We evlute the drift speed using Eqution Becuse the current is constnt, the verge current during ny time intervl is the sme s the constnt current: I vg 5 I. Anlyze The periodic tle of the elements in Appendix C shows tht the molr mss of copper is M g/mol. Recll tht 1 mol of ny sustnce contins Avogdro s numer of toms (N A mol 21 ). Use the molr mss nd the density of copper to find the volume of 1 mole of copper: From the ssumption tht ech copper tom contriutes one free electron to the current, find the electron density in copper: olve Eqution 27.4 for the drift speed nd sustitute for the electron density: V 5 M r n 5 N A V 5 N Ar M v d 5 I vg nqa 5 I nqa 5 IM qan A r A kg/mol2 ustitute numericl vlues: v d C m mol kg/m m/s Finlize This result shows tht typicl drift speeds re very smll. For instnce, electrons trveling with speed of m/s would tke out 75 min to trvel 1 m! You might therefore wonder why light turns on lmost instntneously when its switch is thrown. In conductor, chnges in the electric field tht drives the free electrons ccording to the prticle in field model trvel through the conductor with speed close to tht of light. o, when you flip on light switch, electrons lredy in the filment of the lightul experience electric forces nd egin moving fter time intervl on the order of nnoseconds Resistnce In ection 24.4, we rgued tht the electric field inside conductor is zero. This sttement is true, however, only if the conductor is in sttic equilirium s stted in tht discussion. The purpose of this section is to descrie wht hppens when there is nonzero electric field in the conductor. As we sw in ection 27.1, current exists in the wire in this cse. Consider conductor of cross-sectionl re A crrying current I. The current density J in the conductor is defined s the current per unit re. Becuse the current I 5 nqv d A, the current density is J ; I A 5 nqv d (27.5) WWCurrent density

155 812 Chpter 27 Current nd Resistnce Georg imon Ohm Germn physicist ( ) Ohm, high school techer nd lter professor t the University of Munich, formulted the concept of resistnce nd discovered the proportionlities expressed in Equtions 27.6 nd V A. Bettmnn/Coris A potentil difference V V V mintined cross the conductor sets up n electric field E, nd this field produces current I tht is proportionl to the potentil difference. Figure 27.5 A uniform conductor of length, nd cross-sectionl re A. Pitfll Prevention 27.3 Eqution 27.7 Is Not Ohm s Lw Mny individuls cll Eqution 27.7 Ohm s lw, ut tht is incorrect. This eqution is simply the definition of resistnce, nd it provides n importnt reltionship etween voltge, current, nd resistnce. Ohm s lw is relted to proportionlity of J to E (Eq. 27.6) or, equivlently, of I to DV, which, from Eqution 27.7, indictes tht the resistnce is constnt, independent of the pplied voltge. We will see some devices for which Eqution 27.7 correctly descries their resistnce, ut tht E I V where J hs I units of mperes per meter squred. This expression is vlid only if the current density is uniform nd only if the surfce of cross-sectionl re A is perpendiculr to the direction of the current. A current density nd n electric field re estlished in conductor whenever potentil difference is mintined cross the conductor. In some mterils, the current density is proportionl to the electric field: J 5 se (27.6) where the constnt of proportionlity s is clled the conductivity of the conductor. 1 Mterils tht oey Eqution 27.6 re sid to follow Ohm s lw, nmed fter Georg imon Ohm. More specificlly, Ohm s lw sttes the following: For mny mterils (including most metls), the rtio of the current density to the electric field is constnt s tht is independent of the electric field producing the current. Mterils nd devices tht oey Ohm s lw nd hence demonstrte this simple reltionship etween E nd J re sid to e ohmic. Experimentlly, however, it is found tht not ll mterils nd devices hve this property. Those tht do not oey Ohm s lw re sid to e nonohmic. Ohm s lw is not fundmentl lw of nture; rther, it is n empiricl reltionship vlid only for certin situtions. We cn otin n eqution useful in prcticl pplictions y considering segment of stright wire of uniform cross-sectionl re A nd length, s shown in Figure A potentil difference DV 5 V 2 V is mintined cross the wire, creting in the wire n electric field nd current. If the field is ssumed to e uniform, the mgnitude of the potentil difference cross the wire is relted to the field within the wire through Eqution 25.6, DV 5 E, Therefore, we cn express the current density (Eq. 27.6) in the wire s J 5 s DV, Becuse J 5 I/A, the potentil difference cross the wire is DV 5, s J 5, sa I 5 R I The quntity R 5,/sA is clled the resistnce of the conductor. We define the resistnce s the rtio of the potentil difference cross conductor to the current in the conductor: R ; DV (27.7) I We will use this eqution gin nd gin when studying electric circuits. This result shows tht resistnce hs I units of volts per mpere. One volt per mpere is defined to e one ohm (V): 1 V ; 1 V/A (27.8) Eqution 27.7 shows tht if potentil difference of 1 V cross conductor cuses current of 1 A, the resistnce of the conductor is 1 V. For exmple, if n electricl pplince connected to 120-V source of potentil difference crries current of 6 A, its resistnce is 20 V. Most electric circuits use circuit elements clled resistors to control the current in the vrious prts of the circuit. As with cpcitors in Chpter 26, mny resistors re uilt into integrted circuit chips, ut stnd-lone resistors re still ville nd do not oey Ohm s lw. 1 Do not confuse conductivity s with surfce chrge density, for which the sme symol is used.

156 27.2 Resistnce 813 Tle 27.1 Color Coding for Resistors Color Numer Multiplier Tolernce Blck 0 1 Brown Red Ornge Yellow Green Blue Violet Gry White Gold % ilver % Colorless 20% widely used. Two common types re the composition resistor, which contins cron, nd the wire-wound resistor, which consists of coil of wire. Vlues of resistors in ohms re normlly indicted y color coding s shown in Figure 27.6 nd Tle The first two colors on resistor give the first two digits in the resistnce vlue, with the deciml plce to the right of the second digit. The third color represents the power of 10 for the multiplier of the resistnce vlue. The lst color is the tolernce of the resistnce vlue. As n exmple, the four colors on the resistor t the ottom of Figure 27.6 re yellow (5 4), violet (5 7), lck ( ), nd gold (5 5%), nd so the resistnce vlue is V with tolernce vlue of 5% 5 2 V. The inverse of conductivity is resistivity 2 r: r 5 1 s (27.9) where r hs the units ohm? meters (V? m). Becuse R 5,/sA, we cn express the resistnce of uniform lock of mteril long the length, s R 5 r, A (27.10) Every ohmic mteril hs chrcteristic resistivity tht depends on the properties of the mteril nd on temperture. In ddition, s you cn see from Eqution 27.10, the resistnce of smple of the mteril depends on the geometry of the smple s well s on the resistivity of the mteril. Tle 27.2 (pge 814) gives the resistivities of vriety of mterils t 20 C. Notice the enormous rnge, from very low vlues for good conductors such s copper nd silver to very high vlues for good insultors such s glss nd ruer. An idel conductor would hve zero resistivity, nd n idel insultor would hve infinite resistivity. Eqution shows tht the resistnce of given cylindricl conductor such s wire is proportionl to its length nd inversely proportionl to its cross-sectionl re. If the length of wire is douled, its resistnce doules. If its cross-sectionl re is douled, its resistnce decreses y one hlf. The sitution is nlogous to the flow of liquid through pipe. As the pipe s length is incresed, the resistnce to flow increses. As the pipe s cross-sectionl re is incresed, more liquid crosses given cross section of the pipe per unit time intervl. Therefore, more liquid flows for the sme pressure differentil pplied to the pipe, nd the resistnce to flow decreses. Ohmic mterils nd devices hve liner current potentil difference reltionship over rod rnge of pplied potentil differences (Fig. 27.7, pge 814). The slope of the I-versus-DV curve in the liner region yields vlue for 1/R. Nonohmic The colored nds on this resistor re yellow, violet, lck, nd gold. Figure 27.6 A close-up view of circuit ord shows the color coding on resistor. The gold nd on the left tells us tht the resistor is oriented ckwrd in this view nd we need to red the colors from right to left. W WResistivity is the inverse of conductivity W WResistnce of uniform mteril long the length < dexns/hutterstock.com Pitfll Prevention 27.4 Resistnce nd Resistivity Resistivity is property of sustnce, wheres resistnce is property of n oject. We hve seen similr pirs of vriles efore. For exmple, density is property of sustnce, wheres mss is property of n oject. Eqution reltes resistnce to resistivity, nd Eqution 1.1 reltes mss to density. 2 Do not confuse resistivity r with mss density or chrge density, for which the sme symol is used.

157 814 Chpter 27 Current nd Resistnce I I Figure 27.7 () The current potentil difference curve for n ohmic mteril. The curve is liner, nd the slope is equl to the inverse of the resistnce of the conductor. () A nonliner current potentil difference curve for junction diode. This device does not oey Ohm s lw. Exmple 27.2 Tle 27.2 Resistivities nd Temperture Coefficients of Resistivity for Vrious Mterils Temperture Mteril Resistivity (V? m) Coefficient [( C) 21 ] ilver Copper Gold Aluminum Tungsten Iron Pltinum Led Nichrome c Cron Germnium ilicon d Glss to Hrd ruer, ulfur Qurtz (fused) All vlues t 20 C. All elements in this tle re ssumed to e free of impurities. ee ection c A nickel chromium lloy commonly used in heting elements. The resistivity of Nichrome vries with composition nd rnges etween nd V? m. d The resistivity of silicon is very sensitive to purity. The vlue cn e chnged y severl orders of mgnitude when it is doped with other toms. mterils hve nonliner current potentil difference reltionship. One common semiconducting device with nonliner I-versus-DV chrcteristics is the junction diode (Fig. 27.7). The resistnce of this device is low for currents in one direction (positive DV) nd high for currents in the reverse direction (negtive DV). In fct, most modern electronic devices, such s trnsistors, hve nonliner current potentil difference reltionships; their proper opertion depends on the prticulr wy they violte Ohm s lw. Q uick Quiz 27.2 A cylindricl wire hs rdius r nd length,. If oth r nd, re douled, does the resistnce of the wire () increse, () decrese, or (c) remin the sme? Q uick Quiz 27.3 In Figure 27.7, s the pplied voltge increses, does the resistnce of the diode () increse, () decrese, or (c) remin the sme? The Resistnce of Nichrome Wire The rdius of 22-guge Nichrome wire is 0.32 mm. (A) Clculte the resistnce per unit length of this wire. olution lope = 1 R V V Conceptulize Tle 27.2 shows tht Nichrome hs resistivity two orders of mgnitude lrger thn the est conductors in the tle. Therefore, we expect it to hve some specil prcticl pplictions tht the est conductors my not hve. Ctegorize We model the wire s cylinder so tht simple geometric nlysis cn e pplied to find the resistnce. Anlyze Use Eqution nd the resistivity of Nichrome from Tle 27.2 to find the resistnce per unit length: R, 5 r A 5 r pr V # m V/m 2 p m2

158 27.2 Resistnce continued (B) If potentil difference of 10 V is mintined cross 1.0-m length of the Nichrome wire, wht is the current in the wire? olution Anlyze Use Eqution 27.7 to find the current: I 5 DV R 5 DV 1R/,2, 5 10 V 13.1 V/m m A Finlize Becuse of its high resistivity nd resistnce to oxidtion, Nichrome is often used for heting elements in tosters, irons, nd electric heters. Wht If? Wht if the wire were composed of copper insted of Nichrome? How would the vlues of the resistnce per unit length nd the current chnge? Answer Tle 27.2 shows us tht copper hs resistivity two orders of mgnitude smller thn tht for Nichrome. Therefore, we expect the nswer to prt (A) to e smller nd the nswer to prt (B) to e lrger. Clcultions show tht copper wire of the sme rdius would hve resistnce per unit length of only V/m. A 1.0-m length of copper wire of the sme rdius would crry current of 190 A with n pplied potentil difference of 10 V. Exmple 27.3 The Rdil Resistnce of Coxil Cle Coxil cles re used extensively for cle television nd other electronic pplictions. A coxil cle consists of two concentric cylindricl conductors. The region etween the conductors is completely filled with polyethylene plstic s shown in Figure Current lekge through the plstic, in the rdil direction, is unwnted. (The cle is designed to conduct current long its length, ut tht is not the current eing considered here.) The rdius of the inner conductor is cm, the rdius of the outer conductor is cm, nd the length is L cm. The resistivity of the plstic is V? m. Clculte the resistnce of the plstic etween the two conductors. olution Conceptulize Imgine two currents s suggested in the text of the prolem. The desired current is long the cle, crried within the conductors. The undesired current corresponds to lekge through the plstic, nd its direction is rdil. Ctegorize Becuse the resistivity nd the geometry of the plstic re known, we ctegorize this prolem s one in which we find the resistnce of the plstic from these prmeters. Eqution 27.10, however, represents the resistnce of lock of mteril. We hve more complicted geometry in this sitution. Becuse the re through which the chrges pss depends on the rdil position, we must use integrl clculus to determine the nswer. Polyethylene Inner conductor Outer conductor dr r End view L Current direction Anlyze We divide the plstic into concentric cylindricl shells of infinitesiml thickness dr (Fig. 27.8). Any chrge pssing from the inner to the outer conductor must move rdilly through this shell. Use differentil form of Eqution 27.10, replcing, with dr for the length vrile: dr 5 r dr/a, where dr is the resistnce of shell of plstic of thickness dr nd surfce re A. Figure 27.8 (Exmple 27.3) A coxil cle. () Polyethylene plstic fills the gp etween the two conductors. () End view, showing current lekge. Write n expression for the resistnce of our hollow cylindricl shell of plstic representing the re s the surfce re of the shell: dr 5 r dr A 5 r 2prL dr continued

159 816 Chpter 27 Current nd Resistnce 27.3 continued Integrte this expression from r 5 to r 5 : (1) R 5 3 dr 5 r 2pL 3 dr r 5 r 2pL ln ustitute the vlues given: R V # m 2p m cm ln cm V Finlize Let s compre this resistnce to tht of the inner copper conductor of the cle long the 15.0-cm length. Use Eqution to find the resistnce of the copper cylinder: Wht If? uppose the coxil cle is enlrged to twice the overll dimeter with two possile choices: (1) the rtio / is held fixed, or (2) the difference 2 is held fixed. For which choice does the lekge current etween the inner nd outer conductors increse when the voltge is pplied etween them? Answer For the current to increse, the resistnce must decrese. For choice (1), in which / is held fixed, Equ- R Cu 5 r, A V # m m2 c p m2 d V This resistnce is 18 orders of mgnitude smller thn the rdil resistnce. Therefore, lmost ll the current corresponds to chrge moving long the length of the cle, with very smll frction leking in the rdil direction. tion (1) shows tht the resistnce is unffected. For choice (2), we do not hve n eqution involving the difference 2 to inspect. Looking t Figure 27.8, however, we see tht incresing nd while holding the difference constnt results in chrge flowing through the sme thickness of plstic ut through lrger re perpendiculr to the flow. This lrger re results in lower resistnce nd higher current A Model for Electricl Conduction In this section, we descrie structurl model of electricl conduction in metls tht ws first proposed y Pul Drude ( ) in (ee ection 21.1 for review of structurl models.) This model leds to Ohm s lw nd shows tht resistivity cn e relted to the motion of electrons in metls. Although the Drude model descried here hs limittions, it introduces concepts tht re pplied in more elorte tretments. Following the outline of structurl models from ection 21.1, the Drude model for electricl conduction hs the following properties: 1. Physicl components: Consider conductor s regulr rry of toms plus collection of free electrons, which re sometimes clled conduction electrons. We identify the system s the comintion of the toms nd the conduction electrons. The conduction electrons, lthough ound to their respective toms when the toms re not prt of solid, ecome free when the toms condense into solid. 2. Behvior of the components: () In the sence of n electric field, the conduction electrons move in rndom directions through the conductor (Fig. 27.3). The sitution is similr to the motion of gs molecules confined in vessel. In fct, some scientists refer to conduction electrons in metl s n electron gs. () When n electric field is pplied to the system, the free electrons drift slowly in direction opposite tht of the electric field (Fig. 27.3), with n verge drift speed v d tht is much smller (typiclly m/s) thn their verge speed v vg etween collisions (typiclly 10 6 m/s). (c) The electron s motion fter collision is independent of its motion efore the collision. The excess energy cquired y the electrons due to

160 27.3 A Model for Electricl Conduction 817 the work done on them y the electric field is trnsferred to the toms of the conductor when the electrons nd toms collide. With regrd to property 2(c) ove, the energy trnsferred to the toms cuses the internl energy of the system nd, therefore, the temperture of the conductor to increse. We re now in position to derive n expression for the drift velocity, using severl of our nlysis models. When free electron of mss m e nd chrge q (5 2e) is sujected to n electric field E, it is descried y the prticle in field model nd experiences force F 5 q E. The electron is prticle under net force, nd its ccelertion cn e found from Newton s second lw, g F 5 m : 5 F m 5 q E (27.11) m e Becuse the electric field is uniform, the electron s ccelertion is constnt, so the electron cn e modeled s prticle under constnt ccelertion. If v i is the electron s initil velocity the instnt fter collision (which occurs t time defined s t 5 0), the velocity of the electron t very short time t lter (immeditely efore the next collision occurs) is, from Eqution 4.8, v f 5 v i 1 t 5 v i 1 q E t (27.12) m e Let s now tke the verge vlue of vf for ll the electrons in the wire over ll possile collision times t nd ll possile vlues of vi. Assuming the initil velocities re rndomly distriuted over ll possile directions (property 2() ove), the verge vlue of vi is zero. The verge vlue of the second term of Eqution is 1q E /m e 2t, where t is the verge time intervl etween successive collisions. Becuse the verge vlue of vf is equl to the drift velocity, vf,vg 5 vd 5 q E t (27.13) m e The vlue of t depends on the size of the metl toms nd the numer of electrons per unit volume. We cn relte this expression for drift velocity in Eqution to the current in the conductor. ustituting the mgnitude of the velocity from Eqution into Eqution 27.4, the verge current in the conductor is given y I vg 5 nq qe ta 5 nq 2 E ta (27.14) m e m e Becuse the current density J is the current divided y the re A, J 5 nq 2 E t where n is the numer of electrons per unit volume. Compring this expression with Ohm s lw, J 5 se, we otin the following reltionships for conductivity nd resistivity of conductor: m e s 5 nq 2 t m e (27.15) W WDrift velocity in terms of microscopic quntities W WCurrent density in terms of microscopic quntities W WConductivity in terms of microscopic quntities r 5 1 s 5 m e (27.16) nq 2 t According to this clssicl model, conductivity nd resistivity do not depend on the strength of the electric field. This feture is chrcteristic of conductor oeying Ohm s lw. W WResistivity in terms of microscopic quntities

161 818 Chpter 27 Current nd Resistnce The model shows tht the resistivity cn e clculted from knowledge of the density of the electrons, their chrge nd mss, nd the verge time intervl t etween collisions. This time intervl is relted to the verge distnce etween collisions / vg (the men free pth) nd the verge speed v vg through the expression 3 t 5, vg v vg (27.17) Although this structurl model of conduction is consistent with Ohm s lw, it does not correctly predict the vlues of resistivity or the ehvior of the resistivity with temperture. For exmple, the results of clssicl clcultions for v vg using the idel gs model for the electrons re out fctor of ten smller thn the ctul vlues, which results in incorrect predictions of vlues of resistivity from Eqution Furthermore, ccording to Equtions nd 27.17, the resistivity is predicted to vry with temperture s does v vg, which, ccording to n idel-gs model (Chpter 21, Eq ), is proportionl to "T. This ehvior is in disgreement with the experimentlly oserved liner dependence of resistivity with temperture for pure metls. (ee ection 27.4.) Becuse of these incorrect predictions, we must modify our structurl model. We shll cll the model tht we hve developed so fr the clssicl model for electricl conduction. To ccount for the incorrect predictions of the clssicl model, we develop it further into quntum mechnicl model, which we shll descrie riefly. We discussed two importnt simplifiction models in erlier chpters, the prticle model nd the wve model. Although we discussed these two simplifiction models seprtely, quntum physics tells us tht this seprtion is not so cler-cut. As we shll discuss in detil in Chpter 40, prticles hve wve-like properties. The predictions of some models cn only e mtched to experimentl results if the model includes the wve-like ehvior of prticles. The structurl model for electricl conduction in metls is one of these cses. Let us imgine tht the electrons moving through the metl hve wve-like properties. If the rry of toms in conductor is regulrly spced (tht is, periodic), the wve-like chrcter of the electrons mkes it possile for them to move freely through the conductor nd collision with n tom is unlikely. For n idelized conductor, no collisions would occur, the men free pth would e infinite, nd the resistivity would e zero. Electrons re scttered only if the tomic rrngement is irregulr (not periodic), s result of structurl defects or impurities, for exmple. At low tempertures, the resistivity of metls is dominted y scttering cused y collisions etween the electrons nd impurities. At high tempertures, the resistivity is dominted y scttering cused y collisions etween the electrons nd the toms of the conductor, which re continuously displced s result of therml gittion, destroying the perfect periodicity. The therml motion of the toms mkes the structure irregulr (compred with n tomic rry t rest), therey reducing the electron s men free pth. Although it is eyond the scope of this text to show this modifiction in detil, the clssicl model modified with the wve-like chrcter of the electrons results in predictions of resistivity vlues tht re in greement with mesured vlues nd predicts liner temperture dependence. Quntum notions hd to e introduced in Chpter 21 to understnd the temperture ehvior of molr specific hets of gses. Here we hve nother cse in which quntum physics is necessry for the model to gree with experiment. Although clssicl physics cn explin tremendous rnge of phenomen, we continue to see hints tht quntum physics must e incorported into our models. We shll study quntum physics in detil in Chpters 40 through Recll tht the verge speed of group of prticles depends on the temperture of the group (Chpter 21) nd is not the sme s the drift speed v d.

162 27.5 uperconductors Resistnce nd Temperture Over limited temperture rnge, the resistivity of conductor vries pproximtely linerly with temperture ccording to the expression r 5 r 0 [1 1 (T 2 T 0 )] (27.18) where r is the resistivity t some temperture T (in degrees Celsius), r 0 is the resistivity t some reference temperture T 0 (usully tken to e 20 C), nd is the temperture coefficient of resistivity. From Eqution 27.18, the temperture coefficient of resistivity cn e expressed s 5 1 r 0 Dr DT (27.19) where Dr 5 r 2 r 0 is the chnge in resistivity in the temperture intervl DT 5 T 2 T 0. The temperture coefficients of resistivity for vrious mterils re given in Tle Notice tht the unit for is degrees Celsius 21 [( C) 21 ]. Becuse resistnce is proportionl to resistivity (Eq ), the vrition of resistnce of smple is R 5 R 0 [1 1 (T 2 T 0 )] (27.20) where R 0 is the resistnce t temperture T 0. Use of this property enles precise temperture mesurements through creful monitoring of the resistnce of proe mde from prticulr mteril. For some metls such s copper, resistivity is nerly proportionl to temperture s shown in Figure A nonliner region lwys exists t very low tempertures, however, nd the resistivity usully reches some finite vlue s the temperture pproches solute zero. This residul resistivity ner solute zero is cused primrily y the collision of electrons with impurities nd imperfections in the metl. In contrst, high-temperture resistivity (the liner region) is predominntly chrcterized y collisions etween electrons nd metl toms. Notice tht three of the vlues in Tle 27.2 re negtive, indicting tht the resistivity of these mterils decreses with incresing temperture. This ehvior is indictive of clss of mterils clled semiconductors, first introduced in ection 23.2, nd is due to n increse in the density of chrge crriers t higher tempertures. Becuse the chrge crriers in semiconductor re often ssocited with impurity toms (s we discuss in more detil in Chpter 43), the resistivity of these mterils is very sensitive to the type nd concentrtion of such impurities. Q uick Quiz 27.4 When does n incndescent lightul crry more current, () immeditely fter it is turned on nd the glow of the metl filment is incresing or () fter it hs een on for few milliseconds nd the glow is stedy? WVrition W of r with temperture W Wtemperture coefficient of resistivity 0 r 0 r As T pproches solute zero, the resistivity pproches nonzero vlue. Figure 27.9 Resistivity versus temperture for metl such s copper. The curve is liner over wide rnge of tempertures, nd r increses with incresing temperture. R ( ) T The resistnce drops discontinuously to zero t T c, which is 4.15 K for mercury uperconductors There is clss of metls nd compounds whose resistnce decreses to zero when they re elow certin temperture T c, known s the criticl temperture. These mterils re known s superconductors. The resistnce temperture grph for superconductor follows tht of norml metl t tempertures ove T c (Fig ). When the temperture is t or elow T c, the resistivity drops suddenly to zero. This phenomenon ws discovered in 1911 y Dutch physicist Heike Kmerlingh-Onnes ( ) s he worked with mercury, which is superconductor elow 4.2 K. Mesurements hve shown tht the resistivities of superconductors elow their T c vlues re less thn V? m, or pproximtely times smller thn the resistivity of copper. In prctice, these resistivities re considered to e zero T c T T (K) 4.4 Figure Resistnce versus temperture for smple of mercury (Hg). The grph follows tht of norml metl ove the criticl temperture T c.

163 820 Chpter 27 Current nd Resistnce A smll permnent mgnet levitted ove disk of the superconductor YB 2 Cu 3 O 7, which is in liquid nitrogen t 77 K. The direction of the effective flow of positive chrge is clockwise. V I Figure A circuit consisting of resistor of resistnce R nd ttery hving potentil difference DV cross its terminls. R c d Courtesy of IBM Reserch Lortory Tody, thousnds of superconductors re known, nd s Tle 27.3 illustrtes, the criticl tempertures of recently discovered superconductors re sustntilly higher thn initilly thought possile. Two kinds of superconductors re recognized. The more recently identified ones re essentilly cermics with high criticl tempertures, wheres superconducting mterils such s those oserved y Kmerlingh-Onnes re metls. If room-temperture superconductor is ever identified, its effect on technology could e tremendous. The vlue of T c is sensitive to chemicl composition, pressure, nd moleculr structure. Copper, silver, nd gold, which re excellent conductors, do not exhiit superconductivity. One truly remrkle feture of superconductors is tht once current is set up in them, it persists without ny pplied potentil difference (ecuse R 5 0). tedy currents hve een oserved to persist in superconducting loops for severl yers with no pprent decy! An importnt nd useful ppliction of superconductivity is in the development of superconducting mgnets, in which the mgnitudes of the mgnetic field re pproximtely ten times greter thn those produced y the est norml electromgnets. uch superconducting mgnets re eing considered s mens of storing energy. uperconducting mgnets re currently used in medicl mgnetic resonnce imging, or MRI, units, which produce high-qulity imges of internl orgns without the need for excessive exposure of ptients to x-rys or other hrmful rdition Electricl Power Tle 27.3 Criticl Tempertures for Vrious uperconductors Mteril T c (K) HgB 2 C 2 Cu 3 O Tl B C Cu O 125 Bi r C Cu O 105 YB 2 Cu 3 O 7 92 N 3 Ge 23.2 N 3 n N 9.46 P 7.18 Hg 4.15 n 3.72 Al 1.19 Zn 0.88 In typicl electric circuits, energy T ET is trnsferred y electricl trnsmission from source such s ttery to some device such s lightul or rdio receiver. Let s determine n expression tht will llow us to clculte the rte of this energy trnsfer. First, consider the simple circuit in Figure 27.11, where energy is delivered to resistor. (Resistors re designted y the circuit symol.) Becuse the connecting wires lso hve resistnce, some energy is delivered to the wires nd some to the resistor. Unless noted otherwise, we shll ssume the resistnce of the wires is smll compred with the resistnce of the circuit element so tht the energy delivered to the wires is negligile. Imgine following positive quntity of chrge Q moving clockwise round the circuit in Figure from point through the ttery nd resistor ck to point. We identify the entire circuit s our system. As the chrge moves from to through the ttery, the electric potentil energy of the system increses y n mount Q DV

164 27.6 Electricl Power 821 while the chemicl potentil energy in the ttery decreses y the sme mount. (Recll from Eq tht DU 5 q DV.) As the chrge moves from c to d through the resistor, however, the electric potentil energy of the system decreses due to collisions of electrons with toms in the resistor. In this process, the electric potentil energy is trnsformed to internl energy corresponding to incresed virtionl motion of the toms in the resistor. Becuse the resistnce of the interconnecting wires is neglected, no energy trnsformtion occurs for pths c nd d. When the chrge returns to point, the net result is tht some of the chemicl potentil energy in the ttery hs een delivered to the resistor nd resides in the resistor s internl energy E int ssocited with moleculr virtion. The resistor is normlly in contct with ir, so its incresed temperture results in trnsfer of energy y het Q into the ir. In ddition, the resistor emits therml rdition T ER, representing nother mens of escpe for the energy. After some time intervl hs pssed, the resistor reches constnt temperture. At this time, the input of energy from the ttery is lnced y the output of energy from the resistor y het nd rdition, nd the resistor is nonisolted system in stedy stte. ome electricl devices include het sinks 4 connected to prts of the circuit to prevent these prts from reching dngerously high tempertures. Het sinks re pieces of metl with mny fins. Becuse the metl s high therml conductivity provides rpid trnsfer of energy y het wy from the hot component nd the lrge numer of fins provides lrge surfce re in contct with the ir, energy cn trnsfer y rdition nd into the ir y het t high rte. Let s now investigte the rte t which the electric potentil energy of the system decreses s the chrge Q psses through the resistor: du dt 5 d dq 1Q DV 2 5 DV 5 I DV dt dt where I is the current in the circuit. The system regins this potentil energy when the chrge psses through the ttery, t the expense of chemicl energy in the ttery. The rte t which the potentil energy of the system decreses s the chrge psses through the resistor is equl to the rte t which the system gins internl energy in the resistor. Therefore, the power P, representing the rte t which energy is delivered to the resistor, is P 5 I DV (27.21) We derived this result y considering ttery delivering energy to resistor. Eqution 27.21, however, cn e used to clculte the power delivered y voltge source to ny device crrying current I nd hving potentil difference DV etween its terminls. Using Eqution nd DV 5 IR for resistor, we cn express the power delivered to the resistor in the lterntive forms P 5 I 2 R 5 1DV 2 2 R (27.22) When I is expressed in mperes, DV in volts, nd R in ohms, the I unit of power is the wtt, s it ws in Chpter 8 in our discussion of mechnicl power. The process y which energy is trnsformed to internl energy in conductor of resistnce R is often clled joule heting; 5 this trnsformtion is lso often referred to s n I 2 R loss. Pitfll Prevention 27.5 Chrges Do Not Move All the Wy Around Circuit in hort Time In terms of understnding the energy trnsfer in circuit, it is useful to imgine chrge moving ll the wy round the circuit even though it would tke hours to do so. Pitfll Prevention 27.6 Misconceptions Aout Current everl common misconceptions re ssocited with current in circuit like tht in Figure One is tht current comes out of one terminl of the ttery nd is then used up s it psses through the resistor, leving current in only one prt of the circuit. The current is ctully the sme everywhere in the circuit. A relted misconception hs the current coming out of the resistor eing smller thn tht going in ecuse some of the current is used up. Yet nother misconception hs current coming out of oth terminls of the ttery, in opposite directions, nd then clshing in the resistor, delivering the energy in this mnner. Tht is not the cse; chrges flow in the sme rottionl sense t ll points in the circuit. Pitfll Prevention 27.7 Energy Is Not Dissipted In some ooks, you my see Eqution descried s the power dissipted in resistor, suggesting tht energy disppers. Insted, we sy energy is delivered to resistor. 4 This usge is nother misuse of the word het tht is ingrined in our common lnguge. 5 It is commonly clled joule heting even though the process of het does not occur when energy delivered to resistor ppers s internl energy. It is nother exmple of incorrect usge of the word het tht hs ecome entrenched in our lnguge.

165 822 Chpter 27 Current nd Resistnce Figure These power lines trnsfer energy from the electric compny to homes nd usinesses. The energy is trnsferred t very high voltge, possily hundreds of thousnds of volts in some cses. Even though it mkes power lines very dngerous, the high voltge results in less loss of energy due to resistnce in the wires. Lester Lefkowitz/Txi/Getty Imges e c 30 W 60 W V Figure (Quick Quiz 27.5) Two lightuls connected cross the sme potentil difference. Exmple 27.4 When trnsporting energy y electricity through power lines (Fig ), you should not ssume the lines hve zero resistnce. Rel power lines do indeed hve resistnce, nd power is delivered to the resistnce of these wires. Utility compnies seek to minimize the energy trnsformed to internl energy in the lines nd mximize the energy delivered to the consumer. Becuse P 5 I DV, the sme mount of energy cn e trnsported either t high currents nd low potentil differences or t low currents nd high potentil differences. Utility compnies choose to trnsport energy t low currents nd high potentil differences primrily for economic resons. Copper wire is very expensive, so it is cheper to use high-resistnce wire (tht is, wire hving smll cross-sectionl re; see Eq ). Therefore, in the expression for the power delivered to resistor, P 5 I 2 R, the resistnce of the wire is fixed t reltively high vlue for economic considertions. The I 2 R loss cn e reduced y keeping the current I s low s possile, which mens trnsferring the energy t high voltge. In some instnces, power is trnsported t potentil differences s gret s 765 kv. At the destintion of the energy, the potentil difference is usully reduced to 4 kv y device clled trnsformer. Another trnsformer drops the potentil difference to 240 V for use in your home. Of course, ech time the potentil difference decreses, the current increses y the sme fctor nd the power remins the sme. We shll discuss trnsformers in greter detil in Chpter 33. Q uick Quiz 27.5 For the two lightuls shown in Figure 27.13, rnk the current vlues t points through f from gretest to lest. Power in n Electric Heter An electric heter is constructed y pplying potentil difference of 120 V cross Nichrome wire tht hs totl resistnce of 8.00 V. Find the current crried y the wire nd the power rting of the heter. olution f d Conceptulize As discussed in Exmple 27.2, Nichrome wire hs high resistivity nd is often used for heting elements in tosters, irons, nd electric heters. Therefore, we expect the power delivered to the wire to e reltively high. Ctegorize We evlute the power from Eqution 27.22, so we ctegorize this exmple s sustitution prolem. Use Eqution 27.7 to find the current in the wire: I 5 DV R V 8.00 V A Find the power rting using the expression P 5 I 2 R from Eqution 27.22: P 5 I 2 R A V W kw Wht If? Wht if the heter were ccidentlly connected to 240-V supply? (Tht is difficult to do ecuse the shpe nd orienttion of the metl contcts in 240-V plugs re different from those in 120-V plugs.) How would tht ffect the current crried y the heter nd the power rting of the heter, ssuming the resistnce remins constnt? Answer If the pplied potentil difference were douled, Eqution 27.7 shows tht the current would doule. According to Eqution 27.22, P 5 (DV) 2 /R, the power would e four times lrger.

166 ummry 823 Exmple 27.5 Linking Electricity nd Thermodynmics AM An immersion heter must increse the temperture of 1.50 kg of wter from 10.0 C to 50.0 C in 10.0 min while operting t 110 V. (A) Wht is the required resistnce of the heter? olution Conceptulize An immersion heter is resistor tht is inserted into continer of wter. As energy is delivered to the immersion heter, rising its temperture, energy leves the surfce of the resistor y het, going into the wter. When the immersion heter reches constnt temperture, the rte of energy delivered to the resistnce y electricl trnsmission (T ET ) is equl to the rte of energy delivered y het (Q ) to the wter. Ctegorize This exmple llows us to link our new understnding of power in electricity with our experience with specific het in thermodynmics (Chpter 20). The wter is nonisolted system. Its internl energy is rising ecuse of energy trnsferred into the wter y het from the resistor, so Eqution 8.2 reduces to DE int 5 Q. In our model, we ssume the energy tht enters the wter from the heter remins in the wter. Anlyze To simplify the nlysis, let s ignore the initil period during which the temperture of the resistor increses nd lso ignore ny vrition of resistnce with temperture. Therefore, we imgine constnt rte of energy trnsfer for the entire 10.0 min. et the rte of energy delivered to the resistor equl P 5 1DV Q R Dt to the rte of energy Q entering the wter y het: Use Eqution 20.4, Q 5 mc DT, to relte the energy input y het to the resulting temperture chnge of the wter nd solve for the resistnce: ustitute the vlues given in the sttement of the prolem: (B) Estimte the cost of heting the wter. olution Multiply the power y the time intervl to find the mount of energy trnsferred to the resistor: Find the cost knowing tht energy is purchsed t n estimted price of 11. per kilowtt-hour: 1DV 2 2 mc DT 5 R Dt R 5 R 5 1DV 2 2 Dt mc DT 1110 V s kg J/kg # 8C C C V T ET 5 P Dt 5 1DV V22 Dt 5 R 28.9 V min2 1 h 60.0 min Wh kwh Cost 5 ( kwh)($0.11/kwh) 5 $ Finlize The cost to het the wter is very low, less thn one cent. In relity, the cost is higher ecuse some energy is trnsferred from the wter into the surroundings y het nd electromgnetic rdition while its temperture is incresing. If you hve electricl devices in your home with power rtings on them, use this power rting nd n pproximte time intervl of use to estimte the cost for one use of the device. ummry Definitions The electric current I in conductor is defined s I ; dq (27.2) dt where dq is the chrge tht psses through cross section of the conductor in time intervl dt. The I unit of current is the mpere (A), where 1 A 5 1 C/s. continued

167 824 Chpter 27 Current nd Resistnce The current density J in conductor is the current per unit re: J ; I A (27.5) The resistnce R of conductor is defined s R ; DV (27.7) I where DV is the potentil difference cross the conductor nd I is the current it crries. The I unit of resistnce is volts per mpere, which is defined to e 1 ohm (V); tht is, 1 V 5 1 V/A. Concepts nd Principles The verge current in conductor is relted to the motion of the chrge crriers through the reltionship For uniform lock of mteril of crosssectionl re A nd length,, the resistnce over the length, is R 5 r, A I vg 5 nqv d A (27.4) where n is the density of chrge crriers, q is the chrge on ech crrier, v d is the drift speed, nd A is the crosssectionl re of the conductor. (27.10) where r is the resistivity of the mteril. The resistivity of conductor vries pproximtely linerly with temperture ccording to the expression r 5 r 0 [1 1 (T 2 T 0 )] (27.18) where r 0 is the resistivity t some reference temperture T 0 nd is the temperture coefficient of resistivity. The current density in n ohmic conductor is proportionl to the electric field ccording to the expression In clssicl model of electricl conduction in metls, the electrons re treted s molecules of gs. In the sence of n electric field, the verge velocity of the electrons is zero. When n electric field is pplied, the electrons move (on verge) with drift velocity v d tht is opposite the electric field. The drift velocity is given y v d 5 q E t (27.13) m e where q is the electron s chrge, m e is the mss of the electron, nd t is the verge time intervl etween electron tom collisions. According to this model, the resistivity of the metl is r 5 nq 2 t where n is the numer of free electrons per unit volume. m e J 5 se (27.6) The proportionlity constnt s is clled the conductivity of the mteril of which the conductor is mde. The inverse of s is known s resistivity r (tht is, r 5 1/s). Eqution 27.6 is known s Ohm s lw, nd mteril is sid to oey this lw if the rtio of its current density to its pplied electric field is constnt tht is independent of the pplied field. (27.16) If potentil difference DV is mintined cross circuit element, the power, or rte t which energy is supplied to the element, is P 5 I DV (27.21) Becuse the potentil difference cross resistor is given y DV 5 IR, we cn express the power delivered to resistor s P 5 I 2 R 5 1DV 2 2 (27.22) R The energy delivered to resistor y electricl trnsmission T ET ppers in the form of internl energy E int in the resistor. Ojective Questions 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. Cr tteries re often rted in mpere-hours. Does this informtion designte the mount of () current, () power, (c) energy, (d) chrge, or (e) potentil the ttery cn supply? 2. Two wires A nd B with circulr cross sections re mde of the sme metl nd hve equl lengths, ut the resistnce of wire A is three times greter thn tht of wire B. (i) Wht is the rtio of the cross-sectionl

168 Conceptul Questions 825 re of A to tht of B? () 3 ()!3 (c) 1 (d) 1/!3 (e) 1 3 (ii) Wht is the rtio of the rdius of A to tht of B? Choose from the sme possiilities s in prt (i). 3. A cylindricl metl wire t room temperture is crrying electric current etween its ends. One end is t potentil V A 5 50 V, nd the other end is t potentil V B 5 0 V. Rnk the following ctions in terms of the chnge tht ech one seprtely would produce in the current from the gretest increse to the gretest decrese. In your rnking, note ny cses of equlity. () Mke V A V with V B 5 0 V. () Adjust V A to triple the power with which the wire converts electriclly trnsmitted energy into internl energy. (c) Doule the rdius of the wire. (d) Doule the length of the wire. (e) Doule the Celsius temperture of the wire. 4. A current-crrying ohmic metl wire hs crosssectionl re tht grdully ecomes smller from one end of the wire to the other. The current hs the sme vlue for ech section of the wire, so chrge does not ccumulte t ny one point. (i) How does the drift speed vry long the wire s the re ecomes smller? () It increses. () It decreses. (c) It remins constnt. (ii) How does the resistnce per unit length vry long the wire s the re ecomes smller? Choose from the sme possiilities s in prt (i). 5. A potentil difference of 1.00 V is mintined cross 10.0-V resistor for period of 20.0 s. Wht totl chrge psses y point in one of the wires connected to the resistor in this time intervl? () 200 C () 20.0 C (c) 2.00 C (d) C (e) C 6. Three wires re mde of copper hving circulr cross sections. Wire 1 hs length L nd rdius r. Wire 2 hs length L nd rdius 2r. Wire 3 hs length 2L nd rdius 3r. Which wire hs the smllest resistnce? () wire 1 () wire 2 (c) wire 3 (d) All hve the sme resistnce. (e) Not enough informtion is given to nswer the question. 7. A metl wire of resistnce R is cut into three equl pieces tht re then plced together side y side to form new cle with length equl to one-third the originl length. Wht is the resistnce of this new cle? () 1 9R () 1 3R (c) R (d) 3R (e) 9R Conceptul Questions 1. If you were to design n electric heter using Nichrome wire s the heting element, wht prmeters of the wire could you vry to meet specific power output such s W? 2. Wht fctors ffect the resistnce of conductor? 3. When the potentil difference cross certin conductor is douled, the current is oserved to increse y fctor of 3. Wht cn you conclude out the conductor? 4. Over the time intervl fter difference in potentil is pplied etween the ends of wire, wht would hppen to the drift velocity of the electrons in wire nd to the current in the wire if the electrons could move freely without resistnce through the wire? 8. A metl wire hs resistnce of 10.0 V t temperture of 20.0 C. If the sme wire hs resistnce of 10.6 V t 90.0 C, wht is the resistnce of this wire when its temperture is C? () V () 9.66 V (c) 10.3 V (d) 13.8 V (e) 6.59 V 9. The current-versus-voltge ehvior of certin electricl device is shown in Figure OQ27.9. When the potentil difference cross the device is 2 V, wht is its resistnce? () 1 V () 3 4 V (c) 4 3 V (d) undefined (e) none of those nswers I (A) 3 1. denotes nswer ville in tudent olutions Mnul/tudy Guide V (V) Figure OQ Two conductors mde of the sme mteril re connected cross the sme potentil difference. Conductor A hs twice the dimeter nd twice the length of conductor B. Wht is the rtio of the power delivered to A to the power delivered to B? () 8 () 4 (c) 2 (d) 1 (e) Two conducting wires A nd B of the sme length nd rdius re connected cross the sme potentil difference. Conductor A hs twice the resistivity of conductor B. Wht is the rtio of the power delivered to A to the power delivered to B? () 2 ()!2 (c) 1 (d) 1/!2 (e) Two lightuls oth operte on 120 V. One hs power of 25 W nd the other 100 W. (i) Which lightul hs higher resistnce? () The dim 25-W lightul does. () The right 100-W lightul does. (c) Both re the sme. (ii) Which lightul crries more current? Choose from the sme possiilities s in prt (i). 13. Wire B hs twice the length nd twice the rdius of wire A. Both wires re mde from the sme mteril. If wire A hs resistnce R, wht is the resistnce of wire B? () 4R () 2R (c) R (d) 1 2R (e) 1 4R 5. How does the resistnce for copper nd for silicon chnge with temperture? Why re the ehviors of these two mterils different? 6. Use the tomic theory of mtter to explin why the resistnce of mteril should increse s its temperture increses. 7. If chrges flow very slowly through metl, why does it not require severl hours for light to come on when you throw switch? 8. Newspper rticles often contin sttements such s volts of electricity surged through the victim s ody. Wht is wrong with this sttement?

169 826 Chpter 27 Current nd Resistnce Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign ection 27.1 Electric Current 1. A 200-km-long high-voltge trnsmission line 2.00 cm AMT in dimeter crries stedy current of A. If M the conductor is copper with free chrge density of electrons per cuic meter, how mny yers does it tke one electron to trvel the full length of the cle? 2. A smll sphere tht crries chrge q is whirled in circle t the end of n insulting string. The ngulr frequency of revolution is v. Wht verge current does this revolving chrge represent? 3. An luminum wire hving cross-sectionl re equl W to m 2 crries current of 5.00 A. The density of luminum is 2.70 g/cm 3. Assume ech luminum tom supplies one conduction electron per tom. Find the drift speed of the electrons in the wire. 4. In the Bohr model of the hydrogen tom (which will AMT e covered in detil in Chpter 42), n electron in the lowest energy stte moves t speed of m/s in circulr pth of rdius m. Wht is the effective current ssocited with this oriting electron? 5. A proton em in n ccelertor crries current of 125 ma. If the em is incident on trget, how mny protons strike the trget in period of 23.0 s? 6. A copper wire hs circulr cross section with rdius Q/C of 1.25 mm. () If the wire crries current of 3.70 A, find the drift speed of the electrons in this wire. () All other things eing equl, wht hppens to the drift speed in wires mde of metl hving lrger numer of conduction electrons per tom thn copper? Explin. 7. uppose the current in conductor decreses exponentilly with time ccording to the eqution I(t) 5 I 0 e 2t/t, where I 0 is the initil current (t t 5 0) nd t is constnt hving dimensions of time. Consider fixed oservtion point within the conductor. () How much chrge psses this point etween t 5 0 nd t 5 t? () How much chrge psses this point etween t 5 0 nd t 5 10t? (c) Wht If? How much chrge psses this point etween t 5 0 nd t 5 `? 8. Figure P27.8 represents section of conductor of W nonuniform dimeter crrying current of I A. Q/C The rdius of cross-section A 1 is r cm. () Wht is the mgnitude of the current density cross A 1? The rdius r 2 t A 2 is lrger thn the rdius r 1 t A 1. () Is the current t A 2 lrger, smller, or the sme? (c) Is the current density t A 2 lrger, smller, or the sme? Assume A 2 5 4A 1. pecify the (d) rdius, (e) current, nd (f) current density t A 2. r 1 r 2 A 1 A 2 I Figure P The quntity of chrge q (in couloms) tht hs pssed W through surfce of re 2.00 cm 2 vries with time ccording to the eqution q 5 4t 3 1 5t 1 6, where t is in seconds. () Wht is the instntneous current through the surfce t t s? () Wht is the vlue of the current density? 10. A Vn de Grff genertor produces em of Q/C 2.00-MeV deuterons, which re hevy hydrogen nuclei contining proton nd neutron. () If the em current is 10.0 ma, wht is the verge seprtion of the deuterons? () Is the electricl force of repulsion mong them significnt fctor in em stility? Explin. 11. The electron em emerging from certin highenergy electron ccelertor hs circulr cross section M of rdius 1.00 mm. () The em current is 8.00 ma. Find the current density in the em ssuming it is uniform throughout. () The speed of the electrons is so close to the speed of light tht their speed cn e tken s 300 Mm/s with negligile error. Find the electron density in the em. (c) Over wht time intervl does Avogdro s numer of electrons emerge from the ccelertor? 12. An electric current in conductor vries with time W ccording to the expression I(t) sin (120pt), where I is in mperes nd t is in seconds. Wht is the totl chrge pssing given point in the conductor from t 5 0 to t s? 13. A tepot with surfce re of 700 cm 2 is to e plted W BIO Q/C with silver. It is ttched to the negtive electrode of n electrolytic cell contining silver nitrte (Ag 1 NO 3 2 ). The cell is powered y 12.0-V ttery nd hs

170 Prolems 827 resistnce of 1.80 V. If the density of silver is kg/m 3, over wht time intervl does mm lyer of silver uild up on the tepot? ection 27.2 Resistnce 14. A lightul hs resistnce of 240 V when operting W with potentil difference of 120 V cross it. Wht is the current in the lightul? 15. A wire 50.0 m long nd 2.00 mm in dimeter is connected to source with potentil difference of 9.11 V, M nd the current is found to e 36.0 A. Assume temperture of 20.0 C nd, using Tle 27.2, identify the metl out of which the wire is mde. 16. A V potentil difference is mintined cross 1.50-m length of tungsten wire tht hs crosssectionl re of mm 2. Wht is the current in the wire? 17. An electric heter crries current of 13.5 A when operting t voltge of 120 V. Wht is the resistnce of the heter? 18. Aluminum nd copper wires of equl length re found to hve the sme resistnce. Wht is the rtio of their rdii? 19. uppose you wish to fricte uniform wire from M 1.00 g of copper. If the wire is to hve resistnce of R V nd ll the copper is to e used, wht must e () the length nd () the dimeter of this wire? 20. uppose you wish to fricte uniform wire from mss m of metl with density r m nd resistivity r. If the wire is to hve resistnce of R nd ll the metl is to e used, wht must e () the length nd () the dimeter of this wire? 21. A portion of Nichrome wire of rdius 2.50 mm is to e used in winding heting coil. If the coil must drw current of 9.25 A when voltge of 120 V is pplied cross its ends, find () the required resistnce of the coil nd () the length of wire you must use to wind the coil. ection 27.3 A Model for Electricl Conduction 22. If the current crried y conductor is douled, wht hppens to () the chrge crrier density, () the current density, (c) the electron drift velocity, nd (d) the verge time intervl etween collisions? 23. A current density of A/m 2 exists in the tmosphere t loction where the electric field is 100 V/m. Clculte the electricl conductivity of the Erth s tmosphere in this region. 24. An iron wire hs cross-sectionl re equl to GP m 2. Crry out the following steps to determine Q/C the drift speed of the conduction electrons in the wire if it crries current of 30.0 A. () How mny kilogrms re there in 1.00 mole of iron? () trting with the density of iron nd the result of prt (), compute the molr density of iron (the numer of moles of iron per cuic meter). (c) Clculte the numer density of iron toms using Avogdro s numer. (d) Otin the numer density of conduction electrons given tht there re two conduction electrons per iron tom. (e) Clculte the drift speed of conduction electrons in this wire. 25. If the mgnitude of the drift velocity of free electrons M in copper wire is m/s, wht is the electric field in the conductor? ection 27.4 Resistnce nd Temperture 26. A certin lightul hs tungsten filment with resistnce of 19.0 V when t 20.0 C nd 140 V when hot. Assume the resistivity of tungsten vries linerly with temperture even over the lrge temperture rnge involved here. Find the temperture of the hot filment. 27. Wht is the frctionl chnge in the resistnce of n iron filment when its temperture chnges from 25.0 C to 50.0 C? 28. While tking photogrphs in Deth Vlley on dy when the temperture is 58.0 C, Bill Hiker finds tht certin voltge pplied to copper wire produces current of 1.00 A. Bill then trvels to Antrctic nd pplies the sme voltge to the sme wire. Wht current does he register there if the temperture is C? Assume tht no chnge occurs in the wire s shpe nd size. 29. If certin silver wire hs resistnce of 6.00 V t 20.0 C, wht resistnce will it hve t 34.0 C? 30. Plethysmogrphs re devices used for mesuring BIO chnges in the volume of internl orgns or lims. In one form of this device, ruer cpillry tue with n inside dimeter of 1.00 mm is filled with mercury t 20.0 C. The resistnce of the mercury is mesured with the id of electrodes seled into the ends of the tue. If 100 cm of the tue is wound in helix round ptient s upper rm, the lood flow during hertet cuses the rm to expnd, stretching the length of the tue y cm. From this oservtion nd ssuming cylindricl symmetry, you cn find the chnge in volume of the rm, which gives n indiction of lood flow. Tking the resistivity of mercury to e V? m, clculte () the resistnce of the mercury nd () the frctionl chnge in resistnce during the hertet. Hint: The frction y which the cross-sectionl re of the mercury column decreses is the frction y which the length increses ecuse the volume of mercury is constnt. 31. () A 34.5-m length of copper wire t 20.0 C hs M rdius of 0.25 mm. If potentil difference of 9.00 V is pplied cross the length of the wire, determine the current in the wire. () If the wire is heted to 30.0 C while the 9.00-V potentil difference is mintined, wht is the resulting current in the wire? 32. An engineer needs resistor with zero overll temperture coefficient of resistnce t 20.0 C. he designs pir of circulr cylinders, one of cron nd one of Nichrome s shown in Figure P27.32 (pge 828). The

171 828 Chpter 27 Current nd Resistnce device must hve n overll resistnce of R 1 1 R V independent of temperture nd uniform rdius of r mm. Ignore therml expnsion of the cylinders nd ssume oth re lwys t the sme temperture. () Cn she meet the design gol with this method? () If so, stte wht you cn determine out the lengths, 1 nd, 2 of ech segment. If not, explin. 1 2 Figure P An luminum wire with dimeter of mm hs M uniform electric field of V/m imposed long its entire length. The temperture of the wire is 50.0 C. Assume one free electron per tom. () Use the informtion in Tle 27.2 to determine the resistivity of luminum t this temperture. () Wht is the current density in the wire? (c) Wht is the totl current in the wire? (d) Wht is the drift speed of the conduction electrons? (e) Wht potentil difference must exist etween the ends of 2.00-m length of the wire to produce the stted electric field? 34. Review. An luminum rod hs resistnce of 1.23 V t 20.0 C. Clculte the resistnce of the rod t 120 C y ccounting for the chnges in oth the resistivity nd the dimensions of the rod. The coefficient of liner expnsion for luminum is ( C) At wht temperture will luminum hve resistivity tht is three times the resistivity copper hs t room temperture? ection 27.6 Electricl Power 36. Assume tht glol lightning on the Erth constitutes constnt current of 1.00 ka etween the ground nd n tmospheric lyer t potentil 300 kv. () Find the power of terrestril lightning. () For comprison, find the power of sunlight flling on the Erth. unlight hs n intensity of W/m 2 ove the tmosphere. unlight flls perpendiculrly on the circulr projected re tht the Erth presents to the un. 37. In hydroelectric instlltion, turine delivers hp to genertor, which in turn trnsfers 80.0% of the mechnicl energy out y electricl trnsmission. Under these conditions, wht current does the genertor deliver t terminl potentil difference of V? 38. A Vn de Grff genertor (see Fig ) is operting so tht the potentil difference etween the highpotentil electrode B nd the chrging needles t A is 15.0 kv. Clculte the power required to drive the elt ginst electricl forces t n instnt when the effective current delivered to the high-potentil electrode is 500 ma. 39. A certin wffle iron is rted t 1.00 kw when connected to 120-V source. () Wht current does the wffle iron crry? () Wht is its resistnce? 40. The potentil difference cross resting neuron in the BIO humn ody is out 75.0 mv nd crries current of out ma. How much power does the neuron relese? 41. uppose your portle DVD plyer drws current of 350 ma t 6.00 V. How much power does the plyer require? 42. Review. A well-insulted electric wter heter wrms AMT 109 kg of wter from 20.0 C to 49.0 C in 25.0 min. M Find the resistnce of its heting element, which is connected cross 240-V potentil difference. 43. A 100-W lightul connected to 120-V source experiences voltge surge tht produces 140 V for moment. By wht percentge does its power output increse? Assume its resistnce does not chnge. 44. The cost of energy delivered to residences y electricl trnsmission vries from $0.070/kWh to $0.258/kWh throughout the United ttes; $0.110/kWh is the verge vlue. At this verge price, clculte the cost of () leving 40.0-W porch light on for two weeks while you re on vction, () mking piece of drk tost in 3.00 min with 970-W toster, nd (c) drying lod of clothes in 40.0 min in W dryer. 45. Btteries re rted in terms of mpere-hours (A? h). W For exmple, ttery tht cn produce current of 2.00 A for 3.00 h is rted t 6.00 A? h. () Wht is the totl energy, in kilowtt-hours, stored in 12.0-V ttery rted t 55.0 A? h? () At $0.110 per kilowtt-hour, wht is the vlue of the electricity produced y this ttery? 46. Residentil uilding codes typiclly require the use W of 12-guge copper wire (dimeter cm) for wiring receptcles. uch circuits crry currents s lrge s Q/C 20.0 A. If wire of smller dimeter (with higher guge numer) crried tht much current, the wire could rise to high temperture nd cuse fire. () Clculte the rte t which internl energy is produced in 1.00 m of 12-guge copper wire crrying 20.0 A. () Wht If? Repet the clcultion for 12-guge luminum wire. (c) Explin whether 12-guge luminum wire would e s sfe s copper wire. 47. Assuming the cost of energy from the electric compny M is $0.110/kWh, compute the cost per dy of operting lmp tht drws current of 1.70 A from 110-V line. 48. An 11.0-W energy-efficient fluorescent lightul is designed to produce the sme illumintion s conventionl 40.0-W incndescent lightul. Assuming cost of $0.110/kWh for energy from the electric compny, how much money does the user of the energyefficient ul sve during 100 h of use? 49. A coil of Nichrome wire is 25.0 m long. The wire hs dimeter of mm nd is t 20.0 C. If it crries current of A, wht re () the mgnitude of the electric field in the wire nd () the power delivered to it? (c) Wht If? If the temperture is incresed to 340 C nd the potentil difference cross the wire remins constnt, wht is the power delivered? 50. Review. A rechrgele ttery of mss 15.0 g delivers n verge current of 18.0 ma to portle DVD plyer t 1.60 V for 2.40 h efore the ttery must e

172 Prolems 829 rechrged. The rechrger mintins potentil difference of 2.30 V cross the ttery nd delivers chrging current of 13.5 ma for 4.20 h. () Wht is the efficiency of the ttery s n energy storge device? () How much internl energy is produced in the ttery during one chrge dischrge cycle? (c) If the ttery is surrounded y idel therml insultion nd hs n effective specific het of 975 J/kg? C, y how much will its temperture increse during the cycle? 51. A 500-W heting coil designed to operte from 110 V is mde of Nichrome wire mm in dimeter. () Assuming the resistivity of the Nichrome remins constnt t its 20.0 C vlue, find the length of wire used. () Wht If? Now consider the vrition of resistivity with temperture. Wht power is delivered to the coil of prt () when it is wrmed to C? 52. Why is the following sitution impossile? A politicin is decrying wsteful uses of energy nd decides to focus on energy used to operte plug-in electric clocks in the United ttes. He estimtes there re 270 million of these clocks, pproximtely one clock for ech person in the popultion. The clocks trnsform energy tken in y electricl trnsmission t the verge rte 2.50 W. The politicin gives speech in which he complins tht, t tody s electricl rtes, the ntion is losing $100 million every yer to operte these clocks. 53. A certin toster hs heting element mde of M Nichrome wire. When the toster is first connected to 120-V source (nd the wire is t temperture of 20.0 C), the initil current is 1.80 A. The current decreses s the heting element wrms up. When the toster reches its finl operting temperture, the current is 1.53 A. () Find the power delivered to the toster when it is t its operting temperture. () Wht is the finl temperture of the heting element? 54. Mke n order-of-mgnitude estimte of the cost of one person s routine use of hndheld hir dryer for 1 yer. If you do not use hir dryer yourself, oserve or interview someone who does. tte the quntities you estimte nd their vlues. 55. Review. The heting element of n electric coffee M mker opertes t 120 V nd crries current of 2.00 A. Assuming the wter sors ll the energy delivered to the resistor, clculte the time intervl during which the temperture of kg of wter rises from room temperture (23.0 C) to the oiling point. 56. A 120-V motor hs mechnicl power output of 2.50 hp. It is 90.0% efficient in converting power tht it tkes in y electricl trnsmission into mechnicl power. () Find the current in the motor. () Find the energy delivered to the motor y electricl trnsmission in 3.00 h of opertion. (c) If the electric compny chrges $0.110/kWh, wht does it cost to run the motor for 3.00 h? Additionl Prolems 57. A prticulr wire hs resistivity of V? m M nd cross-sectionl re of m 2. A length of this wire is to e used s resistor tht will receive 48 W of power when connected cross 20-V ttery. Wht length of wire is required? 58. Determine the temperture t which the resistnce of n luminum wire will e twice its vlue t 20.0 C. Assume its coefficient of resistivity remins constnt. 59. A cr owner forgets to turn off the hedlights of his cr while it is prked in his grge. If the 12.0-V ttery in his cr is rted t 90.0 A? h nd ech hedlight requires 36.0 W of power, how long will it tke the ttery to completely dischrge? 60. Lightul A is mrked 25 W 120 V, nd lightul B Q/C is mrked 100 W 120 V. These lels men tht ech lightul hs its respective power delivered to it when it is connected to constnt 120-V source. () Find the resistnce of ech lightul. () During wht time intervl does 1.00 C pss into lightul A? (c) Is this chrge different upon its exit versus its entry into the lightul? Explin. (d) In wht time intervl does 1.00 J pss into lightul A? (e) By wht mechnisms does this energy enter nd exit the lightul? Explin. (f) Find the cost of running lightul A continuously for 30.0 dys, ssuming the electric compny sells its product t $0.110 per kwh. 61. One wire in high-voltge trnsmission line crries W A strting t 700 kv for distnce of 100 mi. If the resistnce in the wire is V/mi, wht is the power loss due to the resistnce of the wire? 62. An experiment is conducted to mesure the electricl resistivity of Nichrome in the form of wires with Q/C different lengths nd cross-sectionl res. For one set of mesurements, student uses 30-guge wire, which hs cross- sectionl re of m 2. The student mesures the potentil difference cross the wire nd the current in the wire with voltmeter nd n mmeter, respectively. () For ech set of mesurements given in the tle tken on wires of three different lengths, clculte the resistnce of the wires nd the corresponding vlues of the resistivity. () Wht is the verge vlue of the resistivity? (c) Explin how this vlue compres with the vlue given in Tle L (m) DV (V) I (A) R (V) r (V? m) A chrge Q is plced on cpcitor of cpcitnce C. The cpcitor is connected into the circuit shown in Figure P27.63, with n open switch, resistor, nd n initilly unchrged cpcitor of cpcitnce 3C. The C Q 3C R Figure P27.63

173 830 Chpter 27 Current nd Resistnce switch is then closed, nd the circuit comes to equilirium. In terms of Q nd C, find () the finl potentil difference etween the pltes of ech cpcitor, () the chrge on ech cpcitor, nd (c) the finl energy stored in ech cpcitor. (d) Find the internl energy ppering in the resistor. 64. Review. An office worker uses n immersion heter to wrm 250 g of wter in light, covered, insulted cup from 20.0 C to 100 C in 4.00 min. The heter is Nichrome resistnce wire connected to 120-V power supply. Assume the wire is t 100 C throughout the 4.00-min time intervl. () pecify reltionship etween dimeter nd length tht the wire cn hve. () Cn it e mde from less thn cm 3 of Nichrome? 65. An x-ry tue used for cncer therpy opertes t BIO 4.00 MV with electrons constituting em current of 25.0 ma striking metl trget. Nerly ll the power in the em is trnsferred to strem of wter flowing through holes drilled in the trget. Wht rte of flow, in kilogrms per second, is needed if the rise in temperture of the wter is not to exceed 50.0 C? 66. An ll-electric cr (not hyrid) is designed to run AMT from nk of 12.0-V tteries with totl energy storge of J. If the electric motor drws 8.00 kw M s the cr moves t stedy speed of 20.0 m/s, () wht is the current delivered to the motor? () How fr cn the cr trvel efore it is out of juice? 67. A stright, cylindricl wire lying long the x xis hs length of m nd dimeter of mm. It is mde of mteril descried y Ohm s lw with resistivity of r V? m. Assume potentil of 4.00 V is mintined t the left end of the wire t x 5 0. Also ssume V 5 0 t x m. Find () the mgnitude nd direction of the electric field in the wire, () the resistnce of the wire, (c) the mgnitude nd direction of the electric current in the wire, nd (d) the current density in the wire. (e) how tht E 5 rj. 68. A stright, cylindricl wire lying long the x xis hs length L nd dimeter d. It is mde of mteril descried y Ohm s lw with resistivity r. Assume potentil V is mintined t the left end of the wire t x 5 0. Also ssume the potentil is zero t x 5 L. In terms of L, d, V, r, nd physicl constnts, derive expressions for () the mgnitude nd direction of the electric field in the wire, () the resistnce of the wire, (c) the mgnitude nd direction of the electric current in the wire, nd (d) the current density in the wire. (e) how tht E 5 rj. 69. An electric utility compny supplies customer s house W from the min power lines (120 V) with two copper wires, ech of which is 50.0 m long nd hs resistnce of V per 300 m. () Find the potentil difference t the customer s house for lod current of 110 A. For this lod current, find () the power delivered to the customer nd (c) the rte t which internl energy is produced in the copper wires. 70. The strin in wire cn e monitored nd computed Q/C y mesuring the resistnce of the wire. Let L i represent the originl length of the wire, A i its originl cross-sectionl re, R i 5 rl i /A i the originl resistnce etween its ends, nd d 5 DL/L i 5 (L 2 L i )/L i the strin resulting from the ppliction of tension. Assume the resistivity nd the volume of the wire do not chnge s the wire stretches. () how tht the resistnce etween the ends of the wire under strin is given y R 5 R i (1 1 2d 1 d 2 ). () If the ssumptions re precisely true, is this result exct or pproximte? Explin your nswer. 71. An ocenogrpher is studying how the ion concentrtion in sewter depends on depth. he mkes mesurement y lowering into the wter pir of concentric metllic cylinders (Fig. P27.71) t the end of cle nd tking dt to determine the resistnce etween these electrodes s function of depth. The wter etween the two cylinders forms cylindricl shell of inner rdius r, outer rdius r, nd length L much lrger thn r. The scientist pplies potentil difference DV etween the inner nd outer surfces, producing n outwrd rdil current I. Let r represent the resistivity of the wter. () Find the resistnce of the wter etween the cylinders in terms of L, r, r, nd r. () Express the resistivity of the wter in terms of the mesured quntities L, r, r, DV, nd I. r r L Figure P Why is the following sitution impossile? An inquisitive physics student tkes 100-W incndescent lightul out of its socket nd mesures its resistnce with n ohmmeter. He mesures vlue of 10.5 V. He is le to connect n mmeter to the lightul socket to correctly mesure the current drwn y the ul while operting. Inserting the ul ck into the socket nd operting the ul from 120-V source, he mesures the current to e 11.4 A. 73. The temperture coefficients of resistivity in Tle 27.2 re sed on reference temperture T 0 of 20.0 C. uppose the coefficients were given the symol 9 nd were sed on T 0 of 0 C. Wht would the coefficient 9 for silver e? Note: The coefficient stisfies r 5 r 0 [1 1 (T 2 T 0 )], where r 0 is the resistivity of the mteril t T C. The coefficient 9 must stisfy the expression r 5 r9 0 [1 1 9T], where r9 0 is the resistivity of the mteril t 0 C. 74. A close nlogy exists etween the flow of energy y Q/C het ecuse of temperture difference (see ection 20.7) nd the flow of electric chrge ecuse of

174 Prolems 831 potentil difference. In metl, energy dq nd electricl chrge dq re oth trnsported y free electrons. Consequently, good electricl conductor is usully good therml conductor s well. Consider thin conducting sl of thickness dx, re A, nd electricl conductivity s, with potentil difference dv etween opposite fces. () how tht the current I 5 dq/dt is given y the eqution on the left: Chrge conduction Therml conduction dq dt 5 sa ` dv dx ` dq dt 5 ka ` dt dx ` In the nlogous therml conduction eqution on the right (Eq ), the rte dq /dt of energy flow y het (in I units of joules per second) is due to temperture grdient dt/dx in mteril of therml conductivity k. () tte nlogous rules relting the direction of the electric current to the chnge in potentil nd relting the direction of energy flow to the chnge in temperture. 75. Review. When stright wire is wrmed, its resistnce is given y R 5 R 0 [1 1 (T 2 T 0 )] ccording to Eqution 27.20, where is the temperture coefficient of resistivity. This expression needs to e modified if we include the chnge in dimensions of the wire due to therml expnsion. For copper wire of rdius mm nd length m, find its resistnce t C, including the effects of oth therml expnsion nd temperture vrition of resistivity. Assume the coefficients re known to four significnt figures. 76. Review. When stright wire is wrmed, its resistnce is given y R 5 R 0 [1 1 (T 2 T 0 )] ccording to Eqution 27.20, where is the temperture coefficient of resistivity. This expression needs to e modified if we include the chnge in dimensions of the wire due to therml expnsion. Find more precise expression for the resistnce, one tht includes the effects of chnges in the dimensions of the wire when it is wrmed. Your finl expression should e in terms of R 0, T, T 0, the temperture coefficient of resistivity, nd the coefficient of liner expnsion Review. A prllel-plte cpcitor consists of squre pltes of edge length, tht re seprted y distnce d, where d,,,. A potentil difference DV is mintined etween the pltes. A mteril of dielectric constnt k fills hlf the spce etween the pltes. The dielectric sl is withdrwn from the cpcitor s shown in Figure P () Find the cpcitnce when d x V the left edge of the dielectric is t distnce x from the center of the cpcitor. () If the dielectric is removed t constnt speed v, wht is the current in the circuit s the dielectric is eing withdrwn? 78. The dielectric mteril etween the pltes of prllelplte cpcitor lwys hs some nonzero conductivity s. Let A represent the re of ech plte nd d the distnce etween them. Let k represent the dielectric constnt of the mteril. () how tht the resistnce R nd the cpcitnce C of the cpcitor re relted y RC 5 kp 0 s () Find the resistnce etween the pltes of 14.0-nF cpcitor with fused qurtz dielectric. 79. Gold is the most ductile of ll metls. For exmple, one grm of gold cn e drwn into wire 2.40 km long. The density of gold is kg/m 3, nd its resistivity is V? m. Wht is the resistnce of such wire t 20.0 C? 80. The current voltge chrcteristic curve for semiconductor diode s function of temperture T is given y I 5 I 0 (e e DV/k B T 2 1) Here the first symol e represents Euler s numer, the se of nturl logrithms. The second e is the mgnitude of the electron chrge, the k B stnds for Boltzmnn s constnt, nd T is the solute temperture. () et up spredsheet to clculte I nd R 5 DV/I for DV V to V in increments of V. Assume I na. () Plot R versus DV for T K, 300 K, nd 320 K. 81. The potentil difference cross the filment of lightul is mintined t constnt vlue while equilirium temperture is eing reched. The stedy-stte current in the ul is only one-tenth of the current drwn y the ul when it is first turned on. If the temperture coefficient of resistivity for the ul t 20.0 C is ( C) 21 nd the resistnce increses linerly with incresing temperture, wht is the finl operting temperture of the filment? Chllenge Prolems 82. A more generl definition of the temperture coefficient of resistivity is 5 1 dr r dt where r is the resistivity t temperture T. () Assuming is constnt, show tht r 5 r 0 e (T 2 T 0 ) where r 0 is the resistivity t temperture T 0. () Using the series expnsion e x < 1 1 x for x,, 1, show tht the resistivity is given pproximtely y the expression r 5 r 0 [1 1 (T 2 T 0 )] for (T 2 T 0 ),, 1 Figure P27.77 v 83. A sphericl shell with inner rdius r nd outer rdius r is formed from mteril of resistivity r. It crries

175 832 Chpter 27 Current nd Resistnce current rdilly, with uniform density in ll directions. how tht its resistnce is R 5 r 4p r r 84. Mteril with uniform resistivity r is formed into wedge s shown in Figure P how tht the resistnce etween fce A nd fce B of this wedge is y 1 R 5 r Fce A L w 1 y 2 2 y 1 2 L Figure P27.84 ln y 2 y 1 w Fce B y A mteril of resistivity r is formed into the shpe of truncted cone of height h s shown in Figure P The ottom end hs rdius, nd the top end hs rdius. Assume the current is distriuted uniformly over ny circulr cross section of the cone so tht the current density does not depend on rdil position. (The current density does vry with position long the xis of the cone.) how tht the resistnce etween the two ends is R 5 r p h Figure P27.85 h

176 Direct-Current Circuits c h p t e r 28 In this chpter, we nlyze simple electric circuits tht contin tteries, resistors, nd cpcitors in vrious comintions. ome circuits contin resistors tht cn e comined using simple rules. The nlysis of more complicted circuits is simplified using Kirchhoff s rules, which follow from the lws of conservtion of energy nd conservtion of electric chrge for isolted systems. Most of the circuits nlyzed re ssumed to e in stedy stte, which mens tht currents in the circuit re constnt in mgnitude nd direction. A current tht is constnt in direction is clled direct current (DC). We will study lternting current (AC), in which the current chnges direction periodiclly, in Chpter 33. Finlly, we discuss electricl circuits in the home Electromotive Force In ection 27.6, we discussed circuit in which ttery produces current. We will generlly use ttery s source of energy for circuits in our discussion. Becuse the potentil difference t the ttery terminls is constnt in prticulr circuit, the current in the circuit is constnt in mgnitude nd direction nd is clled direct current. A ttery is clled either source of electromotive force or, more commonly, source of emf. (The phrse electromotive force is n unfortunte historicl term, descriing not force, ut rther potentil difference in volts.) The emf e of ttery is the mximum possile voltge the ttery cn provide etween its terminls. You cn think of source of emf s chrge pump. When n electric potentil difference exists etween two points, the source moves chrges uphill from the lower potentil to the higher. We shll generlly ssume the connecting wires in circuit hve no resistnce. The positive terminl of ttery is t higher potentil thn the negtive terminl Electromotive Force 28.2 Resistors in eries nd Prllel 28.3 Kirchhoff s Rules 28.4 RC Circuits 28.5 Household Wiring nd Electricl fety A technicin repirs connection on circuit ord from computer. In our lives tody, we use vrious items contining electric circuits, including mny with circuit ords much smller thn the ord shown in the photogrph. These include hndheld gme plyers, cell phones, nd digitl cmers. In this chpter, we study simple types of circuits nd lern how to nlyze them. (Tromx/hutterstock.com) 833

177 834 Chpter 28 Direct-Current Circuits I e IR V r R c d e f 0 e e f R Figure 28.1 () Circuit digrm of source of emf e (in this cse, ttery), of internl resistnce r, connected to n externl resistor of resistnce R. () Grphicl representtion showing how the electric potentil chnges s the circuit in () is trversed clockwise. Pitfll Prevention 28.1 Wht Is Constnt in Bttery? It is common misconception tht ttery is source of constnt current. Eqution 28.3 shows tht is not true. The current in the circuit depends on the resistnce R connected to the ttery. It is lso not true tht ttery is source of constnt terminl voltge s shown y Eqution A ttery is source of constnt emf. c e Ir r d I Becuse rel ttery is mde of mtter, there is resistnce to the flow of chrge within the ttery. This resistnce is clled internl resistnce r. For n idelized ttery with zero internl resistnce, the potentil difference cross the ttery (clled its terminl voltge) equls its emf. For rel ttery, however, the terminl voltge is not equl to the emf for ttery in circuit in which there is current. To understnd why, consider the circuit digrm in Figure We model the ttery s shown in the digrm; it is represented y the dshed rectngle contining n idel, resistnce-free emf e in series with n internl resistnce r. A resistor of resistnce R is connected cross the terminls of the ttery. Now imgine moving through the ttery from to d nd mesuring the electric potentil t vrious loctions. Pssing from the negtive terminl to the positive terminl, the potentil increses y n mount e. As we move through the resistnce r, however, the potentil decreses y n mount Ir, where I is the current in the circuit. Therefore, the terminl voltge of the ttery DV 5 V d 2 V is DV 5 e 2 Ir (28.1) From this expression, notice tht e is equivlent to the open-circuit voltge, tht is, the terminl voltge when the current is zero. The emf is the voltge leled on ttery; for exmple, the emf of D cell is 1.5 V. The ctul potentil difference etween ttery s terminls depends on the current in the ttery s descried y Eqution Figure 28.1 is grphicl representtion of the chnges in electric potentil s the circuit is trversed in the clockwise direction. Figure 28.1 shows tht the terminl voltge DV must equl the potentil difference cross the externl resistnce R, often clled the lod resistnce. The lod resistor might e simple resistive circuit element s in Figure 28.1, or it could e the resistnce of some electricl device (such s toster, electric heter, or lightul) connected to the ttery (or, in the cse of household devices, to the wll outlet). The resistor represents lod on the ttery ecuse the ttery must supply energy to operte the device contining the resistnce. The potentil difference cross the lod resistnce is DV 5 IR. Comining this expression with Eqution 28.1, we see tht e 5 IR 1 Ir (28.2) Figure 28.1 shows grphicl representtion of this eqution. olving for the current gives I 5 (28.3) R 1 r Eqution 28.3 shows tht the current in this simple circuit depends on oth the lod resistnce R externl to the ttery nd the internl resistnce r. If R is much greter thn r, s it is in mny rel-world circuits, we cn neglect r. Multiplying Eqution 28.2 y the current I in the circuit gives Ie = I 2 R 1 I 2 r (28.4) Eqution 28.4 indictes tht ecuse power P 5 I DV (see Eq ), the totl power output Ie ssocited with the emf of the ttery is delivered to the externl lod resistnce in the mount I 2 R nd to the internl resistnce in the mount I 2 r. Q uick Quiz 28.1 To mximize the percentge of the power from the emf of ttery tht is delivered to device externl to the ttery, wht should the internl resistnce of the ttery e? () It should e s low s possile. () It should e s high s possile. (c) The percentge does not depend on the internl resistnce. e Exmple 28.1 Terminl Voltge of Bttery A ttery hs n emf of 12.0 V nd n internl resistnce of V. Its terminls re connected to lod resistnce of 3.00 V.

178 28.1 Electromotive Force continued (A) Find the current in the circuit nd the terminl voltge of the ttery. olution Conceptulize tudy Figure 28.1, which shows circuit consistent with the prolem sttement. The ttery delivers energy to the lod resistor. Ctegorize This exmple involves simple clcultions from this section, so we ctegorize it s sustitution prolem. e Use Eqution 28.3 to find the current in the circuit: I 5 R 1 r V 3.00 V V A Use Eqution 28.1 to find the terminl voltge: DV 5 e 2 Ir V A V V To check this result, clculte the voltge cross the lod resistnce R: DV 5 IR A V V (B) Clculte the power delivered to the lod resistor, the power delivered to the internl resistnce of the ttery, nd the power delivered y the ttery. olution Use Eqution to find the power delivered to the lod resistor: Find the power delivered to the internl resistnce: Find the power delivered y the ttery y dding these quntities: P R 5 I 2 R 5 (3.93 A) 2 (3.00 V) W P r 5 I 2 r 5 (3.93 A) 2 ( V) W P 5 P R 1 P r W W W Wht If? As ttery ges, its internl resistnce increses. uppose the internl resistnce of this ttery rises to 2.00 V towrd the end of its useful life. How does tht lter the ttery s ility to deliver energy? Answer Let s connect the sme 3.00-V lod resistor to the ttery. e Find the new current in the ttery: I 5 R 1 r V 3.00 V V A Find the new terminl voltge: Find the new powers delivered to the lod resistor nd internl resistnce: DV 5 e 2 Ir V 2 (2.40 A)(2.00 V) V P R 5 I 2 R 5 (2.40 A) 2 (3.00 V) W P r 5 I 2 r 5 (2.40 A) 2 (2.00 V) W In this sitution, the terminl voltge is only 60% of the emf. Notice tht 40% of the power from the ttery is delivered to the internl resistnce when r is 2.00 V. When r is V s in prt (B), this percentge is only 1.6%. Consequently, even though the emf remins fixed, the incresing internl resistnce of the ttery significntly reduces the ttery s ility to deliver energy to n externl lod. Exmple 28.2 Mtching the Lod Find the lod resistnce R for which the mximum power is delivered to the lod resistnce in Figure olution Conceptulize Think out vrying the lod resistnce in Figure 28.1 nd the effect on the power delivered to the lod resistnce. When R is lrge, there is very little current, so the power I 2 R delivered to the lod resistor is smll. continued

179 836 Chpter 28 Direct-Current Circuits 28.2 continued When R is smll, let's sy R,, r, the current is lrge nd the power delivered to the internl resistnce is I 2 r.. I 2 R. Therefore, the power delivered to the lod resistor is smll compred to tht delivered to the internl resistnce. For some intermedite vlue of the resistnce R, the power must mximize. Ctegorize We ctegorize this exmple s n nlysis prolem ecuse we must undertke procedure to mximize the power. The circuit is the sme s tht in Exmple The lod resistnce R in this cse, however, is vrile. Anlyze Find the power delivered to the lod resistnce using Eqution 27.22, with I given y Eqution 28.3: Differentite the power with respect to the lod resistnce R nd set the derivtive equl to zero to mximize the power: olve for R : Figure 28.2 (Exmple 28.2) Grph of the power P delivered y ttery to lod resistor of resistnce R s function of R. (1) P 5 I 2 R 5 e2 R 1R 1 r2 2 dp dr 5 d e 2 dr c R 1R 1 r2 d 5 d 2 dr 3e2 R 1R 1 r R 5 r [e 2 (R 1 r) 22 ] 1 [e 2 R(22)(R 1 r) 23 ] 5 0 e 2 1R 1 r2 2 2e2 R 1R 1 r2 3 1R 1 r2 5 e2 1r 2 R R 1 r2 3 Finlize To check this result, let s plot P versus R s in Figure The grph shows tht P reches mximum vlue t R 5 r. Eqution (1) shows tht this mximum vlue is P mx 5 e 2 /4r Resistors in eries nd Prllel r 2r 3r When two or more resistors re connected together s re the incndescent lightuls in Figure 28.3, they re sid to e in series comintion. Figure 28.3 is the circuit digrm for the lightuls, shown s resistors, nd the ttery. Wht if you wnted to replce the series comintion with single resistor tht would drw the sme current from the ttery? Wht would e its vlue? In series connection, if n mount of chrge Q exits resistor R 1, chrge Q must lso enter the second resistor R 2. Otherwise, chrge would ccumulte on the wire etween the resistors. Therefore, the sme mount of chrge psses through oth resistors in given time intervl nd the currents re the sme in oth resistors: I 5 I 1 5 I 2 where I is the current leving the ttery, I 1 is the current in resistor R 1, nd I 2 is the current in resistor R 2. The potentil difference pplied cross the series comintion of resistors divides etween the resistors. In Figure 28.3, ecuse the voltge drop 1 from to equls I 1 R 1 nd the voltge drop from to c equls I 2 R 2, the voltge drop from to c is DV 5 DV 1 1 DV 2 5 I 1 R 1 1 I 2 R 2 The potentil difference cross the ttery is lso pplied to the equivlent resistnce R eq in Figure 28.3c: DV 5 IR eq P mx P R 1 The term voltge drop is synonymous with decrese in electric potentil cross resistor. It is often used y individuls working with electric circuits.

180 28.2 Resistors in eries nd Prllel 837 A pictoril representtion of two resistors connected in series to ttery A circuit digrm showing the two resistors connected in series to ttery A circuit digrm showing the equivlent resistnce of the resistors in series V 1 R 1 R 2 V 2 R 1 R 2 c c I 1 I 2 V 1 V 2 R eq R 1 R 2 V I V V c Figure 28.3 Two lightuls with resistnces R 1 nd R 2 connected in series. All three digrms re equivlent. where the equivlent resistnce hs the sme effect on the circuit s the series comintion ecuse it results in the sme current I in the ttery. Comining these equtions for DV gives IR eq 5 I 1 R 1 1 I 2 R 2 R eq 5 R 1 1 R 2 (28.5) where we hve cnceled the currents I, I 1, nd I 2 ecuse they re ll the sme. We see tht we cn replce the two resistors in series with single equivlent resistnce whose vlue is the sum of the individul resistnces. The equivlent resistnce of three or more resistors connected in series is R eq 5 R 1 1 R 2 1 R 3 1??? (28.6) This reltionship indictes tht the equivlent resistnce of series comintion of resistors is the numericl sum of the individul resistnces nd is lwys greter thn ny individul resistnce. Looking ck t Eqution 28.3, we see tht the denomintor of the right-hnd side is the simple lgeric sum of the externl nd internl resistnces. Tht is consistent with the internl nd externl resistnces eing in series in Figure If the filment of one lightul in Figure 28.3 were to fil, the circuit would no longer e complete (resulting in n open-circuit condition) nd the second lightul would lso go out. This fct is generl feture of series circuit: if one device in the series cretes n open circuit, ll devices re inopertive. Q uick Quiz 28.2 With the switch in the circuit of Figure 28.4 closed, there is no current in R 2 ecuse the current hs n lternte zero-resistnce pth through the switch. There is current in R 1, nd this current is mesured with the mmeter ( device for mesuring current) t the ottom of the circuit. If the switch is opened (Fig. 28.4), there is current in R 2. Wht hppens to the reding on the mmeter when the switch is opened? () The reding goes up. () The reding goes down. (c) The reding does not chnge. R 2 A R 1 R 2 A R 1 I I I W Wthe equivlent resistnce of series comintion of resistors Pitfll Prevention 28.2 Lightuls Don t Burn We will descrie the end of the life of n incndescent lightul y sying the filment fils rther thn y sying the lightul urns out. The word urn suggests comustion process, which is not wht occurs in lightul. The filure of lightul results from the slow sulimtion of tungsten from the very hot filment over the life of the lightul. The filment eventully ecomes very thin ecuse of this process. The mechnicl stress from sudden temperture increse when the lightul is turned on cuses the thin filment to rek. Figure 28.4 (Quick Quiz 28.2) Wht hppens when the switch is opened? Pitfll Prevention 28.3 Locl nd Glol Chnges A locl chnge in one prt of circuit my result in glol chnge throughout the circuit. For exmple, if single resistor is chnged in circuit contining severl resistors nd tteries, the currents in ll resistors nd tteries, the terminl voltges of ll tteries, nd the voltges cross ll resistors my chnge s result.

181 838 Chpter 28 Direct-Current Circuits Figure 28.5 Two lightuls with resistnces R 1 nd R 2 connected in prllel. All three digrms re equivlent. A pictoril representtion of two resistors connected in prllel to ttery A circuit digrm showing the two resistors connected in prllel to ttery A circuit digrm showing the equivlent resistnce of the resistors in prllel R 1 V 1 R 1 I 1 I 1 Pitfll Prevention 28.4 Current Does Not Tke the Pth of Lest Resistnce You my hve herd the phrse current tkes the pth of lest resistnce (or similr wording) in reference to prllel comintion of current pths such tht there re two or more pths for the current to tke. uch wording is incorrect. The current tkes ll pths. Those pths with lower resistnce hve lrger currents, ut even very high resistnce pths crry some of the current. In theory, if current hs choice etween zero-resistnce pth nd finite resistnce pth, ll the current tkes the pth of zero resistnce; pth with zero resistnce, however, is n ideliztion. R V 2 2 R R eq R 1 R 2 I 2 I 2 I I I V Now consider two resistors in prllel comintion s shown in Figure As with the series comintion, wht is the vlue of the single resistor tht could replce the comintion nd drw the sme current from the ttery? Notice tht oth resistors re connected directly cross the terminls of the ttery. Therefore, the potentil differences cross the resistors re the sme: DV 5 DV 1 5 DV 2 where DV is the terminl voltge of the ttery. When chrges rech point in Figure 28.5, they split into two prts, with some going towrd R 1 nd the rest going towrd R 2. A junction is ny such point in circuit where current cn split. This split results in less current in ech individul resistor thn the current leving the ttery. Becuse electric chrge is conserved, the current I tht enters point must equl the totl current leving tht point: I 5 I 1 1 I 2 5 DV 1 1 DV 2 R 1 where I 1 is the current in R 1 nd I 2 is the current in R 2. The current in the equivlent resistnce R eq in Figure 28.5c is I 5 DV where the equivlent resistnce hs the sme effect on the circuit s the two resistors in prllel; tht is, the equivlent resistnce drws the sme current I from the ttery. Comining these equtions for I, we see tht the equivlent resistnce of two resistors in prllel is given y DV R eq 5 DV 1 R 1 1 DV 2 R 2 V V c R eq R 2 1 R eq 5 1 R R 2 (28.7) where we hve cnceled DV, DV 1, nd DV 2 ecuse they re ll the sme. An extension of this nlysis to three or more resistors in prllel gives I This expression shows tht the inverse of the equivlent resistnce of two or more resistors in prllel comintion is equl to the sum of the inverses of the indithe equivlent resistnce of prllel comintion of resistors 1 R eq 5 1 R R R 3 1 c (28.8)

182 28.2 Resistors in eries nd Prllel 839 vidul resistnces. Furthermore, the equivlent resistnce is lwys less thn the smllest resistnce in the group. Household circuits re lwys wired such tht the pplinces re connected in prllel. Ech device opertes independently of the others so tht if one is switched off, the others remin on. In ddition, in this type of connection, ll the devices operte on the sme voltge. Let s consider two exmples of prcticl pplictions of series nd prllel circuits. Figure 28.6 illustrtes how three-wy incndescent lightul is constructed to provide three levels of light intensity. 2 The socket of the lmp is equipped with three-wy switch for selecting different light intensities. The lightul contins two filments. When the lmp is connected to 120-V source, one filment receives 100 W of power nd the other receives 75 W. The three light intensities re mde possile y pplying the 120 V to one filment lone, to the other filment lone, or to the two filments in prllel. When switch 1 is closed nd switch 2 is opened, current exists only in the 75-W filment. When switch 1 is open nd switch 2 is closed, current exists only in the 100-W filment. When oth switches re closed, current exists in oth filments nd the totl power is 175 W. If the filments were connected in series nd one of them were to rek, no chrges could pss through the lightul nd it would not glow, regrdless of the switch position. If, however, the filments were connected in prllel nd one of them (for exmple, the 75-W filment) were to rek, the lightul would continue to glow in two of the switch positions ecuse current exists in the other (100-W) filment. As second exmple, consider strings of incndescent lights tht re used for mny ornmentl purposes such s decorting Christms trees. Over the yers, oth prllel nd series connections hve een used for strings of lights. Becuse series-wired lightuls operte with less energy per ul nd t lower temperture, they re sfer thn prllel-wired lightuls for indoor Christms-tree use. If, however, the filment of single lightul in series-wired string were to fil (or if the lightul were removed from its socket), ll the lights on the string would go out. The populrity of series-wired light strings diminished ecuse trouleshooting filed lightul is tedious, time-consuming chore tht involves trilnd-error sustitution of good lightul in ech socket long the string until the defective one is found. In prllel-wired string, ech lightul opertes t 120 V. By design, the lightuls re righter nd hotter thn those on series-wired string. As result, they re inherently more dngerous (more likely to strt fire, for instnce), ut if one lightul in prllel-wired string fils or is removed, the rest of the lightuls continue to glow. To prevent the filure of one lightul from cusing the entire string to go out, new design ws developed for so-clled miniture lights wired in series. When the filment reks in one of these miniture lightuls, the rek in the filment represents the lrgest resistnce in the series, much lrger thn tht of the intct filments. As result, most of the pplied 120 V ppers cross the lightul with the roken filment. Inside the lightul, smll jumper loop covered y n insulting mteril is wrpped round the filment leds. When the filment fils nd 120 V ppers cross the lightul, n rc urns the insultion on the jumper nd connects the filment leds. This connection now completes the circuit through the lightul even though its filment is no longer ctive (Fig. 28.7, pge 840). When lightul fils, the resistnce cross its terminls is reduced to lmost zero ecuse of the lternte jumper connection mentioned in the preceding prgrph. All the other lightuls not only sty on, ut they glow more rightly ecuse W filment 75-W filment 120 V Figure 28.6 A three-wy incndescent lightul. 2 The three-wy lightul nd other household devices ctully operte on lternting current (AC), to e introduced in Chpter 33.

183 Direct-Current Circuits When the filment is intct, chrges flow in the filment. When the filment reks, chrges flow in the jumper connection. I I Filment I Jumper R2 c ly ee.c Glss insultor. Cengge Lerning/George emple Figure 28.7 () chemtic digrm of modern miniture incndescent holidy lightul, with jumper connection to provide current pth if the filment reks. () A holidy lightul with roken filment. (c) A Christms-tree lightul. om 840 Chpter 28 s. w the totl resistnce of the string is reduced nd consequently the current in ech remining lightul increses. Ech lightul opertes t slightly higher temperture thn efore. As more lightuls fil, the current keeps rising, the filment of ech remining lightul opertes t higher temperture, nd the lifetime of the lightul is reduced. For this reson, you should check for filed (nonglowing) lightuls in such series-wired string nd replce them s soon s possile, therey mximizing the lifetimes of ll the lightuls. R1 ys ic A Q uick Quiz 28.3 With the switch in the circuit of Figure 28.8 open, there is no current in R 2. There is current in R 1, however, nd it is mesured with the mmeter t the right side of the circuit. If the switch is closed (Fig. 28.8), there is current in R 2. Wht hppens to the reding on the mmeter when the switch is closed? () The reding increses. () The reding decreses. (c) The reding does not chnge. r ph R2 R1 Q uick Quiz 28.4 Consider the following choices: () increses, () decreses, (c) remins the sme. From these choices, choose the est nswer for the following situtions. (i) In Figure 28.3, third resistor is dded in series with the first two. Wht hppens to the current in the ttery? (ii) Wht hppens to the terminl voltge of the ttery? (iii) In Figure 28.5, third resistor is dded in prllel with the first two. Wht hppens to the current in the ttery? (iv) Wht hppens to the terminl voltge of the ttery?. sw A w w w Figure 28.8 (Quick Quiz 28.3) Wht hppens when the switch is closed? Conceptul Exmple 28.3 Lndscpe Lights A homeowner wishes to instll low-voltge lndscpe lighting in his ck yrd. To sve money, he purchses inexpensive 18-guge cle, which hs reltively high resistnce per unit length. This cle consists of two side-y-side wires seprted y insultion, like the cord on n pplince. He runs 200-foot length of this cle from the power supply to the frthest point t which he plns to position light fixture. He ttches light fixtures cross the two wires on the cle t 10-foot intervls so tht the light fixtures re in prllel. Becuse of the cle s resistnce, the rightness of the lightuls in the fixtures is not s desired. Which of the following prolems does the homeowner hve? () All the lightuls glow eqully less rightly thn they would if lower-resistnce cle hd een used. () The rightness of the lightuls decreses s you move frther from the power supply.

184 28.2 Resistors in eries nd Prllel continued olution A circuit digrm for the system ppers in Figure The horizontl resistors with letter suscripts (such s R A ) represent the resistnce of the wires in the cle etween the light fixtures, nd the verticl resistors with numer suscripts (such s R 1 ) represent the resistnce of the light fixtures themselves. Prt of the terminl voltge of the power supply is dropped cross resistors R A nd R B. Therefore, the voltge cross light fixture R 1 is less thn the terminl voltge. There is further voltge drop cross resistors R C nd R D. Consequently, the voltge cross light fixture R 2 is smller thn tht cross Power supply R 1. This pttern continues down the line of light fixtures, so the correct choice is (). Ech successive light fixture hs smller voltge cross it nd glows less rightly thn the one efore. Exmple 28.4 Find the Equivlent Resistnce Four resistors re connected s shown in Figure (A) Find the equivlent resistnce etween points nd c. olution Conceptulize Imgine chrges flowing into nd through this comintion from the left. All chrges must pss from to through the first two resistors, ut the chrges split t into two different pths when encountering the comintion of the 6.0-V nd the 3.0-V resistors. Ctegorize Becuse of the simple nture of the comintion of resistors in Figure 28.10, we ctegorize this exmple s one for which we cn use the rules for series nd prllel comintions of resistors. R A R B R C Resistnce of the light fixtures R 1 R D Anlyze The comintion of resistors cn e reduced in steps s shown in Figure Find the equivlent resistnce etween nd of the 8.0-V nd 4.0-V resistors, which re in series (left-hnd red-rown circles): Find the equivlent resistnce etween nd c of the 6.0-V nd 3.0-V resistors, which re in prllel (righthnd red-rown circles): The circuit of equivlent resistnces now looks like Figure The 12.0-V nd 2.0-V resistors re in series (green circles). Find the equivlent resistnce from to c: R 2 Resistnce in the wires of the cle Figure 28.9 (Conceptul Exmple 28.3) The circuit digrm for set of lndscpe light fixtures connected in prllel cross the two wires of two-wire cle. Figure (Exmple 28.4) The originl network of resistors is reduced to single equivlent resistnce. R eq V V V R eq 6.0 V V V c 8.0 R eq V V V 12.0 I R eq V 3 This resistnce is tht of the single equivlent resistor in Figure 28.10c V 6.0 I 1 I c c c (B) Wht is the current in ech resistor if potentil difference of 42 V is mintined etween nd c? continued

185 842 Chpter 28 Direct-Current Circuits 28.4 continued olution The currents in the 8.0-V nd 4.0-V resistors re the sme ecuse they re in series. In ddition, they crry the sme current tht would exist in the 14.0-V equivlent resistor suject to the 42-V potentil difference. Use Eqution 27.7 (R 5 DV/I) nd the result from prt (A) to find the current in the 8.0-V nd 4.0-V resistors: et the voltges cross the resistors in prllel in Figure equl to find reltionship etween the currents: Use I 1 1 I A to find I 1 : Find I 2 : I 5 DV c R eq 5 42 V 14.0 V A DV 1 5 DV 2 (6.0 V)I 1 5 (3.0 V)I 2 I 2 5 2I 1 I 1 1 I A I 1 1 2I A I A I 2 5 2I 1 5 2(1.0 A) A Finlize As finl check of our results, note tht DV c 5 (6.0 V)I 1 5 (3.0 V)I V nd DV 5 (12.0 V)I 5 36 V; therefore, DV c 5 DV 1 DV c 5 42 V, s it must. Exmple 28.5 Three Resistors in Prllel Three resistors re connected in prllel s shown in Figure A potentil difference of 18.0 V is mintined etween points nd. (A) Clculte the equivlent resistnce of the circuit. olution Conceptulize Figure shows tht we re deling with simple prllel comintion of three resistors. Notice tht the current I splits into three currents I 1, I 2, nd I 3 in the three resistors. I 1 I 2 I 3 Ctegorize This prolem cn e solved with rules developed in this section, so we ctegorize it s sustitution prolem. Becuse the three resistors re connected in prllel, we cn use the rule for resistors in prllel, Eqution 28.8, to evlute the equivlent resistnce. Use Eqution 28.8 to find R eq : (B) Find the current in ech resistor V I I 1 I 2 I V 9.00 Figure (Exmple 28.5) () Three resistors connected in prllel. The voltge cross ech resistor is 18.0 V. () Another circuit with three resistors nd ttery. Is it equivlent to the circuit in ()? R eq 3.00 V V V V R eq V V 11 I olution The potentil difference cross ech resistor is 18.0 V. Apply the reltionship DV 5 IR to find the currents: I 1 5 DV R V 3.00 V A I 2 5 DV R V 6.00 V A I 3 5 DV R V 9.00 V A (C) Clculte the power delivered to ech resistor nd the totl power delivered to the comintion of resistors.

186 Kirchhoff s Rules 28.5 c o n t i n u e d olution Apply the reltionship P 5 I 2R to ech resistor using the currents clculted in prt (B): 3.00-V: P 1 5 I12R 1 5 (6.00 A)2(3.00 V) W 6.00-V: P 2 5 I22R 2 5 (3.00 A)2(6.00 V) 5 54 W 9.00-V: P 3 5 I 32R 3 5 (2.00 A)2(9.00 V) 5 36 W om These results show tht the smllest resistor receives the most power. umming the three quntities gives totl power of 198 W. We could hve clculted this finl result from prt (A) y considering the equivlent resistnce s follows: P 5 (DV )2/R eq 5 (18.0 V)2/1.64 V W. W ht if the circuit were s shown in Figure insted of s in Figure 28.11? How would tht ffect the clcultion? W h t I f? ee ly.c Answer There would e no effect on the clcultion. The physicl plcement of the ttery is not importnt. Only the electricl rrngement is importnt. In Figure 28.11, the ttery still mintins potentil difference of 18.0 V etween points nd, so the two circuits in the figure re electriclly identicl. s. w 28.3 Kirchhoff s Rules ys ic As we sw in the preceding section, comintions of resistors cn e simplified nd nlyzed using the expression DV 5 IR nd the rules for series nd prllel comintions of resistors. Very often, however, it is not possile to reduce circuit to single loop using these rules. The procedure for nlyzing more complex circuits is mde possile y using the following two principles, clled Kirchhoff s rules. 1. Junction rule. At ny junction, the sum of the currents must equl zero: r ph I50 (28.9) junction sw 2. Loop rule. The sum of the potentil differences cross ll elements round ny closed circuit loop must e zero: DV 5 0 The mount of chrge flowing out of the rnches on the right must equl the mount flowing into the single rnch on the left. (28.10) I1. closed loop w w w Kirchhoff s first rule is sttement of conservtion of electric chrge. All chrges tht enter given point in circuit must leve tht point ecuse chrge cnnot uild up or dispper t point. Currents directed into the junction re entered into the sum in the junction rule s 1I, wheres currents directed out of junction re entered s 2I. Applying this rule to the junction in Figure gives I1 2 I2 2 I3 5 0 Figure represents mechnicl nlog of this sitution, in which wter flows through rnched pipe hving no leks. Becuse wter does not uild up nywhere in the pipe, the flow rte into the pipe on the left equls the totl flow rte out of the two rnches on the right. Kirchhoff s second rule follows from the lw of conservtion of energy for n isolted system. Let s imgine moving chrge round closed loop of circuit. When the chrge returns to the strting point, the chrge circuit system must hve the sme totl energy s it hd efore the chrge ws moved. The sum of the increses in energy s the chrge psses through some circuit elements must equl the sum of the decreses in energy s it psses through other elements. The potentil energy of the system decreses whenever the chrge moves through potentil drop 2IR cross resistor or whenever it moves in the reverse direction through I2 I3 The mount of wter flowing out of the rnches on the right must equl the mount flowing into the single rnch on the left. Flow in Flow out Figure () Kirchhoff s junction rule. () A mechnicl nlog of the junction rule.

187 844 Chpter 28 Direct-Current Circuits In ech digrm, V V V nd the circuit element is trversed from to, left to right. c d I V IR I V IR e V e e V e Figure Rules for determining the signs of the potentil differences cross resistor nd ttery. (The ttery is ssumed to hve no internl resistnce.) Gustv Kirchhoff Germn Physicist ( ) Kirchhoff, professor t Heidelerg, nd Roert Bunsen invented the spectroscope nd founded the science of spectroscopy, which we shll study in Chpter 42. They discovered the elements cesium nd ruidium nd invented stronomicl spectroscopy. Prints & Photogrphs Division, Lirry of Congress, LC-UZ source of emf. The potentil energy increses whenever the chrge psses through ttery from the negtive terminl to the positive terminl. When pplying Kirchhoff s second rule, imgine trveling round the loop nd consider chnges in electric potentil rther thn the chnges in potentil energy descried in the preceding prgrph. Imgine trveling through the circuit elements in Figure towrd the right. The following sign conventions pply when using the second rule: Chrges move from the high-potentil end of resistor towrd the lowpotentil end, so if resistor is trversed in the direction of the current, the potentil difference DV cross the resistor is 2IR (Fig ). If resistor is trversed in the direction opposite the current, the potentil difference DV cross the resistor is 1IR (Fig ). If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction of the emf (from negtive to positive), the potentil difference DV is 1e (Fig c). If source of emf (ssumed to hve zero internl resistnce) is trversed in the direction opposite the emf (from positive to negtive), the potentil difference DV is 2e (Fig d). There re limits on the numer of times you cn usefully pply Kirchhoff s rules in nlyzing circuit. You cn use the junction rule s often s you need s long s you include in it current tht hs not een used in preceding junction-rule eqution. In generl, the numer of times you cn use the junction rule is one fewer thn the numer of junction points in the circuit. You cn pply the loop rule s often s needed s long s new circuit element (resistor or ttery) or new current ppers in ech new eqution. In generl, to solve prticulr circuit prolem, the numer of independent equtions you need to otin from the two rules equls the numer of unknown currents. Complex networks contining mny loops nd junctions generte gret numer of independent liner equtions nd correspondingly gret numer of unknowns. uch situtions cn e hndled formlly through the use of mtrix lger. Computer softwre cn lso e used to solve for the unknowns. The following exmples illustrte how to use Kirchhoff s rules. In ll cses, it is ssumed the circuits hve reched stedy-stte conditions; in other words, the currents in the vrious rnches re constnt. Any cpcitor cts s n open rnch in circuit; tht is, the current in the rnch contining the cpcitor is zero under stedy-stte conditions. Prolem-olving trtegy Kirchhoff s Rules The following procedure is recommended for solving prolems tht involve circuits tht cnnot e reduced y the rules for comining resistors in series or prllel. 1. Conceptulize. tudy the circuit digrm nd mke sure you recognize ll elements in the circuit. Identify the polrity of ech ttery nd try to imgine the directions in which the current would exist in the tteries. 2. Ctegorize. Determine whether the circuit cn e reduced y mens of comining series nd prllel resistors. If so, use the techniques of ection If not, pply Kirchhoff s rules ccording to the Anlyze step elow. 3. Anlyze. Assign lels to ll known quntities nd symols to ll unknown quntities. You must ssign directions to the currents in ech prt of the circuit. Although the ssignment of current directions is ritrry, you must dhere rigorously to the directions you ssign when you pply Kirchhoff s rules. Apply the junction rule (Kirchhoff s first rule) to ll junctions in the circuit except one. Now pply the loop rule (Kirchhoff s second rule) to s mny loops in

188 28.3 Kirchhoff s Rules 845 Prolem-olving trtegy continued the circuit s re needed to otin, in comintion with the equtions from the junction rule, s mny equtions s there re unknowns. To pply this rule, you must choose direction in which to trvel round the loop (either clockwise or counterclockwise) nd correctly identify the chnge in potentil s you cross ech element. Be creful with signs! olve the equtions simultneously for the unknown quntities. 4. Finlize. Check your numericl nswers for consistency. Do not e lrmed if ny of the resulting currents hve negtive vlue. Tht only mens you hve guessed the direction of tht current incorrectly, ut its mgnitude will e correct. Exmple 28.6 A ingle-loop Circuit A single-loop circuit contins two resistors nd two tteries s shown in Figure (Neglect the internl resistnces of the tteries.) Find the current in the circuit. olution Conceptulize Figure shows the polrities of the tteries nd guess t the direction of the current. The 12-V ttery is the stronger of the two, so the current should e counterclockwise. Therefore, we expect our guess for the direction of the current to e wrong, ut we will continue nd see how this incorrect guess is represented y our finl nswer. Ctegorize We do not need Kirchhoff s rules to nlyze this simple circuit, ut let s use them nywy simply to see how they re pplied. There re no junctions in this single-loop circuit; therefore, the current is the sme in ll elements. Anlyze Let s ssume the current is clockwise s shown in Figure Trversing the circuit in the clockwise direction, strting t, we see tht represents potentil difference of 1e 1, c represents potentil difference of 2IR 1, c d represents potentil difference of 2e 2, nd d represents potentil difference of 2IR 2. Apply Kirchhoff s loop rule to the single loop in the circuit: olve for I nd use the vlues given in Figure 28.14: (1) I 5 e 1 2 e 2 R 1 1 R 2 5 d e V R 2 10 I R e 2 12 V Figure (Exmple 28.6) A series circuit contining two tteries nd two resistors, where the polrities of the tteries re in opposition. o DV 5 0 e 1 2 IR 1 2 e 2 2 IR V 2 12 V 8.0 V 1 10 V A Finlize The negtive sign for I indictes tht the direction of the current is opposite the ssumed direction. The emfs in the numertor sutrct ecuse the tteries in Figure hve opposite polrities. The resistnces in the denomintor dd ecuse the two resistors re in series. Wht If? Wht if the polrity of the 12.0-V ttery were reversed? How would tht ffect the circuit? Answer Although we could repet the Kirchhoff s rules clcultion, let s insted exmine Eqution (1) nd modify it ccordingly. Becuse the polrities of the two tteries re now in the sme direction, the signs of e 1 nd e 2 re the sme nd Eqution (1) ecomes I 5 e 1 1 e 2 R 1 1 R V 1 12 V 8.0 V 1 10 V A c Exmple 28.7 A Multiloop Circuit Find the currents I 1, I 2, nd I 3 in the circuit shown in Figure on pge 846. continued

189 846 Chpter 28 Direct-Current Circuits 28.7 continued olution Conceptulize Imgine physiclly rerrnging the circuit while keeping it electriclly the sme. Cn you rerrnge it so tht it consists of simple series or prllel comintions of resistors? You should find tht you cnnot. (If the 10.0-V ttery were removed nd replced y wire from to the 6.0-V resistor, the circuit would consist of only series nd prllel comintions.) Ctegorize We cnnot simplify the circuit y the rules ssocited with comining resistnces in series nd in prllel. Therefore, this prolem is one in which we must use Kirchhoff s rules. Figure (Exmple 28.7) A circuit contining different rnches. Anlyze We ritrrily choose the directions of the currents s leled in Figure Apply Kirchhoff s junction rule to junction c : (1) I 1 1 I 2 2 I We now hve one eqution with three unknowns: I 1, I 2, nd I 3. There re three loops in the circuit: cd, efc, nd efd. We need only two loop equtions to determine the unknown currents. (The third eqution would give no new informtion.) Let s choose to trverse these loops in the clockwise direction. Apply Kirchhoff s loop rule to loops cd nd efc: Use Eqution (1) to find I 3 : e V 14.0 V cd: (2) 10.0 V 2 (6.0 V)I 1 2 (2.0 V)I efc: 2(4.0 V)I V 1 (6.0 V)I V 5 0 (3) V 1 (6.0 V)I 1 2 (4.0 V)I olve Eqution (1) for I 3 nd sustitute into Eqution (2): 10.0 V 2 (6.0 V)I 1 2 (2.0 V)(I 1 1 I 2 ) 5 0 Multiply ech term in Eqution (3) y 4 nd ech term in Eqution (4) y 3: Add Eqution (6) to Eqution (5) to eliminte I 1 nd find I 2 : (4) 10.0 V 2 (8.0 V)I 1 2 (2.0 V)I (5) V 1 (24.0 V)I 1 2 (16.0 V)I (6) 30.0 V 2 (24.0 V)I 1 2 (6.0 V)I V 2 (22.0 V)I I A Use this vlue of I 2 in Eqution (3) to find I 1 : V 1 (6.0 V)I 1 2 (4.0 V)(23.0 A) V 1 (6.0 V)I V 5 0 I A I 3 5 I 1 1 I A A A Finlize Becuse our vlues for I 2 nd I 3 re negtive, the directions of these currents re opposite those indicted in Figure The numericl vlues for the currents re correct. Despite the incorrect direction, we must continue to use these negtive vlues in susequent clcultions ecuse our equtions were estlished with our originl choice of direction. Wht would hve hppened hd we left the current directions s leled in Figure ut trversed the loops in the opposite direction? I 1 f c d I 2 I RC Circuits o fr, we hve nlyzed direct-current circuits in which the current is constnt. In DC circuits contining cpcitors, the current is lwys in the sme direction ut my vry in mgnitude t different times. A circuit contining series comintion of resistor nd cpcitor is clled n RC circuit.

190 Chrging Cpcitor Figure shows simple series RC circuit. Let s ssume the cpcitor in this circuit is initilly unchrged. There is no current while the switch is open (Fig ). If the switch is thrown to position t t 5 0 (Fig ), however, chrge egins to flow, setting up current in the circuit, nd the cpcitor egins to chrge. 3 Notice tht during chrging, chrges do not jump cross the cpcitor pltes ecuse the gp etween the pltes represents n open circuit. Insted, chrge is trnsferred etween ech plte nd its connecting wires due to the electric field estlished in the wires y the ttery until the cpcitor is fully chrged. As the pltes re eing chrged, the potentil difference cross the cpcitor increses. The vlue of the mximum chrge on the pltes depends on the voltge of the ttery. Once the mximum chrge is reched, the current in the circuit is zero ecuse the potentil difference cross the cpcitor mtches tht supplied y the ttery. To nlyze this circuit quntittively, let s pply Kirchhoff s loop rule to the circuit fter the switch is thrown to position. Trversing the loop in Figure clockwise gives e 2 q 2 ir 5 0 (28.11) C where q/c is the potentil difference cross the cpcitor nd ir is the potentil difference cross the resistor. We hve used the sign conventions discussed erlier for the signs on e nd ir. The cpcitor is trversed in the direction from the positive plte to the negtive plte, which represents decrese in potentil. Therefore, we use negtive sign for this potentil difference in Eqution Note tht lowercse q nd i re instntneous vlues tht depend on time (s opposed to stedy-stte vlues) s the cpcitor is eing chrged. We cn use Eqution to find the initil current I i in the circuit nd the mximum chrge Q mx on the cpcitor. At the instnt the switch is thrown to position (t 5 0), the chrge on the cpcitor is zero. Eqution shows tht the initil current I i in the circuit is mximum nd is given y e I i 5 1current t t 5 02 (28.12) R At this time, the potentil difference from the ttery terminls ppers entirely cross the resistor. Lter, when the cpcitor is chrged to its mximum vlue Q mx, chrges cese to flow, the current in the circuit is zero, nd the potentil difference from the ttery terminls ppers entirely cross the cpcitor. ustituting i 5 0 into Eqution gives the mximum chrge on the cpcitor: Q mx 5 Ce (mximum chrge) (28.13) To determine nlyticl expressions for the time dependence of the chrge nd current, we must solve Eqution 28.11, single eqution contining two vriles q nd i. The current in ll prts of the series circuit must e the sme. Therefore, the current in the resistnce R must e the sme s the current etween ech cpcitor plte nd the wire connected to it. This current is equl to the time rte of chnge of the chrge on the cpcitor pltes. Therefore, we sustitute i 5 dq/dt into Eqution nd rerrnge the eqution: dq e dt 5 R 2 q RC To find n expression for q, we solve this seprle differentil eqution s follows. First comine the terms on the right-hnd side: dq Ce dt 5 RC 2 q RC 5 2 q 2 Ce RC 3 In previous discussions of cpcitors, we ssumed stedy-stte sitution, in which no current ws present in ny rnch of the circuit contining cpcitor. Now we re considering the cse efore the stedy-stte condition is relized; in this sitution, chrges re moving nd current exists in the wires connected to the cpcitor RC Circuits 847 e e C When the switch is thrown to position, the cpcitor egins to chrge up. e C When the switch is thrown to position, the cpcitor dischrges. c i C Figure A cpcitor in series with resistor, switch, nd ttery. i R R R

191 848 Chpter 28 Direct-Current Circuits Chrge s function of time for cpcitor eing chrged Current s function of time for cpcitor eing chrged Multiply this eqution y dt nd divide y q 2 Ce: dq q 2 Ce RC dt Integrte this expression, using q 5 0 t t 5 0: 3 0 q dq q 2 Ce RC 3 t 0 dt ln q 2 Ce 2Ce 5 2 t RC From the definition of the nturl logrithm, we cn write this expression s q(t) 5 Ce(1 2 e 2t/RC ) 5 Q mx (1 2 e 2t/RC ) (28.14) where e is the se of the nturl logrithm nd we hve mde the sustitution from Eqution We cn find n expression for the chrging current y differentiting Eqution with respect to time. Using i 5 dq/dt, we find tht i 1t2 5 e R e2t/rc (28.15) Plots of cpcitor chrge nd circuit current versus time re shown in Figure Notice tht the chrge is zero t t 5 0 nd pproches the mximum vlue Ce s t `. The current hs its mximum vlue I i 5 e/r t t 5 0 nd decys exponentilly to zero s t `. The quntity RC, which ppers in the exponents of Equtions nd 28.15, is clled the time constnt t of the circuit: t 5 RC (28.16) The time constnt represents the time intervl during which the current decreses to 1/e of its initil vlue; tht is, fter time intervl t, the current decreses to i 5 e 21 I i I i. After time intervl 2t, the current decreses to i 5 e 22 I i I i, nd so forth. Likewise, in time intervl t, the chrge increses from zero to Ce[1 2 e 21 ] Ce. The chrge pproches its mximum vlue Ce s t pproches infinity. C 0.632C q e e =RC t The current hs its mximum vlue I i e/r t t 0 nd decys to zero exponentilly s t pproches infinity. I i 0.368I i i I i = R e t After time intervl equl to one time constnt t hs pssed, the chrge is 63.2% of the mximum vlue Ce. After time intervl equl to one time constnt t hs pssed, the current is 36.8% of its initil vlue. Figure () Plot of cpcitor chrge versus time for the circuit shown in Figure () Plot of current versus time for the circuit shown in Figure

192 28.4 RC Circuits 849 The following dimensionl nlysis shows tht t hs units of time: 3t4 5 3RC4 5 c DV I Q DV d 5 c Q Q /Dt d 5 3Dt4 5 T Becuse t 5 RC hs units of time, the comintion t/rc is dimensionless, s it must e to e n exponent of e in Equtions nd The energy supplied y the ttery during the time intervl required to fully chrge the cpcitor is Q mx e 5 Ce 2. After the cpcitor is fully chrged, the energy stored in the cpcitor is 1 2Q mx e 5 1 2Ce 2, which is only hlf the energy output of the ttery. It is left s prolem (Prolem 68) to show tht the remining hlf of the energy supplied y the ttery ppers s internl energy in the resistor. Dischrging Cpcitor Imgine tht the cpcitor in Figure is completely chrged. An initil potentil difference Q i /C exists cross the cpcitor, nd there is zero potentil difference cross the resistor ecuse i 5 0. If the switch is now thrown to position t t 5 0 (Fig c), the cpcitor egins to dischrge through the resistor. At some time t during the dischrge, the current in the circuit is i nd the chrge on the cpcitor is q. The circuit in Figure 28.16c is the sme s the circuit in Figure except for the sence of the ttery. Therefore, we eliminte the emf e from Eqution to otin the pproprite loop eqution for the circuit in Figure 28.16c: 2 q 2 ir 5 0 (28.17) C When we sustitute i 5 dq/dt into this expression, it ecomes 2R dq dt 5 q C dq q RC dt Integrting this expression using q 5 Q i t t 5 0 gives q dq 3 Q i q RC 3 dt 0 ln q 5 2 t Q i RC t q 1t2 5 Q i e 2t/RC (28.18) Differentiting Eqution with respect to time gives the instntneous current s function of time: i 1t2 5 2 Q i RC e2t/rc (28.19) where Q i /RC 5 I i is the initil current. The negtive sign indictes tht s the cpcitor dischrges, the current direction is opposite its direction when the cpcitor ws eing chrged. (Compre the current directions in Figs nd 28.16c.) Both the chrge on the cpcitor nd the current decy exponentilly t rte chrcterized y the time constnt t 5 RC. W WChrge s function of time for dischrging cpcitor W WCurrent s function of time for dischrging cpcitor e R C R Q uick Quiz 28.5 Consider the circuit in Figure nd ssume the ttery hs no internl resistnce. (i) Just fter the switch is closed, wht is the current in the ttery? () 0 () e/2r (c) 2e/R (d) e/r (e) impossile to determine (ii) After very long time, wht is the current in the ttery? Choose from the sme choices. Figure (Quick Quiz 28.5) How does the current vry fter the switch is closed?

193 850 Chpter 28 Direct-Current Circuits Conceptul Exmple 28.8 Intermittent Windshield Wipers Mny utomoiles re equipped with windshield wipers tht cn operte intermittently during light rinfll. How does the opertion of such wipers depend on the chrging nd dischrging of cpcitor? olution The wipers re prt of n RC circuit whose time constnt cn e vried y selecting different vlues of R through multiposition switch. As the voltge cross the cpcitor increses, the cpcitor reches point t which it dischrges nd triggers the wipers. The circuit then egins nother chrging cycle. The time intervl etween the individul sweeps of the wipers is determined y the vlue of the time constnt. Exmple 28.9 Chrging Cpcitor in n RC Circuit An unchrged cpcitor nd resistor re connected in series to ttery s shown in Figure 28.16, where e V, C mf, nd R V. The switch is thrown to position. Find the time constnt of the circuit, the mximum chrge on the cpcitor, the mximum current in the circuit, nd the chrge nd current s functions of time. olution Conceptulize tudy Figure nd imgine throwing the switch to position s shown in Figure Upon doing so, the cpcitor egins to chrge. Ctegorize We evlute our results using equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Evlute the time constnt of the circuit from Eqution 28.16: Evlute the mximum chrge on the cpcitor from Eqution 28.13: Evlute the mximum current in the circuit from Eqution 28.12: Use these vlues in Equtions nd to find the chrge nd current s functions of time: Exmple Dischrging Cpcitor in n RC Circuit t 5 RC 5 ( V)( F) s Q mx 5 Ce 5 (5.00 mf)(12.0 V) mc e I i 5 R V ma V (1) q 1t e 2t/ (2) i 1t e 2t/4.00 In Equtions (1) nd (2), q is in microcouloms, i is in micromperes, nd t is in seconds. Consider cpcitor of cpcitnce C tht is eing dischrged through resistor of resistnce R s shown in Figure 28.16c. (A) After how mny time constnts is the chrge on the cpcitor one-fourth its initil vlue? olution Conceptulize tudy Figure nd imgine throwing the switch to position s shown in Figure 28.16c. Upon doing so, the cpcitor egins to dischrge. Ctegorize We ctegorize the exmple s one involving dischrging cpcitor nd use the pproprite equtions.

194 28.4 RC Circuits continued Anlyze ustitute q(t) 5 Q i /4 into Eqution 28.18: Tke the logrithm of oth sides of the eqution nd solve for t: Use Equtions nd to express the energy stored in the cpcitor t ny time t: ustitute U 1t Q i 2 /2C2 into Eqution (1): Exmple Energy Delivered to Resistor A 5.00-mF cpcitor is chrged to potentil difference of 800 V nd then dischrged through resistor. How much energy is delivered to the resistor in the time intervl required to fully dischrge the cpcitor? olution Q i 4 5 Q ie 2t/RC e 2t/RC 2ln t RC t 5 RC ln RC t (B) The energy stored in the cpcitor decreses with time s the cpcitor dischrges. After how mny time constnts is this stored energy one-fourth its initil vlue? olution Tke the logrithm of oth sides of the eqution nd solve for t: (1) U 1t 2 5 q 2 2C 5 Q i Conceptulize In Exmple 28.10, we considered the energy decrese in dischrging cpcitor to vlue of onefourth the initil energy. In this exmple, the cpcitor fully dischrges. Ctegorize We solve this exmple using two pproches. The first pproch is to model the circuit s n isolted system for energy. Becuse energy in n isolted system is conserved, the initil electric potentil energy U E stored in the continued Q 2 1 i 4 1 AM 2C 5 Q i 4 5 e 22t/RC 2 2C e22t/rc 2ln t RC 2 2C e22t/rc t 5 1 2RC ln RC t Finlize Notice tht ecuse the energy depends on the squre of the chrge, the energy in the cpcitor drops more rpidly thn the chrge on the cpcitor. Wht If? Wht if you wnt to descrie the circuit in terms of the time intervl required for the chrge to fll to one-hlf its originl vlue rther thn y the time constnt t? Tht would give prmeter for the circuit clled its hlflife t 1/2. How is the hlf-life relted to the time constnt? Answer In one hlf-life, the chrge flls from Q i to Q i /2. Therefore, from Eqution 28.18, which leds to Q i 2 5 Q ie 2t 1/2/RC t 1/ t e 2t 1/2/RC The concept of hlf-life will e importnt to us when we study nucler decy in Chpter 44. The rdioctive decy of n unstle smple ehves in mthemticlly similr mnner to dischrging cpcitor in n RC circuit.

195 852 Chpter 28 Direct-Current Circuits continued cpcitor is trnsformed into internl energy E int 5 E R in the resistor. The second pproch is to model the resistor s nonisolted system for energy. Energy enters the resistor y electricl trnsmission from the cpcitor, cusing n increse in the resistor s internl energy. Anlyze We egin with the isolted system pproch. Write the pproprite reduction of the conservtion of energy eqution, Eqution 8.2: ustitute the initil nd finl vlues of the energies: Use Eqution for the electric potentil energy in the cpcitor: ustitute numericl vlues: DU 1 DE int 5 0 (0 2 U E ) 1 (E int 2 0) 5 0 E R 5 U E E R 5 1 2Ce 2 E R F V J The second pproch, which is more difficult ut perhps more instructive, is to note tht s the cpcitor dischrges through the resistor, the rte t which energy is delivered to the resistor y electricl trnsmission is i 2 R, where i is the instntneous current given y Eqution Evlute the energy delivered to the resistor y integrting the power over ll time ecuse it tkes n infinite time intervl for the cpcitor to completely dischrge: ustitute for the power delivered to the resistor: ustitute for the current from Eqution 28.19: ustitute the vlue of the integrl, which is RC/2 (see Prolem 44): P 5 de dt ` E R 5 3 i 2 R dt 0 ` E R Q i RC e2t/rc 2R dt 5 Q i 2 ` RC 2 3 e 22t/RC dt 5 e2 ` R 3 e 22t/RC dt 0 E R 5 e2 R RC ` E R 5 3 P dt Ce 2 Finlize This result grees with tht otined using the isolted system pproch, s it must. We cn use this second pproch to find the totl energy delivered to the resistor t ny time fter the switch is closed y simply replcing the upper limit in the integrl with tht specific vlue of t Household Wiring nd Electricl fety Mny considertions re importnt in the design of n electricl system of home tht will provide dequte electricl service for the occupnts while mximizing their sfety. We discuss some spects of home electricl system in this section. Household Wiring Household circuits represent prcticl ppliction of some of the ides presented in this chpter. In our world of electricl pplinces, it is useful to understnd the power requirements nd limittions of conventionl electricl systems nd the sfety mesures tht prevent ccidents. In conventionl instlltion, the utility compny distriutes electric power to individul homes y mens of pir of wires, with ech home connected in prl- 0 0

196 28.5 Household Wiring nd Electricl fety 853 lel to these wires. One wire is clled the live wire 4 s illustrted in Figure 28.19, nd the other is clled the neutrl wire. The neutrl wire is grounded; tht is, its electric potentil is tken to e zero. The potentil difference etween the live nd neutrl wires is pproximtely 120 V. This voltge lterntes in time, nd the potentil of the live wire oscilltes reltive to ground. Much of wht we hve lerned so fr for the constnt-emf sitution (direct current) cn lso e pplied to the lternting current tht power compnies supply to usinesses nd households. (Alternting voltge nd current re discussed in Chpter 33.) To record household s energy consumption, meter is connected in series with the live wire entering the house. After the meter, the wire splits so tht there re severl seprte circuits in prllel distriuted throughout the house. Ech circuit contins circuit reker (or, in older instlltions, fuse). A circuit reker is specil switch tht opens if the current exceeds the rted vlue for the circuit reker. The wire nd circuit reker for ech circuit re crefully selected to meet the current requirements for tht circuit. If circuit is to crry currents s lrge s 30 A, hevy wire nd n pproprite circuit reker must e selected to hndle this current. A circuit used to power only lmps nd smll pplinces often requires only 20 A. Ech circuit hs its own circuit reker to provide protection for tht prt of the entire electricl system of the house. As n exmple, consider circuit in which toster oven, microwve oven, nd coffee mker re connected (corresponding to R 1, R 2, nd R 3 in Fig ). We cn clculte the current in ech pplince y using the expression P 5 I DV. The toster oven, rted t W, drws current of W/120 V A. The microwve oven, rted t W, drws 10.8 A, nd the coffee mker, rted t 800 W, drws 6.67 A. When the three pplinces re operted simultneously, they drw totl current of 25.8 A. Therefore, the circuit must e wired to hndle t lest this much current. If the rting of the circuit reker protecting the circuit is too smll sy, 20 A the reker will e tripped when the third pplince is turned on, preventing ll three pplinces from operting. To void this sitution, the toster oven nd coffee mker cn e operted on one 20-A circuit nd the microwve oven on seprte 20-A circuit. Mny hevy-duty pplinces such s electric rnges nd clothes dryers require 240 V for their opertion. The power compny supplies this voltge y providing third wire tht is 120 V elow ground potentil (Fig ). The potentil difference etween this live wire nd the other live wire (which is 120 V ove ground potentil) is 240 V. An pplince tht opertes from 240-V line requires hlf s much current compred with operting it t 120 V; therefore, smller wires cn e used in the higher-voltge circuit without overheting. Electricl fety When the live wire of n electricl outlet is connected directly to ground, the circuit is completed nd short-circuit condition exists. A short circuit occurs when lmost zero resistnce exists etween two points t different potentils, nd the result is very lrge current. When tht hppens ccidentlly, properly operting circuit reker opens the circuit nd no dmge is done. A person in contct with ground, however, cn e electrocuted y touching the live wire of fryed cord or other exposed conductor. An exceptionlly effective (nd dngerous!) ground contct is mde when the person either touches wter pipe (normlly t ground potentil) or stnds on the ground with wet feet. The ltter sitution represents effective ground contct ecuse norml, nondistilled wter is conductor due to the lrge numer of ions ssocited with impurities. This sitution should e voided t ll cost. The electricl meter mesures the power in wtts. Live Neutrl R 1 W Electricl meter Circuit reker R 2 R V 0 V Figure Wiring digrm for household circuit. The resistnces represent pplinces or other electricl devices tht operte with n pplied voltge of 120 V. 120 V. Cengge Lerning/George emple 120 V Figure () An outlet for connection to 240-V supply. () The connections for ech of the openings in 240-V outlet. 4 Live wire is common expression for conductor whose electric potentil is ove or elow ground potentil.

197 854 Chpter 28 Direct-Current Circuits Figure () A digrm of the circuit for n electric drill with only two connecting wires. The norml current pth is from the live wire through the motor connections nd ck to ground through the neutrl wire. () This shock cn e voided y connecting the drill cse to ground through third ground wire. The wire colors represent electricl stndrds in the United ttes: the hot wire is lck, the ground wire is green, nd the neutrl wire is white (shown s gry in the figure). In the sitution shown, the live wire hs come into contct with the drill cse. As result, the person holding the drill cts s current pth to ground nd receives n electric shock. Ouch! Motor I Hot Wll outlet Circuit reker Neutrl In this sitution, the drill cse remins t ground potentil nd no current exists in the person. I Motor I I Hot 3-wire outlet 120 V Ground Electric shock cn result in ftl urns or cn cuse the muscles of vitl orgns such s the hert to mlfunction. The degree of dmge to the ody depends on the mgnitude of the current, the length of time it cts, the prt of the ody touched y the live wire, nd the prt of the ody in which the current exists. Currents of 5 ma or less cuse senstion of shock, ut ordinrily do little or no dmge. If the current is lrger thn out 10 ma, the muscles contrct nd the person my e unle to relese the live wire. If the ody crries current of out 100 ma for only few seconds, the result cn e ftl. uch lrge current prlyzes the respirtory muscles nd prevents rething. In some cses, currents of pproximtely 1 A cn produce serious (nd sometimes ftl) urns. In prctice, no contct with live wires is regrded s sfe whenever the voltge is greter thn 24 V. Mny 120-V outlets re designed to ccept three-pronged power cord. (This feture is required in ll new electricl instlltions.) One of these prongs is the live wire t nominl potentil of 120 V. The second is the neutrl wire, nominlly t 0 V, which crries current to ground. Figure shows connection to n electric drill with only these two wires. If the live wire ccidentlly mkes contct with the csing of the electric drill (which cn occur if the wire insultion wers off), current cn e crried to ground y wy of the person, resulting in n electric shock. The third wire in three-pronged power cord, the round prong, is sfety ground wire tht normlly crries no current. It is oth grounded nd connected directly to the csing of the pplince. If the live wire is ccidentlly shorted to the csing in this sitution, most of the current tkes the low-resistnce pth through the pplince to ground s shown in Figure pecil power outlets clled ground-fult circuit interrupters, or GFCIs, re used in kitchens, throoms, sements, exterior outlets, nd other hzrdous res of homes. These devices re designed to protect persons from electric shock y sensing smll currents (, 5 ma) leking to ground. (The principle of their opertion I Neutrl I Ground Circuit reker 120 V Ground I

198 Ojective Questions 855 is descried in Chpter 31.) When n excessive lekge current is detected, the current is shut off in less thn 1 ms. ummry Definition The emf of ttery is equl to the voltge cross its terminls when the current is zero. Tht is, the emf is equivlent to the open-circuit voltge of the ttery. Concepts nd Principles The equivlent resistnce of set of resistors connected in series comintion is R eq 5 R 1 1 R 2 1 R 3 1??? (28.6) The equivlent resistnce of set of resistors connected in prllel comintion is found from the reltionship 1 R eq 5 1 R R R 3 1 c (28.8) If cpcitor is chrged with ttery through resistor of resistnce R, the chrge on the cpcitor nd the current in the circuit vry in time ccording to the expressions q(t) 5 Q mx (1 2 e 2t/RC ) (28.14) i 1t 2 5 e R e2t/rc (28.15) where Q mx 5 Ce is the mximum chrge on the cpcitor. The product RC is clled the time constnt t of the circuit. Circuits involving more thn one loop re conveniently nlyzed with the use of Kirchhoff s rules: 1. Junction rule. At ny junction, the sum of the currents must equl zero: junction I 5 0 (28.9) 2. Loop rule. The sum of the potentil differences cross ll elements round ny circuit loop must e zero: closed loop DV 5 0 (28.10) When resistor is trversed in the direction of the current, the potentil difference DV cross the resistor is 2IR. When resistor is trversed in the direction opposite the current, DV 5 1IR. When source of emf is trversed in the direction of the emf (negtive terminl to positive terminl), the potentil difference is 1e. When source of emf is trversed opposite the emf (positive to negtive), the potentil difference is 2e. If chrged cpcitor of cpcitnce C is dischrged through resistor of resistnce R, the chrge nd current decrese exponentilly in time ccording to the expressions q(t) 5 Q i e 2t/RC (28.18) i 1t2 5 2 Q i RC e2t/rc (28.19) where Q i is the initil chrge on the cpcitor nd Q i /RC is the initil current in the circuit. Ojective Questions 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. Is circuit reker wired () in series with the device it is protecting, () in prllel, or (c) neither in series or in prllel, or (d) is it impossile to tell? 2. A ttery hs some internl resistnce. (i) Cn the potentil difference cross the terminls of the ttery e equl to its emf? () no () yes, if the ttery is soring energy y electricl trnsmission (c) yes, if more thn one wire is connected to ech terminl (d) yes, if the current in the ttery is zero (e) yes, with no specil condition required. (ii) Cn the terminl voltge exceed the emf? Choose your nswer from the sme possiilities s in prt (i).

199 856 Chpter 28 Direct-Current Circuits 3. The terminls of ttery re connected cross two resistors in series. The resistnces of the resistors re not the sme. Which of the following sttements re correct? Choose ll tht re correct. () The resistor with the smller resistnce crries more current thn the other resistor. () The resistor with the lrger resistnce crries less current thn the other resistor. (c) The current in ech resistor is the sme. (d) The potentil difference cross ech resistor is the sme. (e) The potentil difference is gretest cross the resistor closest to the positive terminl. 4. When operting on 120-V circuit, n electric heter receives W of power, toster receives W, nd n electric oven receives W. If ll three pplinces re connected in prllel on 120-V circuit nd turned on, wht is the totl current drwn from n externl source? () 24.0 A () 32.0 A (c) 40.0 A (d) 48.0 A (e) none of those nswers 5. If the terminls of ttery with zero internl resistnce re connected cross two identicl resistors in series, the totl power delivered y the ttery is 8.00 W. If the sme ttery is connected cross the sme resistors in prllel, wht is the totl power delivered y the ttery? () 16.0 W () 32.0 W (c) 2.00 W (d) 4.00 W (e) none of those nswers 6. everl resistors re connected in series. Which of the following sttements is correct? Choose ll tht re correct. () The equivlent resistnce is greter thn ny of the resistnces in the group. () The equivlent resistnce is less thn ny of the resistnces in the group. (c) The equivlent resistnce depends on the voltge pplied cross the group. (d) The equivlent resistnce is equl to the sum of the resistnces in the group. (e) None of those sttements is correct. 7. Wht is the time constnt of the circuit shown in Figure OQ28.7? Ech of the five resistors hs resistnce R, nd ech of the five cpcitors hs cpcitnce C. The internl resistnce of the ttery is negligile. () RC () 5RC (c) 10RC (d) 25RC (e) none of those nswers R R R R R C C C C C difference () current (c) power delivered (d) chrge entering ech resistor in given time intervl (e) none of those nswers 10. The terminls of ttery re connected cross two resistors in prllel. The resistnces of the resistors re not the sme. Which of the following sttements is correct? Choose ll tht re correct. () The resistor with the lrger resistnce crries more current thn the other resistor. () The resistor with the lrger resistnce crries less current thn the other resistor. (c) The potentil difference cross ech resistor is the sme. (d) The potentil difference cross the lrger resistor is greter thn the potentil difference cross the smller resistor. (e) The potentil difference is greter cross the resistor closer to the ttery. 11. Are the two hedlights of cr wired () in series with ech other, () in prllel, or (c) neither in series nor in prllel, or (d) is it impossile to tell? 12. In the circuit shown in Figure OQ28.12, ech ttery is delivering energy to the circuit y electricl trnsmission. All the resistors hve equl resistnce. (i) Rnk the electric potentils t points,, c, d, nd e from highest to lowest, noting ny cses of equlity in the rnking. (ii) Rnk the mgnitudes of the currents t the sme points from gretest to lest, noting ny cses of equlity. 12 V 9 V Figure OQ28.12 V Figure OQ When resistors with different resistnces re connected in series, which of the following must e the sme for ech resistor? Choose ll correct nswers. () potentil difference () current (c) power delivered (d) chrge entering ech resistor in given time intervl (e) none of those nswers 9. When resistors with different resistnces re connected in prllel, which of the following must e the sme for ech resistor? Choose ll correct nswers. () potentil 13. everl resistors re connected in prllel. Which of the following sttements re correct? Choose ll tht re correct. () The equivlent resistnce is greter thn ny of the resistnces in the group. () The equivlent resistnce is less thn ny of the resistnces in the group. (c) The equivlent resistnce depends on the voltge pplied cross the group. (d) The equivlent resistnce is equl to the sum of the resistnces in the group. (e) None of those sttements is correct. 14. A circuit consists of three iden- C A ticl lmps connected to ttery s in Figure OQ The ttery hs some internl resistnce. The switch, originlly open, is closed. (i) Wht then hppens to the rightness of lmp B? () It increses. () It decreses somewht. (c) It does e c not chnge. (d) It drops to zero. For prts (ii) to (vi), choose from the sme possiilities () through (d). (ii) Wht hppens to the rightness of lmp C? (iii) Wht hppens to the current in the ttery? (iv) Wht hppens to the potentil difference cross lmp A? (v) Wht hppens to the potentil difference d B Figure OQ28.14

200 Prolems 857 cross lmp C? (vi) Wht hppens to the totl power delivered to the lmps y the ttery? 15. A series circuit consists of three identicl lmps connected to ttery s shown in Figure OQ The switch, originlly open, is closed. (i) Wht then hppens to the rightness of lmp B? () It increses. () It decreses somewht. (c) It does not chnge. (d) It drops to zero. For prts (ii) to (vi), choose from the sme possiilities () through (d). (ii) Wht hppens to the rightness of lmp C? (iii) Wht hppens to the current in the ttery? (iv) Wht hppens to the potentil difference cross Conceptul Questions 1. uppose prchutist lnds on high-voltge wire nd grs the wire s she prepres to e rescued. () Will she e electrocuted? () If the wire then reks, should she continue to hold onto the wire s she flls to the ground? Explin. 2. A student clims tht the second of two lightuls in series is less right thn the first ecuse the first lightul uses up some of the current. How would you respond to this sttement? 3. Why is it possile for ird to sit on high-voltge wire without eing electrocuted? 4. Given three lightuls nd ttery, sketch s mny different electric circuits s you cn. 5. A ski resort consists of few chirlifts nd severl interconnected downhill runs on the side of mountin, with lodge t the ottom. The chirlifts re nlogous to tteries, nd the runs re nlogous to resistors. Descrie how two runs cn e in series. Descrie how three runs cn e in prllel. ketch junction etween one chirlift nd two runs. tte Kirchhoff s junction rule for ski resorts. One of the skiers hppens to e crrying skydiver s ltimeter. he never tkes the sme set of chirlifts nd runs twice, ut keeps pssing you t the fixed loction where you re working. tte Kirchhoff s loop rule for ski resorts. Prolems The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide lmp A? (v) Wht hppens to the potentil difference cross lmp C? (vi) Wht hppens to the totl power delivered to the lmps y the ttery? e 1. denotes nswer ville in tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign A 6. Referring to Figure CQ28.6, C descrie wht hppens to the lightul fter the switch is closed. Assume the cpcitor hs lrge cpcitnce nd is initilly unchrged. Also ssume the light illumintes when connected directly Figure CQ28.6 cross the ttery terminls. 7. o tht your grndmother cn listen to A Pririe Home Compnion, you tke her edside rdio to the hospitl where she is stying. You re required to hve mintennce worker test the rdio for electricl sfety. Finding tht it develops 120 V on one of its knos, he does not let you tke it to your grndmother s room. Your grndmother complins tht she hs hd the rdio for mny yers nd noody hs ever gotten shock from it. You end up hving to uy new plstic rdio. () Why is your grndmother s old rdio dngerous in hospitl room? () Will the old rdio e sfe ck in her edroom? 8. () Wht dvntge does 120-V opertion offer over 240 V? () Wht disdvntges does it hve? 9. Is the direction of current in ttery lwys from the negtive terminl to the positive terminl? Explin. 10. Compre series nd prllel resistors to the series nd prllel rods in Figure on pge 610. How re the situtions similr? B Figure OQ28.15 BIO Q/C C ection 28.1 Electromotive Force 1. A ttery hs n emf of 15.0 V. The terminl voltge M of the ttery is 11.6 V when it is delivering 20.0 W of power to n externl lod resistor R. () Wht is the vlue of R? () Wht is the internl resistnce of the ttery?

201 858 Chpter 28 Direct-Current Circuits 2. Two 1.50-V tteries with their positive terminls AMT in the sme direction re inserted in series into flshlight. One ttery hs n internl resistnce of V, nd the other hs n internl resistnce of V. When the switch is closed, the ul crries current of 600 ma. () Wht is the ul s resistnce? () Wht frction of the chemicl energy trnsformed ppers s internl energy in the tteries? 3. An utomoile ttery hs n emf of 12.6 V nd W n internl resistnce of V. The hedlights together hve n equivlent resistnce of 5.00 V (ssumed constnt). Wht is the potentil difference cross the hedlight uls () when they re the only lod on the ttery nd () when the strter motor is operted, with 35.0 A of current in the motor? 4. As in Exmple 28.2, consider power supply with Q/C fixed emf e nd internl resistnce r cusing current in lod resistnce R. In this prolem, R is fixed nd r is vrile. The efficiency is defined s the energy delivered to the lod divided y the energy delivered y the emf. () When the internl resistnce is djusted for mximum power trnsfer, wht is the efficiency? () Wht should e the internl resistnce for mximum possile efficiency? (c) When the electric compny sells energy to customer, does it hve gol of high efficiency or of mximum power trnsfer? Explin. (d) When student connects loudspeker to n mplifier, does she most wnt high efficiency or high power trnsfer? Explin. ection 28.2 Resistors in eries nd Prllel 5. Three 100-V resistors re connected s shown in Figure P28.5. The mximum power tht cn sfely e W delivered to ny one resistor is 25.0 W. () Wht is the mximum potentil difference tht cn e pplied to the terminls nd? () For the voltge determined in prt (), wht is the power delivered to ech resistor? (c) Wht is the totl power delivered to the comintion of resistors? Figure P A lightul mrked 75 W [t] 120 V is screwed into Q/C socket t one end of long extension cord, in which ech of the two conductors hs resistnce V. The other end of the extension cord is plugged into 120-V outlet. () Explin why the ctul power delivered to the lightul cnnot e 75 W in this sitution. () Drw circuit digrm. (c) Find the ctul power delivered to the lightul in this circuit. 7. Wht is the equivlent resistnce of the comintion of identicl resistors etween points nd in Figure P28.7? 8. Consider the two circuits shown in Figure P28.8 in Q/C which the tteries re identicl. The resistnce of ech lightul is R. Neglect the internl resistnces of the tteries. () Find expressions for the currents in ech lightul. () How does the rightness of B compre with tht of C? Explin. (c) How does the rightness of A compre with tht of B nd of C? Explin. A B C e Figure P Consider the circuit shown in Figure P28.9. Find M () the current in the 20.0-V resistor nd () the potentil difference etween points nd V Figure P28.9 e () You need 45-V resistor, ut the stockroom hs Q/C only 20-V nd 50-V resistors. How cn the desired resistnce e chieved under these circumstnces? () Wht cn you do if you need 35-V resistor? 11. A ttery with e V nd no internl resistnce supplies current to the circuit shown in Figure P When the doule-throw switch is open s shown in the figure, the current in the ttery is 1.00 ma. When the switch is closed in position, the current in the e R R 1 R R 2 2 R R R Figure P28.7 Figure P28.11 Prolems 11 nd 12. R R 3

202 Prolems 859 ttery is 1.20 ma. When the switch is closed in position, the current in the ttery is 2.00 ma. Find the resistnces () R 1, () R 2, nd (c) R A ttery with emf e nd no internl resistnce supplies current to the circuit shown in Figure P When the doule-throw switch is open s shown in the figure, the current in the ttery is I 0. When the switch is closed in position, the current in the ttery is I. When the switch is closed in position, the current in the ttery is I. Find the resistnces () R 1, () R 2, nd (c) R () Find the equivlent resistnce etween points nd M in Figure P () Clculte the current in ech resistor if potentil difference of 34.0 V is pplied etween points nd Figure P () When the switch in the circuit of Figure P28.14 Q/C is closed, will the equivlent resistnce etween points nd increse or decrese? tte your resoning. () Assume the equivlent resistnce drops y 50.0% when the switch is closed. Determine the vlue of R. R Figure P Two resistors connected in series hve n equivlent resistnce of 690 V. When they re connected in prllel, their equivlent resistnce is 150 V. Find the resistnce of ech resistor. 16. Four resistors re connected to ttery s shown in Q/C Figure P () Determine the potentil difference cross ech resistor in terms of e. () Determine the current in ech resistor in terms of I. (c) Wht If? If R 3 is incresed, explin wht hppens to the current in ech of the resistors. (d) In the limit tht R 3 `, wht re the new vlues of the current in ech resistor in terms of I, the originl current in the ttery? 17. Consider the comintion of resistors shown in Figure P () Find the equivlent resistnce etween points nd. () If voltge of 35.0 V is pplied etween points nd, find the current in ech resistor Figure P For the purpose of mesuring the electric resistnce BIO of shoes through the ody of the werer stnding on metl ground plte, the Americn Ntionl tndrds Institute (ANI) specifies the circuit shown in Figure P The potentil difference DV cross the 1.00-MV resistor is mesured with n idel voltmeter. () how tht the resistnce of the footwer is 50.0 V 2 DV R shoes 5 DV () In medicl test, current through the humn ody should not exceed 150 ma. Cn the current delivered y the ANI- specified circuit exceed 150 ma? To decide, consider person stnding refoot on the ground plte V 1.00 M V Figure P Clculte the power delivered to ech resistor in the W circuit shown in Figure P V Figure P R 1 = R e I Figure P28.16 R 2 = 2R R 4 = 3R R 3 = 4R 20. Why is the following sitution impossile? A technicin is testing circuit tht contins resistnce R. He relizes tht etter design for the circuit would include resistnce 7 3R rther thn R. He hs three dditionl resistors, ech with resistnce R. By comining these dditionl resistors in certin comintion tht is then plced in series with the originl resistor, he chieves the desired resistnce. 21. Consider the circuit shown in Figure P28.21 on pge 860. () Find the voltge cross the 3.00-V resistor. () Find the current in the 3.00-V resistor.

203 860 Chpter 28 Direct-Current Circuits 2.00 ection 28.3 Kirchhoff s Rules 22. In Figure P28.22, show how to dd just enough mmeters to mesure every different current. how how to dd just enough voltmeters to mesure the potentil difference cross ech resistor nd cross ech ttery V V 3.00 Figure P V Figure P28.22 Prolems 22 nd The circuit shown in Figure P28.22 is connected for M 2.00 min. () Determine the current in ech rnch of Q/C the circuit. () Find the energy delivered y ech ttery. (c) Find the energy delivered to ech resistor. (d) Identify the type of energy storge trnsformtion tht occurs in the opertion of the circuit. (e) Find the totl mount of energy trnsformed into internl energy in the resistors. 24. For the circuit shown in Figure 12.0 V P28.24, clculte () the current in the 2.00-V resistor nd () the potentil difference etween points nd. 25. Wht re the expected redings of () the idel mmeter M V nd () the idel voltmeter in Figure P28.25? Figure P The following equtions descrie n electric circuit: 2I 1 (220 V) V 2 I 2 (370 V) 5 0 1I 2 (370 V) 1 I 3 (150 V) V 5 0 I 1 1 I 3 2 I () Drw digrm of the circuit. () Clculte the unknowns nd identify the physicl mening of ech unknown. 27. Tking R kv nd e V in Figure P28.27, determine the direction nd mgnitude of the current in the horizontl wire etween nd e. e R 4R c 3R 2R Figure P28.27 d Jumper cles re connected from fresh ttery in Q/C one cr to chrge ded ttery in nother cr. Figure P28.28 shows the circuit digrm for this sitution. While the cles re connected, the ignition switch of the cr with the ded ttery is closed nd the strter is ctivted to strt the engine. Determine the current in () the strter nd () the ded ttery. (c) Is the ded ttery eing chrged while the strter is operting? V Live ttery 12 V Ded ttery Figure P28.28 Ignition switch e e 0.06 trter 29. The mmeter shown in Figure P28.29 reds 2.00 A. W Find () I 1, () I 2, nd (c) e. A V V 6.00 Figure P V I 1 I V A e Figure P In the circuit of Figure P28.30, determine () the current in ech resistor nd () the potentil difference W cross the 200-V resistor.

204 Prolems Using Kirchhoff s rules, () find the current in ech M resistor shown in Figure P28.31 nd () find the potentil difference etween points c nd f. 1 2 R V 60.0 V 80.0 V R k R 1 f e c 3.00 k 4.00 k e e e Figure P In the circuit of Figure P28.32, the current I A Q/C nd the vlues of e for the idel ttery nd R re unknown. Wht re the currents () I 2 nd () I 3? (c) Cn you find the vlues of e nd R? If so, find their vlues. If not, explin. I V 360 V 80.0 V R Figure P28.30 e 24.0 V 3.00 I Figure P In Figure P28.33, find () the current in ech resistor nd () the power delivered to ech resistor V 12.0 V I 1 d I 1 loop, nd (c) the junction on the left side. In ech cse, suppress units for clrity nd simplify, comining the terms. (d) olve the junction eqution for I 3. (e) Using the eqution found in prt (d), eliminte I 3 from the eqution found in prt (). (f) olve the equtions found in prts () nd (e) simultneously for the two unknowns I 1 nd I 2. (g) ustitute the nswers found in prt (f) into the junction eqution found in prt (d), solving for I 3. (h) Wht is the significnce of the negtive nswer for I 2? V 11.0 I I V I 3 Figure P V 35. Find the potentil difference cross ech resistor in M Figure P V 3.00 V Figure P V () Cn the circuit shown in Figure P28.36 e reduced Q/C to single resistor connected to ttery? Explin. Clculte the currents () I 1, (c) I 2, nd (d) I V 12.0 V I 1 I I I 2 Figure P For the circuit shown in Figure P28.34, we wish to GP find the currents I 1, I 2, nd I 3. Use Kirchhoff s rules to Q/C otin equtions for () the upper loop, () the lower I 3 ection 28.4 RC Circuits Figure P An unchrged cpcitor nd resistor re connected in series to source of emf. If e V, C mf, nd R V, find () the time constnt of the circuit, () the mximum chrge on the cpcitor, nd (c) the chrge on the cpcitor t time equl to one time constnt fter the ttery is connected.

205 862 Chpter 28 Direct-Current Circuits 38. Consider series RC circuit s in Figure P28.38 for W which R MV, C mf, nd e V. Find () the time constnt of the circuit nd () the mximum chrge on the cpcitor fter the switch is thrown closed. (c) Find the current in the resistor 10.0 s fter the switch is closed V mf C e R Figure P28.38 Prolems 38, 67, nd A 2.00-nF cpcitor with n initil chrge of 5.10 mc W is dischrged through 1.30-kV resistor. () Clculte the current in the resistor 9.00 ms fter the resistor is connected cross the terminls of the cpcitor. () Wht chrge remins on the cpcitor fter 8.00 ms? (c) Wht is the mximum current in the resistor? 40. A 10.0-mF cpcitor is chrged y 10.0-V ttery through resistnce R. The cpcitor reches potentil difference of 4.00 V in time intervl of 3.00 s fter chrging egins. Find R. 41. In the circuit of Figure P28.41, the switch hs een W open for long time. It is then suddenly closed. Tke e V, R kv, R kv, nd C mf. Determine the time constnt () efore the switch is closed nd () fter the switch is closed. (c) Let the switch e closed t t 5 0. Determine the current in the switch s function of time. e R 1 R 2 Figure P28.41 Prolems 41 nd In the circuit of Figure P28.41, the switch hs een open for long time. It is then suddenly closed. Determine the time constnt () efore the switch is closed nd () fter the switch is closed. (c) Let the switch e closed t t 5 0. Determine the current in the switch s function of time. 43. The circuit in Figure P28.43 hs een connected for M long time. () Wht is the potentil difference cross the cpcitor? () If the ttery is disconnected from the circuit, over wht time intervl does the cpcitor dischrge to one-tenth its initil voltge? C 44. how tht the integrl e ` 0 e22t/rc dt in Exmple hs the vlue 1 2 RC. 45. A chrged cpcitor is connected to resistor nd switch s in Figure P The circuit hs time constnt of 1.50 s. oon fter the switch is closed, the chrge on the cpcitor is 75.0% of its initil chrge. () Find the time intervl required for the cpcitor to rech this chrge. () If R kv, wht is the vlue of C? C +Q Q Figure P28.43 Figure P28.45 ection 28.5 Household Wiring nd Electricl fety 46. An electric heter is rted t W, toster M t 750 W, nd n electric grill t W. The Q/C three pplinces re connected to common 120-V household circuit. () How much current does ech drw? () If the circuit is protected with 25.0-A circuit reker, will the circuit reker e tripped in this sitution? Explin your nswer. 47. A heting element in stove is designed to receive M W when connected to 240 V. () Assuming the resistnce is constnt, clculte the current in the heting element if it is connected to 120 V. () Clculte the power it receives t tht voltge. 48. Turn on your desk lmp. Pick up the cord, with your thum nd index finger spnning the width of the cord. () Compute n order-of-mgnitude estimte for the current in your hnd. Assume the conductor inside the lmp cord next to your thum is t potentil, 10 2 V t typicl instnt nd the conductor next to your index finger is t ground potentil (0 V). The resistnce of your hnd depends strongly on the thickness nd the moisture content of the outer lyers of your skin. Assume the resistnce of your hnd etween fingertip nd thum tip is, 10 4 V. You my model the cord s hving ruer insultion. tte the other quntities you mesure or estimte nd their vlues. Explin your resoning. () uppose your ody is isolted from ny other chrges or currents. In orderof-mgnitude terms, estimte the potentil difference etween your thum where it contcts the cord nd your finger where it touches the cord. R

206 Prolems 863 Additionl Prolems 49. Assume you hve ttery of emf e nd three identicl lightuls, ech hving constnt resistnce R. Wht is the totl power delivered y the ttery if the lightuls re connected () in series nd () in prllel? (c) For which connection will the lightuls shine the rightest? 50. Find the equivlent resistnce etween points nd in Figure P Figure P Four 1.50-V AA tteries in series re used to power smll rdio. If the tteries cn move chrge of 240 C, how long will they lst if the rdio hs resistnce of 200 V? 52. Four resistors re connected in prllel cross 9.20-V Q/C ttery. They crry currents of 150 ma, 45.0 ma, 14.0 ma, nd 4.00 ma. If the resistor with the lrgest resistnce is replced with one hving twice the resistnce, () wht is the rtio of the new current in the ttery to the originl current? () Wht If? If insted the resistor with the smllest resistnce is replced with one hving twice the resistnce, wht is the rtio of the new totl current to the originl current? (c) On Ferury night, energy leves house y severl energy leks, including W y conduction through the ceiling, 450 W y infiltrtion (irflow) round the windows, 140 W y conduction through the sement wll ove the foundtion sill, nd 40.0 W y conduction through the plywood door to the ttic. To produce the iggest sving in heting ills, which one of these energy trnsfers should e reduced first? Explin how you decide. Clifford wrtz suggested the ide for this prolem. 53. The circuit in Figure P28.53 hs een connected for severl seconds. Find the current () in the 4.00-V t V 6.00 mf I I 1 3 c d h 8.00 V 3.00 I = 0 g I f 5.00 I 3 e 4.00 V tery, () in the 3.00-V resistor, (c) in the 8.00-V ttery, nd (d) in the 3.00-V ttery. (e) Find the chrge on the cpcitor. 54. The circuit in Figure P28.54 consists of three resistors Q/C nd one ttery with no internl resistnce. () Find the current in the 5.00-V resistor. () Find the power delivered to the 5.00-V resistor. (c) In ech of the circuits in Figures P28.54, P28.54c, nd P28.54d, n dditionl 15.0-V ttery hs een inserted into the circuit. Which digrm or digrms represent circuit tht requires the use of Kirchhoff s rules to find the currents? Explin why. (d) In which of these three new circuits is the smllest mount of power delivered to the 10.0-V resistor? (You need not clculte the power in ech circuit if you explin your nswer.) V c V 15.0 V V 15.0 V d Figure P V 15.0 V 55. For the circuit shown in Figure P28.55, the idel volt meter reds 6.00 V nd the idel mmeter reds 3.00 ma. Find () the vlue of R, () the emf of the ttery, nd (c) the voltge cross the 3.00-kV resistor. V R e A 3.00 k Figure P28.55 I 1 Figure P The resistnce etween terminls nd in Figure P28.56 is 75.0 V. If the resistors leled R hve the sme vlue, determine R. R Figure P28.56 R 5.00

207 864 Chpter 28 Direct-Current Circuits 57. () Clculte the potentil difference etween points nd in Figure P28.57 nd () identify which point is t the higher potentil. cpcitor? () How much chrge remins on the 2.00-mF cpcitor? (c) Wht is the current in the resistor t this time? 12.0 V V µ F 2.00 µ F 10.0 Figure P Why is the following sitution impossile? A ttery hs n emf of e V nd n internl resistnce of r V. A resistnce R is connected cross the ttery nd extrcts from it power of P W. 59. A rechrgele ttery hs n emf of 13.2 V nd n M internl resistnce of V. It is chrged y 14.7-V power supply for time intervl of 1.80 h. After chrging, the ttery returns to its originl stte s it delivers constnt current to lod resistor over 7.30 h. Find the efficiency of the ttery s n energy storge device. (The efficiency here is defined s the energy delivered to the lod during dischrge divided y the energy delivered y the 14.7-V power supply during the chrging process.) 60. Find () the equivlent resistnce of the circuit in Figure P28.60, () the potentil difference cross ech resistor, (c) ech current indicted in Figure P28.60, nd (d) the power delivered to ech resistor V I 1 I 3 c e I 2 I 4 I d Figure P When two unknown resistors re connected in series with ttery, the ttery delivers 225 W nd crries totl current of 5.00 A. For the sme totl current, 50.0 W is delivered when the resistors re connected in prllel. Determine the vlue of ech resistor. 62. When two unknown resistors re connected in series with ttery, the ttery delivers totl power P s nd crries totl current of I. For the sme totl current, totl power P p is delivered when the resistors re connected in prllel. Determine the vlue of ech resistor. 63. The pir of cpcitors in Figure P28.63 re fully chrged y 12.0-V ttery. The ttery is disconnected, nd the switch is then closed. After 1.00 ms hs elpsed, () how much chrge remins on the 3.00-mF f 500 Figure P A power supply hs n open-circuit voltge of 40.0 V nd n internl resistnce of 2.00 V. It is used to chrge two storge tteries connected in series, ech hving n emf of 6.00 V nd internl resistnce of V. If the chrging current is to e 4.00 A, () wht dditionl resistnce should e dded in series? At wht rte does the internl energy increse in () the supply, (c) in the tteries, nd (d) in the dded series resistnce? (e) At wht rte does the chemicl energy increse in the tteries? 65. The circuit in Figure P28.65 contins two resistors, R kv nd R kv, nd two cpcitors, C mf nd C mf, connected to ttery with emf e V. If there re no chrges on the cpcitors efore switch is closed, determine the chrges on cpcitors () C 1 nd () C 2 s functions of time, fter the switch is closed. R 1 C 1 R 2 e C 2 Figure P Two resistors R 1 nd R 2 re in prllel with ech other. Together they crry totl current I. () Determine the current in ech resistor. () Prove tht this division of the totl current I etween the two resistors results in less power delivered to the comintion thn ny other division. It is generl principle tht current in direct current circuit distriutes itself so tht the totl power delivered to the circuit is minimum. 67. The vlues of the components in simple series RC circuit contining switch (Fig. P28.38) re C mf, AMT M R V, nd e V. At the instnt 10.0 s fter the switch is closed, clculte () the chrge on the cpcitor, () the current in the resistor, (c) the rte t which energy is eing stored in the cpcitor, nd (d) the rte t which energy is eing delivered y the ttery.

208 Prolems A ttery is used to chrge cpcitor through resistor s shown in Figure P how tht hlf the energy supplied y the ttery ppers s internl energy in the resistor nd hlf is stored in the cpcitor. 69. A young mn owns cnister vcuum clener mrked 535 W [t] 120 V nd Volkswgen Beetle, which he wishes to clen. He prks the cr in his prtment prking lot nd uses n inexpensive extension cord 15.0 m long to plug in the vcuum clener. You my ssume the clener hs constnt resistnce. () If the resistnce of ech of the two conductors in the extension cord is V, wht is the ctul power delivered to the clener? () If insted the power is to e t lest 525 W, wht must e the dimeter of ech of two identicl copper conductors in the cord he uys? (c) Repet prt () ssuming the power is to e t lest 532 W. 70. () Determine the equilirium chrge on the cpcitor in the circuit of Figure P28.70 s function of R. Q/C () Evlute the chrge when R V. (c) Cn the chrge on the cpcitor e zero? If so, for wht vlue of R? (d) Wht is the mximum possile mgnitude of the chrge on the cpcitor? For wht vlue of R is it chieved? (e) Is it experimentlly meningful to tke R 5 `? Explin your nswer. If so, wht chrge mgnitude does it imply? 5.00 V Figure P mf witch shown in Figure P28.71 hs een closed for long time, nd the electric circuit crries constnt current. Tke C mf, C mf, R kv, nd R kv. The power delivered to R 2 is 2.40 W. () Find the chrge on C 1. () Now the switch is opened. After mny milliseconds, y how much hs the chrge on C 2 chnged? C 1 R 2 Figure P Three identicl 60.0-W, 120-V lightuls re connected M cross 120-V power source s shown in Figure P Assuming the resistnce of ech lightul is constnt (even though in relity the resistnce might increse mrkedly with current), find () the totl power supplied y the power source nd () the potentil difference cross ech lightul. R R 1 C 2 R V R 2 R 3 Figure P A regulr tetrhedron is pyrmid with tringulr se nd tringulr sides s shown in Figure P Imgine the six stright lines in Figure P28.73 re ech 10.0-V resistors, with junctions t the four vertices. A 12.0-V ttery is connected to ny two of the vertices. Find () the equivlent resistnce of the tetrhedron etween these vertices nd () the current in the ttery. Figure P An idel voltmeter connected cross certin fresh Q/C 9-V ttery reds 9.30 V, nd n idel mmeter riefly connected cross the sme ttery reds 3.70 A. We sy the ttery hs n open-circuit voltge of 9.30 V nd short- circuit current of 3.70 A. Model the ttery s source of emf e in series with n internl resistnce r s in Figure Determine oth () e nd () r. An experimenter connects two of these identicl tteries together s shown in Figure P Find (c) the open-circuit voltge nd (d) the short-circuit current of the pir of connected tteries. (e) The experimenter connects 12.0-V resistor etween the exposed terminls of the connected tteries. Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects second identicl resistor in prllel with the first. Find the power delivered to ech resistor. (h) Becuse the sme pir of tteries is connected cross oth resistors s ws connected cross the single resistor, why is the power in prt (g) not the sme s tht in prt (f)? Figure P In Figure P28.75 on pge 866, suppose the switch hs een closed for time intervl sufficiently long for the cpcitor to ecome fully chrged. Find () the

209 866 Chpter 28 Direct-Current Circuits stedy-stte current in ech resistor nd () the chrge Q mx on the cpcitor. (c) The switch is now opened t t 5 0. Write n eqution for the current in R 2 s function of time nd (d) find the time intervl required for the chrge on the cpcitor to fll to one-fifth its initil vlue V 12.0 k R 2 = 15.0 k Figure P mf 3.00 k 76. Figure P28.76 shows circuit model for the trnsmission of n electricl signl such s cle TV to lrge numer of suscriers. Ech suscrier connects lod resistnce R L etween the trnsmission line nd the ground. The ground is ssumed to e t zero potentil nd le to crry ny current etween ny ground connections with negligile resistnce. The resistnce of the trnsmission line etween the connection points of different suscriers is modeled s the constnt resistnce R T. how tht the equivlent resistnce cross the signl source is R eq R TR L 1 R T2 2 1/2 1 R T 4 uggestion: Becuse the numer of suscriers is lrge, the equivlent resistnce would not chnge noticely if the first suscrier cnceled the service. Consequently, the equivlent resistnce of the section of the circuit to the right of the first lod resistor is nerly equl to R eq. ignl source R T R T R T R L R L R L Figure P The student engineer of cmpus rdio sttion wishes to verify the effectiveness of the lightning rod on the ntenn mst (Fig. P28.77). The unknown resistnce A R y C R x B R y R x is etween points C nd E. Point E is true ground, ut it is inccessile E Figure P28.77 for direct mesurement ecuse this strtum is severl meters elow the Erth s surfce. Two identicl rods re driven into the ground t A nd B, introducing n unknown resistnce R y. The procedure is s follows. Mesure resistnce R 1 etween points A nd B, then connect A nd B with hevy conducting wire nd mesure resistnce R 2 etween points A nd C. () Derive n eqution for R x in terms of the oservle resistnces, R 1 nd R 2. () A stisfctory ground resistnce would e R x, 2.00 V. Is the grounding of the sttion dequte if mesurements give R V nd R V? Explin. 78. The circuit shown in Figure P28.78 is set up in the lortory to mesure n unknown cpcitnce C in series with resistnce R MV powered y ttery whose emf is 6.19 V. The dt given in the tle re the mesured voltges cross the cpcitor s function of time, where t 5 0 represents the instnt t which the switch is thrown to position. () Construct grph of ln (e/dv) versus t nd perform liner lest-squres fit to the dt. () From the slope of your grph, otin vlue for the time constnt of the circuit nd vlue for the cpcitnce. Dv (V) t (s) ln (e/dv) e Figure P An electric tekettle hs multiposition switch nd two heting coils. When only one coil is switched on, the well-insulted kettle rings full pot of wter to oil over the time intervl Dt. When only the other coil is switched on, it tkes time intervl of 2 Dt to oil the sme mount of wter. Find the time intervl required to oil the sme mount of wter if oth coils re switched on () in prllel connection nd () in series connection. 80. A voltge DV is pplied to series configurtion of n resistors, ech of resistnce R. The circuit components re reconnected in prllel configurtion, nd voltge DV is gin pplied. how tht the power delivered to the series configurtion is 1/n 2 times the power delivered to the prllel configurtion. 81. In plces such s hospitl operting rooms or fctories for BIO electronic circuit ords, electric sprks must e voided. A person stnding on grounded floor nd touching nothing else cn typiclly hve ody cpcitnce of 150 pf, in prllel with foot cpcitnce of 80.0 pf produced y the dielectric soles of his or her shoes. The person cquires sttic electric chrge from interctions with his or her surroundings. The sttic chrge flows to ground through the equivlent resistnce of the two V C R

210 Prolems 867 shoe soles in prllel with ech other. A pir of ruersoled street shoes cn present n equivlent resistnce of MV. A pir of shoes with specil stticdissiptive soles cn hve n equivlent resistnce of 1.00 MV. Consider the person s ody nd shoes s forming n RC circuit with the ground. () How long does it tke the ruer-soled shoes to reduce person s potentil from V to 100 V? () How long does it tke the sttic-dissiptive shoes to do the sme thing? Chllenge Prolems 82. The switch in Figure P28.82 closes when DV c. 2 3 DV nd opens when DV c, 1 3 DV. The idel voltmeter reds Voltgecontrolled switch C R 1 R 2 V V c V V c V 2V 3 V 3 Figure P28.82 potentil difference s plotted in Figure P Wht is the period T of the wveform in terms of R 1, R 2, nd C? 83. The resistor R in Figure P28.83 receives 20.0 W of power. Determine the vlue of R. T V 30.0 Figure P28.83 t 40.0 R

211 chpter 29 Mgnetic Fields 29.1 Anlysis Model: Prticle in Field (Mgnetic) 29.2 Motion of Chrged Prticle in Uniform Mgnetic Field 29.3 Applictions Involving Chrged Prticles Moving in Mgnetic Field 29.4 Mgnetic Force Acting on Current-Crrying Conductor 29.5 Torque on Current Loop in Uniform Mgnetic Field 29.6 The Hll Effect An engineer performs test on the electronics ssocited with one of the superconducting mgnets in the Lrge Hdron Collider t the Europen Lortory for Prticle Physics, operted y the Europen Orgniztion for Nucler Reserch (CERN). The mgnets re used to control the motion of chrged prticles in the ccelertor. We will study the effects of mgnetic fields on moving chrged prticles in this chpter. (CERN) 868 Mny historins of science elieve tht the compss, which uses mgnetic needle, ws used in Chin s erly s the 13th century BC, its invention eing of Aric or Indin origin. The erly Greeks knew out mgnetism s erly s 800 BC. They discovered tht the stone mgnetite (Fe 3 O 4 ) ttrcts pieces of iron. Legend scries the nme mgnetite to the shepherd Mgnes, the nils of whose shoes nd the tip of whose stff stuck fst to chunks of mgnetite while he pstured his flocks. In 1269, Pierre de Mricourt of Frnce found tht the directions of needle ner sphericl nturl mgnet formed lines tht encircled the sphere nd pssed through two points dimetriclly opposite ech other, which he clled the poles of the mgnet. usequent experiments showed tht every mgnet, regrdless of its shpe, hs two poles, clled north (N) nd south () poles, tht exert forces on other mgnetic poles similr to the wy electric chrges exert forces on one nother. Tht is, like poles (N N or ) repel ech other, nd opposite poles (N ) ttrct ech other.

212 29.1 Anlysis Model: Prticle in Field (Mgnetic) 869 The poles received their nmes ecuse of the wy mgnet, such s tht in compss, ehves in the presence of the Erth s mgnetic field. If r mgnet is suspended from its midpoint nd cn swing freely in horizontl plne, it will rotte until its north pole points to the Erth s geogrphic North Pole nd its south pole points to the Erth s geogrphic outh Pole. 1 In 1600, Willim Gilert ( ) extended de Mricourt s experiments to vriety of mterils. He knew tht compss needle orients in preferred directions, so he suggested tht the Erth itself is lrge, permnent mgnet. In 1750, experimenters used torsion lnce to show tht mgnetic poles exert ttrctive or repulsive forces on ech other nd tht these forces vry s the inverse squre of the distnce etween intercting poles. Although the force etween two mgnetic poles is otherwise similr to the force etween two electric chrges, electric chrges cn e isolted (witness the electron nd proton), wheres single mgnetic pole hs never een isolted. Tht is, mgnetic poles re lwys found in pirs. All ttempts thus fr to detect n isolted mgnetic pole hve een unsuccessful. No mtter how mny times permnent mgnet is cut in two, ech piece lwys hs north nd south pole. 2 The reltionship etween mgnetism nd electricity ws discovered in 1819 when, during lecture demonstrtion, Hns Christin Oersted found tht n electric current in wire deflected nery compss needle. 3 In the 1820s, further connections etween electricity nd mgnetism were demonstrted independently y Frdy nd Joseph Henry ( ). They showed tht n electric current cn e produced in circuit either y moving mgnet ner the circuit or y chnging the current in nery circuit. These oservtions demonstrte tht chnging mgnetic field cretes n electric field. Yers lter, theoreticl work y Mxwell showed tht the reverse is lso true: chnging electric field cretes mgnetic field. This chpter exmines the forces tht ct on moving chrges nd on current-crrying wires in the presence of mgnetic field. The source of the mgnetic field is descried in Chpter Anlysis Model: Prticle in Field (Mgnetic) In our study of electricity, we descried the interctions etween chrged ojects in terms of electric fields. Recll tht n electric field surrounds ny electric chrge. In ddition to contining n electric field, the region of spce surrounding ny moving electric chrge lso contins mgnetic field. A mgnetic field lso surrounds mgnetic sustnce mking up permnent mgnet. Historiclly, the symol B hs een used to represent mgnetic field, nd we use this nottion in this ook. The direction of the mgnetic field B t ny loction is the direction in which compss needle points t tht loction. As with the electric field, we cn represent the mgnetic field y mens of drwings with mgnetic field lines. Figure 29.1 shows how the mgnetic field lines of r mgnet cn e trced with the id of compss. Notice tht the mgnetic field lines outside the mgnet Hns Christin Oersted Dnish Physicist nd Chemist ( ) Oersted is est known for oserving tht compss needle deflects when plced ner wire crrying current. This importnt discovery ws the first evidence of the connection etween electric nd mgnetic phenomen. Oersted ws lso the first to prepre pure luminum. North Wind/North Wind Picture Archives -- All rights reserved. N Figure 29.1 Compss needles cn e used to trce the mgnetic field lines in the region outside r mgnet. 1 The Erth s geogrphic North Pole is mgneticlly south pole, wheres the Erth s geogrphic outh Pole is mgneticlly north pole. Becuse opposite mgnetic poles ttrct ech other, the pole on mgnet tht is ttrcted to the Erth s geogrphic North Pole is the mgnet s north pole nd the pole ttrcted to the Erth s geogrphic outh Pole is the mgnet s south pole. 2 There is some theoreticl sis for speculting tht mgnetic monopoles isolted north or south poles my exist in nture, nd ttempts to detect them re n ctive experimentl field of investigtion. 3 The sme discovery ws reported in 1802 y n Itlin jurist, Gin Domenico Romgnosi, ut ws overlooked, proly ecuse it ws pulished in n oscure journl.

213 870 Chpter 29 Mgnetic Fields Figure 29.2 Mgnetic field ptterns cn e displyed with iron filings sprinkled on pper ner mgnets. Mgnetic field pttern surrounding r mgnet Mgnetic field pttern etween opposite poles (N ) of two r mgnets Mgnetic field pttern etween like poles (N N) of two r mgnets c point wy from the north pole nd towrd the south pole. One cn disply mgnetic field ptterns of r mgnet using smll iron filings s shown in Figure When we spek of compss mgnet hving north pole nd south pole, it is more proper to sy tht it hs north-seeking pole nd south-seeking pole. This wording mens tht the north-seeking pole points to the north geogrphic pole of the Erth, wheres the south-seeking pole points to the south geogrphic pole. Becuse the north pole of mgnet is ttrcted towrd the north geogrphic pole of the Erth, the Erth s south mgnetic pole is locted ner the north geogrphic pole nd the Erth s north mgnetic pole is locted ner the south geogrphic pole. In fct, the configurtion of the Erth s mgnetic field, pictured in Figure 29.3, is very much like the one tht would e chieved y urying gigntic r mgnet deep in the Erth s interior. If compss needle is supported y erings tht llow it to rotte in the verticl plne s well s in the horizontl plne, the needle is horizontl with respect to the Erth s surfce only ner the equtor. As the compss is moved northwrd, the needle rottes so tht it points more nd more towrd the Erth s surfce. Finlly, t point ner Hudson By in Cnd, the north pole of the needle points directly downwrd. This site, first found in 1832, is considered to e the loction of the south mgnetic pole of the Erth. It is pproximtely mi from the Erth s geogrphic Mgnetic xis A south mgnetic Axis of pole is ner the rottion Erth s north outh mgnetic North geogrphic pole. pole geogrphic 11 pole Geogrphic equtor Henry Lep nd Jim Lehmn Mgnetic equtor N Figure 29.3 The Erth s mgnetic field lines. outh geogrphic pole North mgnetic pole A north mgnetic pole is ner the Erth s south geogrphic pole.

214 29.1 Anlysis Model: Prticle in Field (Mgnetic) 871 North Pole, nd its exct position vries slowly with time. imilrly, the north mgnetic pole of the Erth is out mi wy from the Erth s geogrphic outh Pole. Although the Erth s mgnetic field pttern is similr to the one tht would e set up y r mgnet deep within the Erth, it is esy to understnd why the source of this mgnetic field cnnot e lrge msses of permnently mgnetized mteril. The Erth does hve lrge deposits of iron ore deep eneth its surfce, ut the high tempertures in the Erth s core prevent the iron from retining ny permnent mgnetiztion. cientists consider it more likely tht the source of the Erth s mgnetic field is convection currents in the Erth s core. Chrged ions or electrons circulting in the liquid interior could produce mgnetic field just like current loop does, s we shll see in Chpter 30. There is lso strong evidence tht the mgnitude of plnet s mgnetic field is relted to the plnet s rte of rottion. For exmple, Jupiter rottes fster thn the Erth, nd spce proes indicte tht Jupiter s mgnetic field is stronger thn the Erth s. Venus, on the other hnd, rottes more slowly thn the Erth, nd its mgnetic field is found to e weker. Investigtion into the cuse of the Erth s mgnetism is ongoing. The direction of the Erth s mgnetic field hs reversed severl times during the lst million yers. Evidence for this reversl is provided y slt, type of rock tht contins iron. Bslt forms from mteril spewed forth y volcnic ctivity on the ocen floor. As the lv cools, it solidifies nd retins picture of the Erth s mgnetic field direction. The rocks re dted y other mens to provide time line for these periodic reversls of the mgnetic field. We cn quntify the mgnetic field B y using our model of prticle in field, like the model discussed for grvity in Chpter 13 nd for electricity in Chpter 23. The existence of mgnetic field t some point in spce cn e determined y mesuring the mgnetic force F B exerted on n pproprite test prticle plced t tht point. This process is the sme one we followed in defining the electric field in Chpter 23. If we perform such n experiment y plcing prticle with chrge q in the mgnetic field, we find the following results tht re similr to those for experiments on electric forces: The mgnetic force is proportionl to the chrge q of the prticle. The mgnetic force on negtive chrge is directed opposite to the force on positive chrge moving in the sme direction. The mgnetic force is proportionl to the mgnitude of the mgnetic field vector B. We lso find the following results, which re totlly different from those for experiments on electric forces: The mgnetic force is proportionl to the speed v of the prticle. If the velocity vector mkes n ngle u with the mgnetic field, the mgnitude of the mgnetic force is proportionl to sin u. When chrged prticle moves prllel to the mgnetic field vector, the mgnetic force on the chrge is zero. When chrged prticle moves in direction not prllel to the mgnetic field vector, the mgnetic force cts in direction perpendiculr to oth v nd B; tht is, the mgnetic force is perpendiculr to the plne formed y v nd B. These results show tht the mgnetic force on prticle is more complicted thn the electric force. The mgnetic force is distinctive ecuse it depends on the velocity of the prticle nd ecuse its direction is perpendiculr to oth v nd B. Figure 29.4 (pge 872) shows the detils of the direction of the mgnetic force on chrged

215 872 Chpter 29 Mgnetic Fields Figure 29.4 () The direction of the mgnetic force F B cting on chrged prticle moving with velocity v in the presence of mgnetic field B. () Mgnetic forces on positive nd negtive chrges. The dshed lines show the pths of the prticles, which re investigted in ection Vector expression for the mgnetic force on chrged prticle moving in mgnetic field Figure 29.5 Two right-hnd rules for determining the direction of the mgnetic force F B 5 q v 3 B cting on prticle with chrge q moving with velocity v in mgnetic field B. () In this rule, the mgnetic force is in the direction in which your thum points. () In this rule, the mgnetic force is in the direction of your plm, s if you re pushing the prticle with your hnd. The mgnetic force is perpendiculr to oth v nd B. F B u v B prticle. Despite this complicted ehvior, these oservtions cn e summrized in compct wy y writing the mgnetic force in the form F B FB 5 q v 3 B (29.1) which y definition of the cross product (see ection 11.1) is perpendiculr to oth v nd B. We cn regrd this eqution s n opertionl definition of the mgnetic field t some point in spce. Tht is, the mgnetic field is defined in terms of the force cting on moving chrged prticle. Eqution 29.1 is the mthemticl representtion of the mgnetic version of the prticle in field nlysis model. Figure 29.5 reviews two right-hnd rules for determining the direction of the cross product v 3 B nd determining the direction of F B. The rule in Figure 29.5 depends on our right-hnd rule for the cross product in Figure Point the four fingers of your right hnd long the direction of v with the plm fcing B nd curl them towrd B. Your extended thum, which is t right ngle to your fingers, points in the direction of v 3 B. Becuse F B 5 q v 3 B, F B is in the direction of your thum if q is positive nd is opposite the direction of your thum if q is negtive. (If you need more help understnding the cross product, you should review ection 11.1, including Fig ) An lterntive rule is shown in Figure Here the thum points in the direction of v nd the extended fingers in the direction of B. Now, the force F B on positive chrge extends outwrd from the plm. The dvntge of this rule is tht the force on the chrge is in the direction you would push on something with your (1) Point your fingers in the direction of v nd then curl them towrd the direction of B. (2) Your upright thum shows the direction of the mgnetic force on positive prticle. v B F B F B v v F B B (1) Point your fingers in the direction of B, with v coming out of your thum. B v The mgnetic forces on oppositely chrged prticles moving t the sme velocity in mgnetic field re in opposite directions. (2) The mgnetic force on positive prticle is in the direction you would push with your plm.

216 29.1 Anlysis Model: Prticle in Field (Mgnetic) 873 hnd: outwrd from your plm. The force on negtive chrge is in the opposite direction. You cn use either of these two right-hnd rules. The mgnitude of the mgnetic force on chrged prticle is F B 5 q vb sin u (29.2) where u is the smller ngle etween v nd B. From this expression, we see tht F B is zero when v is prllel or ntiprllel to B (u 5 0 or 1808) nd mximum when v is perpendiculr to B (u 5 908). Let s compre the importnt differences etween the electric nd mgnetic versions of the prticle in field model: The electric force vector is long the direction of the electric field, wheres the mgnetic force vector is perpendiculr to the mgnetic field. The electric force cts on chrged prticle regrdless of whether the prticle is moving, wheres the mgnetic force cts on chrged prticle only when the prticle is in motion. The electric force does work in displcing chrged prticle, wheres the mgnetic force ssocited with stedy mgnetic field does no work when prticle is displced ecuse the force is perpendiculr to the displcement of its point of ppliction. From the lst sttement nd on the sis of the work kinetic energy theorem, we conclude tht the kinetic energy of chrged prticle moving through mgnetic field cnnot e ltered y the mgnetic field lone. The field cn lter the direction of the velocity vector, ut it cnnot chnge the speed or kinetic energy of the prticle. From Eqution 29.2, we see tht the I unit of mgnetic field is the newton per coulom-meter per second, which is clled the tesl (T): N 1 T 5 1 C # m/s Becuse coulom per second is defined to e n mpere, N 1 T 5 1 A # m A non-i mgnetic-field unit in common use, clled the guss (G), is relted to the tesl through the conversion 1 T G. Tle 29.1 shows some typicl vlues of mgnetic fields. Q uick Quiz 29.1 An electron moves in the plne of this pper towrd the top of the pge. A mgnetic field is lso in the plne of the pge nd directed towrd the right. Wht is the direction of the mgnetic force on the electron? () towrd the top of the pge () towrd the ottom of the pge (c) towrd the left edge of the pge (d) towrd the right edge of the pge (e) upwrd out of the pge (f) downwrd into the pge W WMgnitude of the mgnetic force on chrged prticle moving in mgnetic field WWThe tesl Tle 29.1 ome Approximte Mgnetic Field Mgnitudes ource of Field Field Mgnitude (T) trong superconducting lortory mgnet 30 trong conventionl lortory mgnet 2 Medicl MRI unit 1.5 Br mgnet urfce of the un urfce of the Erth Inside humn rin (due to nerve impulses)

217 874 Chpter 29 Mgnetic Fields Anlysis Model Prticle in Field (Mgnetic) Imgine some source (which we will investigte lter) estlishes mgnetic field B throughout spce. Now imgine prticle with chrge q is plced in tht field. The prticle intercts with the mgnetic field so tht the prticle experiences mgnetic force given y FB 5 q v 3 B (29.1) Exmple 29.1 An Electron Moving in Mgnetic Field z An electron in n old-style television picture tue moves towrd the front of the tue with speed of m/s long the x xis (Fig. 29.6). urrounding the neck of the tue re coils of wire tht crete mgnetic field of mgnitude T, directed t n ngle of 608 to the x xis nd lying in e the xy plne. Clculte the mgnetic force on the electron. 60 Figure 29.6 (Exmple 29.1) v olution The mgnetic force FB cting on the electron is in the negtive z direction when v nd B x Conceptulize Recll tht the mgnetic force on chrged prticle is perpendiculr to the plne formed y the velocity lie in the xy plne. F B nd mgnetic field vectors. Use one of the right-hnd rules in Figure 29.5 to convince yourself tht the direction of the force on the electron is downwrd in Figure Ctegorize We evlute the mgnetic force using the mgnetic version of the prticle in field model. Anlyze Use Eqution 29.2 to find the mgnitude of the mgnetic force: x v z q F B q v B B y Exmples: n ion moves in circulr pth in the mgnetic field of mss spectrometer (ection 29.3) coil in motor rottes in response to the mgnetic field in the motor (Chpter 31) mgnetic field is used to seprte prticles emitted y rdioctive sources (Chpter 44) in ule chmer, prticles creted in collisions follow curved pths in mgnetic field, llowing the prticles to e identified (Chpter 46) AM F B 5 q vb sin u 5 ( C)( m/s)(0.025 T)(sin 608) N Finlize For prctice using the vector product, evlute this force in vector nottion using Eqution The mgnitude of the mgnetic force my seem smll to you, ut rememer tht it is cting on very smll prticle, the electron. To convince yourself tht this is sustntil force for n electron, clculte the initil ccelertion of the electron due to this force Motion of Chrged Prticle in Uniform Mgnetic Field Before we continue our discussion, some explntion of the nottion used in this ook is in order. To indicte the direction of B in illustrtions, we sometimes present perspective views such s those in Figure If B lies in the plne of the pge or is present in perspective drwing, we use green vectors or green field lines with rrowheds. In nonperspective illustrtions, we depict mgnetic field perpendiculr to nd directed out of the pge with series of green dots, which represent the tips of rrows coming towrd you (see Fig. 29.7). In this cse, the field is leled B y

218 29.2 Motion of Chrged Prticle in Uniform Mgnetic Field 875 Mgnetic field lines coming out of the pper re indicted y dots, representing the tips of rrows coming outwrd. B out Mgnetic field lines going into the pper re indicted y crosses, representing the fethers of rrows going inwrd. B in Figure 29.7 Representtions of mgnetic field lines perpendiculr to the pge. B out. If B is directed perpendiculrly into the pge, we use green crosses, which represent the fethered tils of rrows fired wy from you, s in Figure In this cse, the field is leled B in, where the suscript in indictes into the pge. The sme nottion with crosses nd dots is lso used for other quntities tht might e perpendiculr to the pge such s forces nd current directions. In ection 29.1, we found tht the mgnetic force cting on chrged prticle moving in mgnetic field is perpendiculr to the prticle s velocity nd consequently the work done y the mgnetic force on the prticle is zero. Now consider the specil cse of positively chrged prticle moving in uniform mgnetic field with the initil velocity vector of the prticle perpendiculr to the field. Let s ssume the direction of the mgnetic field is into the pge s in Figure The prticle in field model tells us tht the mgnetic force on the prticle is perpendiculr to oth the mgnetic field lines nd the velocity of the prticle. The fct tht there is force on the prticle tells us to pply the prticle under net force model to the prticle. As the prticle chnges the direction of its velocity in response to the mgnetic force, the mgnetic force remins perpendiculr to the velocity. As we found in ection 6.1, if the force is lwys perpendiculr to the velocity, the pth of the prticle is circle! Figure 29.8 shows the prticle moving in circle in plne perpendiculr to the mgnetic field. Although mgnetism nd mgnetic forces my e new nd unfmilir to you now, we see mgnetic effect tht results in something with which we re fmilir: the prticle in uniform circulr motion model! The prticle moves in circle ecuse the mgnetic force FB is perpendiculr to v nd B nd hs constnt mgnitude qvb. As Figure 29.8 illustrtes, the The mgnetic force F B cting on the chrge is lwys directed towrd the center of the circle. B in v q F B q r F B F B q v v Figure 29.8 When the velocity of chrged prticle is perpendiculr to uniform mgnetic field, the prticle moves in circulr pth in plne perpendiculr to B.

219 876 Chpter 29 Mgnetic Fields rottion is counterclockwise for positive chrge in mgnetic field directed into the pge. If q were negtive, the rottion would e clockwise. We use the prticle under net force model to write Newton s second lw for the prticle: o F 5 F B 5 m Becuse the prticle moves in circle, we lso model it s prticle in uniform circulr motion nd we replce the ccelertion with centripetl ccelertion: F B 5 qvb 5 mv 2 r This expression leds to the following eqution for the rdius of the circulr pth: r 5 mv qb y (29.3) Tht is, the rdius of the pth is proportionl to the liner momentum mv of the prticle nd inversely proportionl to the mgnitude of the chrge on the prticle nd to the mgnitude of the mgnetic field. The ngulr speed of the prticle (from Eq ) is v 5 v r 5 qb m (29.4) The period of the motion (the time intervl the prticle requires to complete one revolution) is equl to the circumference of the circle divided y the speed of the prticle: T 5 2pr v 5 2p v 5 2pm qb (29.5) These results show tht the ngulr speed of the prticle nd the period of the circulr motion do not depend on the speed of the prticle or on the rdius of the orit. The ngulr speed v is often referred to s the cyclotron frequency ecuse chrged prticles circulte t this ngulr frequency in the type of ccelertor clled cyclotron, which is discussed in ection If chrged prticle moves in uniform mgnetic field with its velocity t some ritrry ngle with respect to B, its pth is helix. For exmple, if the field is directed in the x direction s shown in Figure 29.9, there is no component of force in the x direction. As result, x 5 0, nd the x component of velocity remins constnt. The chrged prticle is prticle in equilirium in this direction. The mgnetic force qv 3 B cuses the components v y nd v z to chnge in time, however, nd the resulting motion is helix whose xis is prllel to the mgnetic field. The projection of the pth onto the yz plne (viewed long the x xis) is circle. (The projections of the pth onto the xy nd xz plnes re sinusoids!) Equtions 29.3 to 29.5 still pply provided v is replced y v ' 5!v y2 1 v z2. q Helicl pth Figure 29.9 A chrged prticle hving velocity vector tht hs component prllel to uniform mgnetic field moves in helicl pth. z B x

220 29.2 Motion of Chrged Prticle in Uniform Mgnetic Field 877 Q uick Quiz 29.2 A chrged prticle is moving perpendiculr to mgnetic field in circle with rdius r. (i) An identicl prticle enters the field, with v perpendiculr to B, ut with higher speed thn the first prticle. Compred with the rdius of the circle for the first prticle, is the rdius of the circulr pth for the second prticle () smller, () lrger, or (c) equl in size? (ii) The mgnitude of the mgnetic field is incresed. From the sme choices, compre the rdius of the new circulr pth of the first prticle with the rdius of its initil pth. Exmple 29.2 A Proton Moving Perpendiculr to Uniform Mgnetic Field A proton is moving in circulr orit of rdius 14 cm in uniform 0.35-T mgnetic field perpendiculr to the velocity of the proton. Find the speed of the proton. olution Conceptulize From our discussion in this section, we know the proton follows circulr pth when moving perpendiculr to uniform mgnetic field. In Chpter 39, we will lern tht the highest possile speed for prticle is the speed of light, m/s, so the speed of the prticle in this prolem must come out to e smller thn tht vlue. Ctegorize The proton is descried y oth the prticle in field model nd the prticle in uniform circulr motion model. These models led to Eqution Anlyze olve Eqution 29.3 for the speed of the prticle: ustitute numericl vlues: v 5 qbr Finlize The speed is indeed smller thn the speed of light, s required. Exmple 29.3 Bending n Electron Bem m p AM v C T m kg m/s Wht If? Wht if n electron, rther thn proton, moves in direction perpendiculr to the sme mgnetic field with this sme speed? Will the rdius of its orit e different? Answer An electron hs much smller mss thn proton, so the mgnetic force should e le to chnge its velocity much more esily thn tht for the proton. Therefore, we expect the rdius to e smller. Eqution 29.3 shows tht r is proportionl to m with q, B, nd v the sme for the electron s for the proton. Consequently, the rdius will e smller y the sme fctor s the rtio of msses m e /m p. In n experiment designed to mesure the mgnitude of uniform mgnetic field, electrons re ccelerted from rest through potentil difference of 350 V nd then enter uniform mgnetic field tht is perpendiculr to the velocity vector of the electrons. The electrons trvel long curved pth ecuse of the mgnetic force exerted on them, nd the rdius of the pth is mesured to e 7.5 cm. (uch curved em of electrons is shown in Fig ) (A) Wht is the mgnitude of the mgnetic field? AM continued Henry Lep nd Jim Lehmn Figure (Exmple 29.3) The ending of n electron em in mgnetic field.

221 878 Chpter 29 Mgnetic Fields 29.3 continued olution Conceptulize This exmple involves electrons ccelerting from rest due to n electric force nd then moving in circulr pth due to mgnetic force. With the help of Figures 29.8 nd 29.10, visulize the circulr motion of the electrons. Ctegorize Eqution 29.3 shows tht we need the speed v of the electron to find the mgnetic field mgnitude, nd v is not given. Consequently, we must find the speed of the electron sed on the potentil difference through which it is ccelerted. To do so, we ctegorize the first prt of the prolem y modeling n electron nd the electric field s n isolted system in terms of energy. Once the electron enters the mgnetic field, we ctegorize the second prt of the prolem s one involving prticle in field nd prticle in uniform circulr motion, s we hve done in this section. Anlyze Write the pproprite reduction of the conservtion of energy eqution, Eqution 8.2, for the electron electric field system: (B) Wht is the ngulr speed of the electrons? olution DK 1 DU 5 0 ustitute the pproprite initil nd finl energies: 1 1 2m e v q DV2 5 0 olve for the speed of the electron: ustitute numericl vlues: Now imgine the electron entering the mgnetic field with this speed. olve Eqution 29.3 for the mgnitude of the mgnetic field: ustitute numericl vlues: Use Eqution 10.10: v 5 Å 22q DV Finlize The ngulr speed cn e represented s v 5 ( rd/s)(1 rev/2p rd) rev/s. The electrons trvel round the circle 24 million times per second! This nswer is consistent with the very high speed found in prt (A). Wht If? Wht if sudden voltge surge cuses the cel nd the ngulr speed remins the sme. Eqution 29.4 ccelerting voltge to increse to 400 V? How does tht ffect the ngulr speed of the electrons, ssuming the mgnetic field remins constnt? is n expression for the cyclotron frequency, which is the sme s the ngulr speed of the electrons. The cyclotron frequency depends only on the chrge q, the mgnetic field B, nd the mss m e, none of which hve chnged. Answer The increse in ccelerting voltge DV cuses the Therefore, the voltge surge hs no effect on the ngulr speed. (In relity, however, the voltge surge my lso electrons to enter the mgnetic field with higher speed v. This higher speed cuses them to trvel in circle with increse the mgnetic field if the mgnetic field is powered y the sme source s the ccelerting voltge. In tht lrger rdius r. The ngulr speed is the rtio of v to r. Both v nd r increse y the sme fctor, so the effects cn- cse, the ngulr speed increses ccording to Eq ) m e C V2 v 5 Å kg B 5 m ev er m/s B kg m/s T C m2 v 5 v r m/s rd/s m When chrged prticles move in nonuniform mgnetic field, the motion is complex. For exmple, in mgnetic field tht is strong t the ends nd wek in the middle such s tht shown in Figure 29.11, the prticles cn oscillte etween two positions. A chrged prticle strting t one end spirls long the field lines until it reches the other end, where it reverses its pth nd spirls ck. This configur

222 29.3 Applictions Involving Chrged Prticles Moving in Mgnetic Field 879 The mgnetic force exerted on the prticle ner either end of the ottle hs component tht cuses the prticle to spirl ck towrd the center. Pth of prticle Figure A chrged prticle moving in nonuniform mgnetic field ( mgnetic ottle) spirls out the field nd oscilltes etween the endpoints. Figure The Vn Allen elts re mde up of chrged prticles trpped y the Erth s nonuniform mgnetic field. The mgnetic field lines re in green, nd the prticle pths re dshed lck lines. tion is known s mgnetic ottle ecuse chrged prticles cn e trpped within it. The mgnetic ottle hs een used to confine plsm, gs consisting of ions nd electrons. uch plsm-confinement scheme could fulfill crucil role in the control of nucler fusion, process tht could supply us in the future with n lmost endless source of energy. Unfortuntely, the mgnetic ottle hs its prolems. If lrge numer of prticles re trpped, collisions etween them cuse the prticles to eventully lek from the system. The Vn Allen rdition elts consist of chrged prticles (mostly electrons nd protons) surrounding the Erth in doughnut-shped regions (Fig ). The prticles, trpped y the Erth s nonuniform mgnetic field, spirl round the field lines from pole to pole, covering the distnce in only few seconds. These prticles originte minly from the un, ut some come from strs nd other hevenly ojects. For this reson, the prticles re clled cosmic rys. Most cosmic rys re deflected y the Erth s mgnetic field nd never rech the tmosphere. ome of the prticles ecome trpped, however, nd it is these prticles tht mke up the Vn Allen elts. When the prticles re locted over the poles, they sometimes collide with toms in the tmosphere, cusing the toms to emit visile light. uch collisions re the origin of the eutiful uror orelis, or northern lights, in the northern hemisphere nd the uror ustrlis in the southern hemisphere. Aurors re usully confined to the polr regions ecuse the Vn Allen elts re nerest the Erth s surfce there. Occsionlly, though, solr ctivity cuses lrger numers of chrged prticles to enter the elts nd significntly distort the norml mgnetic field lines ssocited with the Erth. In these situtions, n uror cn sometimes e seen t lower ltitudes Applictions Involving Chrged Prticles Moving in Mgnetic Field A chrge moving with velocity v in the presence of oth n electric field E nd mgnetic field B is descried y two prticle in field models. It experiences oth n electric force q E nd mgnetic force qv 3 B. The totl force (clled the Lorentz force) cting on the chrge is F 5 q E 1 q v 3 B (29.6)

223 880 Chpter 29 Mgnetic Fields B in F B ource E v F e lit Figure A velocity selector. When positively chrged prticle is moving with velocity v in the presence of mgnetic field directed into the pge nd n electric field directed to the right, it experiences n electric force q E to the right nd mgnetic force qv 3 B to the left. Velocity elector In mny experiments involving moving chrged prticles, it is importnt tht ll prticles move with essentilly the sme velocity, which cn e chieved y pplying comintion of n electric field nd mgnetic field oriented s shown in Figure A uniform electric field is directed to the right (in the plne of the pge in Fig ), nd uniform mgnetic field is pplied in the direction perpendiculr to the electric field (into the pge in Fig ). If q is positive nd the velocity v is upwrd, the mgnetic force q v 3 B is to the left nd the electric force q E is to the right. When the mgnitudes of the two fields re chosen so tht qe 5 qvb, the forces cncel. The chrged prticle is modeled s prticle in equilirium nd moves in stright verticl line through the region of the fields. From the expression qe 5 qvb, we find tht v 5 E B (29.7) Only those prticles hving this speed pss undeflected through the mutully perpendiculr electric nd mgnetic fields. The mgnetic force exerted on prticles moving t speeds greter thn tht is stronger thn the electric force, nd the prticles re deflected to the left. Those moving t slower speeds re deflected to the right. The Mss pectrometer A mss spectrometer seprtes ions ccording to their mss-to-chrge rtio. In one version of this device, known s the Binridge mss spectrometer, em of ions first psses through velocity selector nd then enters second uniform mgnetic field B 0 tht hs the sme direction s the mgnetic field in the selector (Fig ). Upon entering the second mgnetic field, the ions re descried y the prticle in uniform circulr motion model. They move in semicircle of rdius r efore striking detector rry t P. If the ions re positively chrged, the em deflects to the left s Figure shows. If the ions re negtively chrged, the em deflects to the right. From Eqution 29.3, we cn express the rtio m/q s m q 5 rb 0 v P Velocity selector r Detector rry B in B 0, in E v q Figure A mss spectrometer. Positively chrged prticles re sent first through velocity selector nd then into region where the mgnetic field B0 cuses the prticles to move in semicirculr pth nd strike detector rry t P.

224 29.3 Applictions Involving Chrged Prticles Moving in Mgnetic Field 881 Electrons re ccelerted from the cthode, pss through two slits, nd re deflected y oth n electric field (formed y the chrged deflection pltes) nd mgnetic field (directed perpendiculr to the electric field). The em of electrons then strikes fluorescent screen. Cthode Using Eqution 29.7 gives lits Deflection pltes m q 5 rb 0B E Mgnetic field coil Fluorescent coting Deflected electron em Undeflected electron em Figure () Thomson s pprtus for mesuring e/m e. () J. J. Thomson (left) in the Cvendish Lortory, University of Cmridge. The mn on the right, Frnk Bldwin Jewett, is distnt reltive of John W. Jewett, Jr., couthor of this text. (29.8) Therefore, we cn determine m/q y mesuring the rdius of curvture nd knowing the field mgnitudes B, B 0, nd E. In prctice, one usully mesures the msses of vrious isotopes of given ion, with the ions ll crrying the sme chrge q. In this wy, the mss rtios cn e determined even if q is unknown. A vrition of this technique ws used y J. J. Thomson ( ) in 1897 to mesure the rtio e/m e for electrons. Figure shows the sic pprtus he used. Electrons re ccelerted from the cthode nd pss through two slits. They then drift into region of perpendiculr electric nd mgnetic fields. The mgnitudes of the two fields re first djusted to produce n undeflected em. When the mgnetic field is turned off, the electric field produces mesurle em deflection tht is recorded on the fluorescent screen. From the size of the deflection nd the mesured vlues of E nd B, the chrge-to-mss rtio cn e determined. The results of this crucil experiment represent the discovery of the electron s fundmentl prticle of nture. The Cyclotron A cyclotron is device tht cn ccelerte chrged prticles to very high speeds. The energetic prticles produced re used to omrd tomic nuclei nd therey produce nucler rections of interest to reserchers. A numer of hospitls use cyclotron fcilities to produce rdioctive sustnces for dignosis nd tretment. Both electric nd mgnetic forces ply key roles in the opertion of cyclotron, schemtic drwing of which is shown in Figure (pge 882). The chrges move inside two semicirculr continers D 1 nd D 2, referred to s dees ecuse of their shpe like the letter D. A high-frequency lternting potentil difference is pplied to the dees, nd uniform mgnetic field is directed perpendiculr to them. A positive ion relesed t P ner the center of the mgnet in one dee moves in semicirculr pth (indicted y the dshed lck line in the drwing) nd rrives ck t the gp in time intervl T/2, where T is the time intervl needed to mke one complete trip round the two dees, given y Eqution The frequency Lucent Technologies Bell Lortory, courtesy AIP Emilio egre Visul Archives Pitfll Prevention 29.1 The Cyclotron Is Not the Only Type of Prticle Accelertor The cyclotron is importnt historiclly ecuse it ws the first prticle ccelertor to produce prticles with very high speeds. Cyclotrons still ply importnt roles in medicl pplictions nd some reserch ctivities. Mny other reserch ctivities mke use of different type of ccelertor clled synchrotron.

225 882 Chpter 29 Mgnetic Fields The lck, dshed, curved lines represent the pth of the prticles. B P Alternting V D 1 D 2 After eing ccelerted, the prticles exit here. North pole of mgnet Figure () A cyclotron consists of n ion source t P, two dees D 1 nd D 2 cross which n lternting potentil difference is pplied, nd uniform mgnetic field. (The south pole of the mgnet is not shown.) () The first cyclotron, invented y E. O. Lwrence nd M.. Livingston in of the pplied potentil difference is djusted so tht the polrity of the dees is reversed in the sme time intervl during which the ion trvels round one dee. If the pplied potentil difference is djusted such tht D 1 is t lower electric potentil thn D 2 y n mount DV, the ion ccelertes cross the gp to D 1 nd its kinetic energy increses y n mount q DV. It then moves round D 1 in semicirculr pth of greter rdius (ecuse its speed hs incresed). After time intervl T/2, it gin rrives t the gp etween the dees. By this time, the polrity cross the dees hs gin een reversed nd the ion is given nother kick cross the gp. The motion continues so tht for ech hlf-circle trip round one dee, the ion gins dditionl kinetic energy equl to q DV. When the rdius of its pth is nerly tht of the dees, the energetic ion leves the system through the exit slit. The cyclotron s opertion depends on T eing independent of the speed of the ion nd of the rdius of the circulr pth (Eq. 29.5). We cn otin n expression for the kinetic energy of the ion when it exits the cyclotron in terms of the rdius R of the dees. From Eqution 29.3, we know tht v 5 qbr/m. Hence, the kinetic energy is K 5 1 2mv 2 5 q 2 B 2 R 2 2m (29.9) When the energy of the ions in cyclotron exceeds out 20 MeV, reltivistic effects come into ply. (uch effects re discussed in Chpter 39.) Oservtions show tht T increses nd the moving ions do not remin in phse with the pplied potentil difference. ome ccelertors overcome this prolem y modifying the period of the pplied potentil difference so tht it remins in phse with the moving ions Mgnetic Force Acting on Current- Crrying Conductor If mgnetic force is exerted on single chrged prticle when the prticle moves through mgnetic field, it should not surprise you tht current-crrying wire lso experiences force when plced in mgnetic field. The current is collection of mny chrged prticles in motion; hence, the resultnt force exerted y the field on the wire is the vector sum of the individul forces exerted on ll the chrged prticles mking up the current. The force exerted on the prticles is trnsmitted to the wire when the prticles collide with the toms mking up the wire. Lwrence Berkeley Ntionl L

226 29.4 Mgnetic Force Acting on Current-Crrying Conductor 883 When there is no current in the wire, the wire remins verticl. When the current is upwrd, the wire deflects to the left. When the current is downwrd, the wire deflects to the right. Figure () A wire suspended verticlly etween the poles of mgnet. () (d) The setup shown in () s seen looking t the south pole of the mgnet so tht the mgnetic field (green crosses) is directed into the pge. N B in I 0 B in c d One cn demonstrte the mgnetic force cting on current-crrying conductor y hnging wire etween the poles of mgnet s shown in Figure For ese in visuliztion, prt of the horseshoe mgnet in prt () is removed to show the end fce of the south pole in prts () through (d) of Figure The mgnetic field is directed into the pge nd covers the region within the shded squres. When the current in the wire is zero, the wire remins verticl s in Figure When the wire crries current directed upwrd s in Figure 29.17c, however, the wire deflects to the left. If the current is reversed s in Figure 29.17d, the wire deflects to the right. Let s quntify this discussion y considering stright segment of wire of length L nd cross-sectionl re A crrying current I in uniform mgnetic field B s in Figure According to the mgnetic version of the prticle in field model, the mgnetic force exerted on chrge q moving with drift velocity vd is q vd 3 B. To find the totl force cting on the wire, we multiply the force q vd 3 B exerted on one chrge y the numer of chrges in the segment. Becuse the volume of the segment is AL, the numer of chrges in the segment is nal, where n is the numer of moile chrge crriers per unit volume. Hence, the totl mgnetic force on the segment of wire of length L is FB 5 1q vd 3 B 2nAL We cn write this expression in more convenient form y noting tht, from Eqution 27.4, the current in the wire is I 5 nqv d A. Therefore, I B in I The verge mgnetic force exerted on chrge moving in the wire is qv d B. F B q I B in v d The mgnetic force on the wire segment of length L is I L B. L Figure A segment of current-crrying wire in mgnetic field B. A FB 5 I L 3 B (29.10) where L is vector tht points in the direction of the current I nd hs mgnitude equl to the length L of the segment. This expression pplies only to stright segment of wire in uniform mgnetic field. Now consider n ritrrily shped wire segment of uniform cross section in mgnetic field s shown in Figure (pge 884). It follows from Eqution tht the mgnetic force exerted on smll segment of vector length d s in the presence of field B is d F B 5 I d s 3 B (29.11) W WForce on segment of current-crrying wire in uniform mgnetic field

227 884 Chpter 29 Mgnetic Fields Figure A wire segment of ritrry shpe crrying current I in mgnetic field B experiences mgnetic force. The mgnetic force on ny segment d s is I d s B nd is directed out of the pge. B d s I Exmple 29.4 Force on emicirculr Conductor A wire ent into semicircle of rdius R forms closed circuit nd crries current I. The wire lies in the xy plne, nd uniform mgnetic field is directed long the positive y xis s in Figure Find the mgnitude nd direction of the mgnetic force cting on the stright portion of the wire nd on the curved portion. olution where d F B is directed out of the pge for the directions of B nd d s in Figure Eqution cn e considered s n lterntive definition of B. Tht is, we cn define the mgnetic field B in terms of mesurle force exerted on current element, where the force is mximum when B is perpendiculr to the element nd zero when B is prllel to the element. To clculte the totl force F B cting on the wire shown in Figure 29.19, we integrte Eqution over the length of the wire: Conceptulize Using the right-hnd rule for cross products, we see tht the force F1 on the stright portion of the wire is out of the pge nd the force F2 on the curved portion is into the pge. Is F2 lrger in mgnitude thn F1 ecuse the length of the curved portion is longer thn tht of the stright portion? Ctegorize Becuse we re deling with current-crrying wire in mgnetic field rther thn single chrged prticle, we must use Eqution to find the totl force on ech portion of the wire. FB 5 I 3 d s 3 B (29.12) where nd represent the endpoints of the wire. When this integrtion is crried out, the mgnitude of the mgnetic field nd the direction the field mkes with the vector d s my differ t different points. Q uick Quiz 29.3 A wire crries current in the plne of this pper towrd the top of the pge. The wire experiences mgnetic force towrd the right edge of the pge. Is the direction of the mgnetic field cusing this force () in the plne of the pge nd towrd the left edge, () in the plne of the pge nd towrd the ottom edge, (c) upwrd out of the pge, or (d) downwrd into the pge? I y R d u u I u d s Figure (Exmple 29.4) The mgnetic force on the stright portion of the loop is directed out of the pge, nd the mgnetic force on the curved portion is directed into the pge. B x Anlyze Notice tht d s is perpendiculr to B everywhere on the stright portion of the wire. Use Eqution to find the force on this portion: F 1 5 I 3 d s 3 R B 5 I 3 B dx k^ 5 2IRB k^ 2R

228 29.5 Torque on Current Loop in Uniform Mgnetic Field continued To find the mgnetic force on the curved prt, first write n expression for the mgnetic force d F 2 on the element d s in Figure 29.20: (1) d F 2 5 Id s 3 B 5 2IB sin u ds k^ From the geometry in Figure 29.20, write n expression for ds: ustitute Eqution (2) into Eqution (1) nd integrte over the ngle u from 0 to p: 29.5 Torque on Current Loop in Uniform Mgnetic Field In ection 29.4, we showed how mgnetic force is exerted on current-crrying conductor plced in mgnetic field. With tht s strting point, we now show tht torque is exerted on current loop plced in mgnetic field. Consider rectngulr loop crrying current I in the presence of uniform mgnetic field directed prllel to the plne of the loop s shown in Figure No mgnetic forces ct on sides y nd c ecuse these wires re prllel to the field; hence, L 3 B 5 0 for these sides. Mgnetic forces do, however, ct on sides x nd v ecuse these sides re oriented perpendiculr to the field. The mgnitude of these forces is, from Eqution 29.10, B No mgnetic forces ct on sides y nd c ecuse these sides re prllel to B. x I y c I I I v ides x nd v re perpendiculr to the mgnetic field nd experience forces. F 2 5 F 4 5 IB The mgnetic forces F 2 nd F 4 exerted on sides x nd v crete torque tht tends to rotte the loop clockwise. x B (2) ds 5 R du F p p IRB sin u du k^ 5 2IRB 3 sin u du k^ 5 2IRB 32cos u 4 p 0 k^ F IRB 1cos p 2 cos 02k^ 5 IRB k^ 5 22IRB k^ Finlize Two very importnt generl sttements follow from this exmple. First, the force on the curved portion is the sme in mgnitude s the force on stright wire etween the sme two points. In generl, the mgnetic force on curved current-crrying wire in uniform mgnetic field is equl to tht on stright wire connecting the endpoints nd crrying the sme current. Furthermore, F 1 1 F is lso generl result: the net mgnetic force cting on ny closed current loop in uniform mgnetic field is zero. O 2 F 4 v 0 Figure () Overhed view of rectngulr current loop in uniform mgnetic field. () Edge view of the loop sighting down sides x nd v. The purple dot in the left circle represents current in wire x coming towrd you; the purple cross in the right circle represents current in wire v moving wy from you.

229 886 Chpter 29 Mgnetic Fields The direction of F2, the mgnetic force exerted on wire x, is out of the pge in the view shown in Figure nd tht of F4, the mgnetic force exerted on wire v, is into the pge in the sme view. If we view the loop from side c nd sight long sides x nd v, we see the view shown in Figure 29.21, nd the two mgnetic forces F2 nd F4 re directed s shown. Notice tht the two forces point in opposite directions ut re not directed long the sme line of ction. If the loop is pivoted so tht it cn rotte out point O, these two forces produce out O torque tht rottes the loop clockwise. The mgnitude of this torque t mx is t mx 5 F F IB IB2 2 5 IB where the moment rm out O is /2 for ech force. Becuse the re enclosed y the loop is A 5, we cn express the mximum torque s t mx 5 IAB (29.13) This mximum-torque result is vlid only when the mgnetic field is prllel to the plne of the loop. The sense of the rottion is clockwise when viewed from side c s indicted in Figure If the current direction were reversed, the force directions would lso reverse nd the rottionl tendency would e counterclockwise. Now suppose the uniform mgnetic field mkes n ngle u, 908 with line perpendiculr to the plne of the loop s in Figure For convenience, let s ssume B is perpendiculr to sides x nd v. In this cse, the mgnetic forces F 1 nd F 3 exerted on sides y nd c cncel ech other nd produce no torque ecuse they ct long the sme line. The mgnetic forces F 2 nd F 4 cting on sides x nd v, however, produce torque out ny point. Referring to the edge view shown in Figure 29.22, we see tht the moment rm of F 2 out the point O is equl to (/2) sin u. Likewise, the moment rm of F 4 out O is lso equl to (/2) sin u. Becuse F 2 5 F 4 5 IB, the mgnitude of the net torque out O is t 5 F 2 2 sin u 1 F 4 2 sin u 5 IB 2 sin u 1 IB sin u 5 IB sin u 2 5 IAB sin u where A 5 is the re of the loop. This result shows tht the torque hs its mximum vlue IAB when the field is perpendiculr to the norml to the plne of the loop (u 5 908) s discussed with regrd to Figure nd is zero when the field is prllel to the norml to the plne of the loop (u 5 0). F 2 x 2 A u O sin u 2 u v B F 4 Figure An edge view of the loop in Figure with the norml to the loop t n ngle u with respect to the mgnetic field. When the norml to the loop mkes n ngle u with the mgnetic field, the moment rm for the torque is (/2) sin u.

230 29.5 Torque on Current Loop in Uniform Mgnetic Field 887 (1) Curl your fingers in the direction of the current round the loop. m A (2) Your thum points in the direction of A nd m. Figure Right-hnd rule for determining the direction of the vector A for current loop. The direction of the mgnetic moment m is the sme s the direction of A. A convenient vector expression for the torque exerted on loop plced in uniform mgnetic field B is t 5 I A 3 B (29.14) where A, the vector shown in Figure 29.22, is perpendiculr to the plne of the loop nd hs mgnitude equl to the re of the loop. To determine the direction of A, use the right-hnd rule descried in Figure When you curl the fingers of your right hnd in the direction of the current in the loop, your thum points in the direction of A. Figure shows tht the loop tends to rotte in the direction of decresing vlues of u (tht is, such tht the re vector A rottes towrd the direction of the mgnetic field). The product I A is defined to e the mgnetic dipole moment m (often simply clled the mgnetic moment ) of the loop: I m ; I A (29.15) The I unit of mgnetic dipole moment is the mpere-meter 2 (A? m 2 ). If coil of wire contins N loops of the sme re, the mgnetic moment of the coil is m coil 5 NI A (29.16) Using Eqution 29.15, we cn express the torque exerted on current-crrying loop in mgnetic field B s t 5 m 3 B (29.17) This result is nlogous to Eqution 26.18, t 5 p 3 E, for the torque exerted on n electric dipole in the presence of n electric field E, where p is the electric dipole moment. Although we otined the torque for prticulr orienttion of B with respect to the loop, the eqution t 5 m 3 B is vlid for ny orienttion. Furthermore, lthough we derived the torque expression for rectngulr loop, the result is vlid for loop of ny shpe. The torque on n N-turn coil is given y Eqution y using Eqution for the mgnetic moment. In ection 26.6, we found tht the potentil energy of system of n electric dipole in n electric field is given y U E 5 2p? E. This energy depends on the orienttion of the dipole in the electric field. Likewise, the potentil energy of system of mgnetic dipole in mgnetic field depends on the orienttion of the dipole in the mgnetic field nd is given y U B 5 2m? B (29.18) W WTorque on current loop in mgnetic field W WMgnetic dipole moment of current loop W WTorque on mgnetic moment in mgnetic field Potentil energy of system of mgnetic moment in mgnetic field

231 888 Chpter 29 Mgnetic Fields Exmple 29.5 This expression shows tht the system hs its lowest energy U min 5 2mB when m points in the sme direction s B. The system hs its highest energy U mx 5 1mB when m points in the direction opposite B. Imgine the loop in Figure is pivoted t point O on sides nd, so tht it is free to rotte. If the loop crries current nd the mgnetic field is turned on, the loop is modeled s rigid oject under net torque, with the torque given y Eqution The torque on the current loop cuses the loop to rotte; this effect is exploited prcticlly in motor. Energy enters the motor y electricl trnsmission, nd the rotting coil cn do work on some device externl to the motor. For exmple, the motor in cr s electricl window system does work on the windows, pplying force on them nd moving them up or down through some displcement. We will discuss motors in more detil in ection Q uick Quiz 29.4 (i) Rnk the mgnitudes of the torques cting on the rectngulr loops (), (), nd (c) shown edge-on in Figure from highest to lowest. All loops re identicl nd crry the sme current. (ii) Rnk the mgnitudes of the net forces cting on the rectngulr loops shown in Figure from highest to lowest. Figure (Quick Quiz 29.4) Which current loop (seen edge-on) experiences the gretest torque, (), (), or (c)? Which experiences the gretest net force? The Mgnetic Dipole Moment of Coil A rectngulr coil of dimensions 5.40 cm cm consists of 25 turns of wire nd crries current of 15.0 ma. A T mgnetic field is pplied prllel to the plne of the coil. (A) Clculte the mgnitude of the mgnetic dipole moment of the coil. olution Conceptulize The mgnetic moment of the coil is independent of ny mgnetic field in which the loop resides, so it depends only on the geometry of the loop nd the current it crries. Ctegorize We evlute quntities sed on equtions developed in this section, so we ctegorize this exmple s sustitution prolem. Use Eqution to clculte the mgnetic moment ssocited with coil consisting of N turns: (B) Wht is the mgnitude of the torque cting on the loop? olution m coil 5 NIA 5 (25)( A)( m)( m) A # m 2 Use Eqution 29.17, noting tht B is perpendiculr to m coil: t 5 m coil B 5 ( A? m 2 )(0.350 T) N # m c

232 29.5 orque on urrent Loop in Uniform Mgnetic Field 889 Exmple 29.6 Rotting Coil Consider the loop of wire in Figure Imgine it is pivoted long side, which is prllel to the xis nd fs tened so tht side remins fixed nd the rest of the loop hngs verticlly in the grvittionl field of the Erth ut cn rotte round side (Fig ). The mss of the loop is 50.0 g, nd the sides re of lengths m nd m. The loop crries current of 3.50 A nd is immersed in verticl uniform mgnetic field of mgnitude T in the positive direction (Fig c). Wht ngle does the plne of the loop mke with the verticl? olu ion Conceptulize In the edge view of Figure 29.25, notice tht the mg netic moment of the loop is to the left. Therefore, when the loop is in the mgnetic field, the mgnetic torque on the loop cuses it to rotte in clockwise direction round side which we choose s the rottion xis. Imgine the loop mking this clock wise rottion so tht the plne of the loop is t some ngle to the verticl s in Figure 29.25c. The grvittionl force on the loop exerts torque tht would cuse rottion in the counter clockwise direction if the mgnetic field were turned off. Anlyze Evlute the mgnetic torque on the loop out side from Eqution 29.17: Evlute the grvittionl torque on the loop, noting tht the grvittionl force cn e modeled to ct t the center of the loop: Figure (Exmple 29.6) () The dimensions of rectngulr current loop. () Edge view of the loop sighting down sides nd. (c) An edge view of the loop in () rotted through n ngle with respect to the horizontl when it is plced in mgnetic field. Ctegorize At some ngle of the loop, the two torques descried in the Conceptulize step re equl in mgnitude nd the loop is t rest. We therefore model the loop s rigid oject in equilirium. 5 2m sin u 5 2IAB cos k 5 2IB cos k mg sin k From the rigid ody in equilirium model, dd the torques nd set the net torque equl to zero: 5 2IB cos mg sin k olve for IB cos u 5 mg sin tn u 5 IB mg u 5 tn IB mg ustitute numericl vlues: u 5 tn 3.50 A m T kg m 1.64 Finlize The ngle is reltively smll, so the loop still hngs lmost verticlly. If the current or the mgnetic field is incresed, however, the ngle increses s the mgnetic torque ecomes stronger.

233 890 Chpter 29 Mgnetic Fields When I is in the x direction nd B in the y direction, oth positive nd negtive chrge crriers re deflected upwrd in the mgnetic field. d t v d B F B z c F B v d Figure To oserve the Hll effect, mgnetic field is pplied to current-crrying conductor. The Hll voltge is mesured etween points nd c. B y I x 29.6 The Hll Effect When current-crrying conductor is plced in mgnetic field, potentil difference is generted in direction perpendiculr to oth the current nd the mgnetic field. This phenomenon, first oserved y Edwin Hll ( ) in 1879, is known s the Hll effect. The rrngement for oserving the Hll effect consists of flt conductor crrying current I in the x direction s shown in Figure A uniform mgnetic field B is pplied in the y direction. If the chrge crriers re electrons moving in the negtive x direction with drift velocity v d, they experience n upwrd mgnetic force F B 5 q v d 3 B, re deflected upwrd, nd ccumulte t the upper edge of the flt conductor, leving n excess of positive chrge t the lower edge (Fig ). This ccumultion of chrge t the edges estlishes n electric field in the conductor nd increses until the electric force on crriers remining in the ulk of the conductor lnces the mgnetic force cting on the crriers. The electrons cn now e descried y the prticle in equilirium model, nd they re no longer deflected upwrd. A sensitive voltmeter connected cross the smple s shown in Figure cn mesure the potentil difference, known s the Hll voltge DV H, generted cross the conductor. If the chrge crriers re positive nd hence move in the positive x direction (for rightwrd current) s shown in Figures nd 29.27, they lso experience n upwrd mgnetic force q v d 3 B, which produces uildup of positive chrge on the upper edge nd leves n excess of negtive chrge on the lower edge. Hence, the sign of the Hll voltge generted in the smple is opposite the sign of the Hll voltge resulting from the deflection of electrons. The sign of the chrge crriers cn therefore e determined from mesuring the polrity of the Hll voltge. In deriving n expression for the Hll voltge, first note tht the mgnetic force exerted on the crriers hs mgnitude qv d B. In equilirium, this force is lnced y the electric force qe H, where E H is the mgnitude of the electric field due to the chrge seprtion (sometimes referred to s the Hll field). Therefore, When the chrge crriers re negtive, the upper edge of the conductor ecomes negtively chrged nd c is t lower electric potentil thn. qv d B 5 qe H E H 5 v d B If d is the width of the conductor, the Hll voltge is DV H 5 E H d 5 v d Bd (29.19) The chrge crriers re no longer deflected when the edges ecome sufficiently chrged tht there is lnce etween the electric force nd the mgnetic force. When the chrge crriers re positive, the upper edge of the conductor ecomes positively chrged nd c is t higher potentil thn. I B c q v d B v d q E H I V H =+=+1.50 V I B c q v d B q E H v d I V H =+=+2.50 V Figure The sign of the Hll voltge depends on the sign of the chrge crriers.

234 29.6 The Hll Effect 891 Therefore, the mesured Hll voltge gives vlue for the drift speed of the chrge crriers if d nd B re known. We cn otin the chrge-crrier density n y mesuring the current in the smple. From Eqution 27.4, we cn express the drift speed s v d 5 I nqa (29.20) where A is the cross-sectionl re of the conductor. ustituting Eqution into Eqution gives DV H 5 IBd (29.21) nqa Becuse A 5 td, where t is the thickness of the conductor, we cn lso express Eqution s DV H 5 IB nqt 5 R HIB (29.22) t where R H 5 1/nq is clled the Hll coefficient. This reltionship shows tht properly clirted conductor cn e used to mesure the mgnitude of n unknown mgnetic field. Becuse ll quntities in Eqution other thn nq cn e mesured, vlue for the Hll coefficient is redily otinle. The sign nd mgnitude of R H give the sign of the chrge crriers nd their numer density. In most metls, the chrge crriers re electrons nd the chrge-crrier density determined from Hlleffect mesurements is in good greement with clculted vlues for such metls s lithium (Li), sodium (N), copper (Cu), nd silver (Ag), whose toms ech give up one electron to ct s current crrier. In this cse, n is pproximtely equl to the numer of conducting electrons per unit volume. This clssicl model, however, is not vlid for metls such s iron (Fe), ismuth (Bi), nd cdmium (Cd) or for semiconductors. These discrepncies cn e explined only y using model sed on the quntum nture of solids. Exmple 29.7 The Hll Effect for Copper WWThe Hll voltge A rectngulr copper strip 1.5 cm wide nd 0.10 cm thick crries current of 5.0 A. Find the Hll voltge for 1.2-T mgnetic field pplied in direction perpendiculr to the strip. olution Conceptulize tudy Figures nd crefully nd mke sure you understnd tht Hll voltge is developed etween the top nd ottom edges of the strip. Ctegorize We evlute the Hll voltge using n eqution developed in this section, so we ctegorize this exmple s sustitution prolem. Assuming one electron per tom is ville for conduction, find the chrgecrrier density in terms of the molr mss M nd density r of copper: ustitute this result into Eqution 29.22: n 5 N A V 5 N Ar M DV H 5 IB nqt 5 MIB N A rqt kg/mol A T2 ustitute numericl vlues: DV H mol kg/m C m mv continued

235 892 Chpter 29 Mgnetic Fields 29.7 continued uch n extremely smll Hll voltge is expected in good conductors. (Notice tht the width of the conductor is not needed in this clcultion.) Wht If? Wht if the strip hs the sme dimensions ut is mde of semiconductor? Will the Hll voltge e smller or lrger? Answer In semiconductors, n is much smller thn it is in metls tht contriute one electron per tom to the current; hence, the Hll voltge is usully lrger ecuse it vries s the inverse of n. Currents on the order of 0.1 ma re generlly used for such mterils. Consider piece of silicon tht hs the sme dimensions s the copper strip in this exmple nd whose vlue for n is electrons/m 3. Tking B T nd I ma, we find tht DV H mv. A potentil difference of this mgnitude is redily mesured. ummry Definition The mgnetic dipole moment m of loop crrying current I is m ; I A (29.15) where the re vector A is perpendiculr to the plne of the loop nd 0 A 0 is equl to the re of the loop. The I unit of m is A? m 2. Concepts nd Principles If chrged prticle moves in uniform mgnetic field so tht its initil velocity is perpendiculr to the field, the prticle moves in circle, the plne of which is perpendiculr to the mgnetic field. The rdius of the circulr pth is r 5 mv qb where m is the mss of the prticle nd q is its chrge. The ngulr speed of the chrged prticle is If stright conductor of length L crries current I, the force exerted on tht conductor when it is plced in uniform mgnetic field B is FB 5 I L 3 B (29.10) where the direction of L is in the direction of the current nd 0 L 0 5 L. v 5 qb m (29.3) (29.4) If n ritrrily shped wire crrying current I is plced in mgnetic field, the mgnetic force exerted on very smll segment d s is d F B 5 I d s 3 B (29.11) To determine the totl mgnetic force on the wire, one must integrte Eqution over the wire, keeping in mind tht oth B nd d s my vry t ech point. The torque t on current loop plced in uniform mgnetic field B is t 5 m 3 B (29.17) The potentil energy of the system of mgnetic dipole in mgnetic field is U B 5 2m? B (29.18)

236 Ojective Questions 893 Anlysis Models for Prolem olving Prticle in Field (Mgnetic) A source (to e discussed in Chpter 30) estlishes mgnetic field B throughout spce. When prticle with chrge q nd moving with velocity v is plced in tht field, it experiences mgnetic force given y FB 5 q v 3 B (29.1) The direction of this mgnetic force is perpendiculr oth to the velocity of the prticle nd to the mgnetic field. The mgnitude of this force is Ojective Questions Ojective Questions 3, 4, nd 6 in Chpter 11 cn e ssigned with this chpter s review for the vector product. 1. denotes nswer ville in tudent olutions Mnul/tudy Guide 1. A sptilly uniform mgnetic field cnnot exert mgnetic force on prticle in which of the following circumstnces? There my e more thn one correct sttement. () The prticle is chrged. () The prticle moves perpendiculr to the mgnetic field. (c) The prticle moves prllel to the mgnetic field. (d) The mgnitude of the mgnetic field chnges with time. (e) The prticle is t rest. 2. Rnk the mgnitudes of the forces exerted on the following prticles from lrgest to smllest. In your rnking, disply ny cses of equlity. () n electron moving t 1 Mm/s perpendiculr to 1-mT mgnetic field () n electron moving t 1 Mm/s prllel to 1-mT mgnetic field (c) n electron moving t 2 Mm/s perpendiculr to 1-mT mgnetic field (d) proton moving t 1 Mm/s perpendiculr to 1-mT mgnetic field (e) proton moving t 1 Mm/s t 458 ngle to 1-mT mgnetic field 3. A prticle with electric chrge is fired into region of spce where the electric field is zero. It moves in stright line. Cn you conclude tht the mgnetic field in tht region is zero? () Yes, you cn. () No; the field might e perpendiculr to the prticle s velocity. (c) No; the field might e prllel to the prticle s velocity. (d) No; the prticle might need to hve chrge of the opposite sign to hve force exerted on it. (e) No; n oservtion of n oject with electric chrge gives no informtion out mgnetic field. 4. A proton moving horizontlly enters region where uniform mgnetic field is directed perpendiculr to the proton s velocity s shown in Figure OQ29.4. After the proton enters the field, does it () deflect downwrd, with its speed remining constnt; () deflect upwrd, moving in semicirculr pth with constnt speed, nd exit the field moving to the left; (c) continue to move in the horizontl direction with constnt velocity; (d) move in circulr orit nd ecome trpped y the field; or (e) deflect out of the plne of the pper? F B 5 0q 0 vb sin u (29.2) where u is the smller ngle etween v nd B. The I unit of B is the tesl (T), where 1 T 5 1 N/A m. v x v Figure OQ29.4 z q F B q v B 5. At certin instnt, proton is moving in the positive x direction through mgnetic field in the negtive z direction. Wht is the direction of the mgnetic force exerted on the proton? () positive z direction () negtive z direction (c) positive y direction (d) negtive y direction (e) The force is zero. 6. A thin copper rod 1.00 m long hs mss of 50.0 g. Wht is the minimum current in the rod tht would llow it to levitte ove the ground in mgnetic field of mgnitude T? () 1.20 A () 2.40 A (c) 4.90 A (d) 9.80 A (e) none of those nswers 7. Electron A is fired horizontlly with speed 1.00 Mm/s into region where verticl mgnetic field exists. Electron B is fired long the sme pth with speed 2.00 Mm/s. (i) Which electron hs lrger mgnetic force exerted on it? () A does. () B does. (c) The forces hve the sme nonzero mgnitude. (d) The forces re oth zero. (ii) Which electron hs pth tht curves more shrply? () A does. () B does. (c) The prticles follow the sme curved pth. (d) The prticles continue to go stright. 8. Clssify ech of the following sttements s chrcteristic () of electric forces only, () of mgnetic forces only, (c) of oth electric nd mgnetic forces, or (d) of neither electric nor mgnetic forces. (i) The force is proportionl to the mgnitude of the field exerting it. (ii) The force is proportionl to the mgnitude of the chrge of the oject on which the force is exerted. (iii) The force exerted on negtively chrged oject is opposite in direction to the force on positive chrge. (iv) The force exerted on sttionry chrged oject is nonzero. (v) The force exerted on moving chrged B y

237 894 Chpter 29 Mgnetic Fields oject is zero. (vi) The force exerted on chrged oject is proportionl to its speed. (vii) The force exerted on chrged oject cnnot lter the oject s speed. (viii) The mgnitude of the force depends on the chrged oject s direction of motion. 9. An electron moves horizontlly cross the Erth s equtor t speed of m/s nd in direction N of E. At this point, the Erth s mgnetic field hs direction due north, is prllel to the surfce, nd hs vlue of T. Wht is the force cting on the electron due to its interction with the Erth s mgnetic field? () N due west () N towrd the Erth s surfce (c) N towrd the Erth s surfce (d) N wy from the Erth s surfce (e) N wy from the Erth s surfce 10. A chrged prticle is trveling through uniform mgnetic field. Which of the following sttements re true of the mgnetic field? There my e more thn one correct sttement. () It exerts force on the prticle prllel to the field. () It exerts force on the prticle long the direction of its motion. (c) It increses the kinetic energy of the prticle. (d) It exerts force tht is perpendiculr to the direction of motion. (e) It does not chnge the mgnitude of the momentum of the prticle. 11. In the velocity selector shown in Figure 29.13, electrons with speed v 5 E/B follow stright pth. Electrons moving significntly fster thn this speed through the sme selector will move long wht kind of pth? () Conceptul Questions 1. Cn constnt mgnetic field set into motion n electron initilly t rest? Explin your nswer. 2. Explin why it is not possile to determine the chrge nd the mss of chrged prticle seprtely y mesuring ccelertions produced y electric nd mgnetic forces on the prticle. 3. Is it possile to orient current loop in uniform mgnetic field such tht the loop does not tend to rotte? Explin. 4. How cn the motion of moving chrged prticle e used to distinguish etween mgnetic field nd n Prolems circle () prol (c) stright line (d) more complicted trjectory 12. Answer ech question yes or no. Assume the motions nd currents mentioned re long the x xis nd fields re in the y direction. () Does n electric field exert force on sttionry chrged oject? () Does mgnetic field do so? (c) Does n electric field exert force on moving chrged oject? (d) Does mgnetic field do so? (e) Does n electric field exert force on stright current-crrying wire? (f) Does mgnetic field do so? (g) Does n electric field exert force on em of moving electrons? (h) Does mgnetic field do so? 13. A mgnetic field exerts torque on ech of the currentcrrying single loops of wire shown in Figure OQ The loops lie in the xy plne, ech crrying the sme mgnitude current, nd the uniform mgnetic field points in the positive x direction. Rnk the loops y the mgnitude of the torque exerted on them y the field from lrgest to smllest. y (m) 1. denotes nswer ville in tudent olutions Mnul/tudy Guide A B B C Figure OQ29.13 x (m) electric field? Give specific exmple to justify your rgument. 5. How cn current loop e used to determine the presence of mgnetic field in given region of spce? 6. Chrged prticles from outer spce, clled cosmic rys, strike the Erth more frequently ner the poles thn ner the equtor. Why? 7. Two chrged prticles re projected in the sme direction into mgnetic field perpendiculr to their velocities. If the prticles re deflected in opposite directions, wht cn you sy out them? The prolems found in this chpter my e ssigned online in Enhnced WeAssign 1. strightforwrd; 2. intermedite; 3. chllenging 1. full solution ville in the tudent olutions Mnul/tudy Guide AMT Anlysis Model tutoril ville in Enhnced WeAssign GP Guided Prolem M Mster It tutoril ville in Enhnced WeAssign W Wtch It video solution ville in Enhnced WeAssign BIO Q/C

238 Prolems 895 ection 29.1 Anlysis Model: Prticle in Field (Mgnetic) Prolems 1 4, 6 7, nd 10 in Chpter 11 cn e ssigned with this section s review for the vector product. 1. At the equtor, ner the surfce of the Erth, the mgnetic field is pproximtely 50.0 mt northwrd, nd the electric field is out 100 N/C downwrd in fir wether. Find the grvittionl, electric, nd mgnetic forces on n electron in this environment, ssuming tht the electron hs n instntneous velocity of m/s directed to the est. 2. Determine the initil direction of the deflection of W chrged prticles s they enter the mgnetic fields shown in Figure P29.2. B right B in B up B t c d Figure P Find the direction of the mgnetic field cting on positively chrged prticle moving in the vrious situtions shown in Figure P29.3 if the direction of the mgnetic force cting on it is s indicted. F B v F B v (out) c Figure P29.3 F B v (in) 4. Consider n electron ner the Erth s equtor. In which direction does it tend to deflect if its velocity is () directed downwrd? () Directed northwrd? (c) Directed westwrd? (d) Directed southestwrd? 5. A proton is projected into mgnetic field tht is directed long the positive x xis. Find the direction of the mgnetic force exerted on the proton for ech of the following directions of the proton s velocity: () the positive y direction, () the negtive y direction, (c) the positive x direction. 6. A proton moving t m/s through mgnetic field of mgnitude 1.70 T experiences mgnetic M force of mgnitude N. Wht is the ngle etween the proton s velocity nd the field? 7. An electron is ccelerted through V from W rest nd then enters uniform 1.70-T mgnetic field. Wht re () the mximum nd () the minimum vlues of the mgnetic force this prticle experiences? 8. A proton moves with velocity of v 5 W 12 i^ 2 4 j^ 1 k^ 2 m/s in region in which the mgnetic field is B 5 1i^ 1 2 j^ 2 k^ 2 T. Wht is the mgnitude of the mgnetic force this prticle experiences? 9. A proton trvels with speed of m/s in AMT direction tht mkes n ngle of with the direction of mgnetic field of mgnitude T in the positive x direction. Wht re the mgnitudes of () the mgnetic force on the proton nd () the proton s ccelertion? 10. A lortory electromgnet produces mgnetic field Q/C of mgnitude 1.50 T. A proton moves through this field with speed of m/s. () Find the mgnitude of the mximum mgnetic force tht could e exerted on the proton. () Wht is the mgnitude of the mximum ccelertion of the proton? (c) Would the field exert the sme mgnetic force on n electron moving through the field with the sme speed? (d) Would the electron experience the sme ccelertion? Explin. 11. A proton moves perpendiculr to uniform mgnetic M field B t speed of m/s nd experiences n ccelertion of m/s 2 in the positive x direction when its velocity is in the positive z direction. Determine the mgnitude nd direction of the field. 12. Review. A chrged prticle of mss 1.50 g is moving t speed of m/s. uddenly, uniform mgnetic field of mgnitude mt in direction perpendiculr to the prticle s velocity is turned on nd then turned off in time intervl of 1.00 s. During this time intervl, the mgnitude nd direction of the velocity of the prticle undergo negligile chnge, ut the prticle moves y distnce of m in direction perpendiculr to the velocity. Find the chrge on the prticle. ection 29.2 Motion of Chrged Prticle in Uniform Mgnetic Field 13. An electron moves in circulr pth perpendiculr to uniform mgnetic field with mgnitude of 2.00 mt. If the speed of the electron is m/s, determine () the rdius of the circulr pth nd () the time intervl required to complete one revolution. 14. An ccelerting voltge of V is pplied to n Q/C electron gun, producing em of electrons originlly trveling horizontlly north in vcuum towrd the center of viewing screen 35.0 cm wy. Wht re () the mgnitude nd () the direction of the deflection on

239 896 Chpter 29 Mgnetic Fields the screen cused y the Erth s grvittionl field? Wht re (c) the mgnitude nd (d) the direction of the deflection on the screen cused y the verticl component of the Erth s mgnetic field, tken s 20.0 mt down? (e) Does n electron in this verticl mgnetic field move s projectile, with constnt vector ccelertion perpendiculr to constnt northwrd component of velocity? (f) Is it good pproximtion to ssume it hs this projectile motion? Explin. 15. A proton (chrge 1e, mss m p ), deuteron (chrge 1e, mss 2m p ), nd n lph prticle (chrge 12e, mss 4m p ) re ccelerted from rest through common potentil difference DV. Ech of the prticles enters uniform mgnetic field B, with its velocity in direction perpendiculr to B. The proton moves in circulr pth of rdius r p. In terms of r p, determine () the rdius r d of the circulr orit for the deuteron nd () the rdius r for the lph prticle. 16. A prticle with chrge q nd kinetic energy K trvels in uniform mgnetic field of mgnitude B. If the prticle moves in circulr pth of rdius R, find expressions for () its speed nd () its mss. 17. Review. One electron collides elsticlly with second AMT electron initilly t rest. After the collision, the rdii of their trjectories re 1.00 cm nd 2.40 cm. The trjectories re perpendiculr to uniform mgnetic field of mgnitude T. Determine the energy (in kev) of the incident electron. 18. Review. One electron collides elsticlly with second electron initilly t rest. After the collision, the rdii of their trjectories re r 1 nd r 2. The trjectories re perpendiculr to uniform mgnetic field of mgnitude B. Determine the energy of the incident electron. 19. Review. An electron moves in circulr pth perpendiculr to constnt mgnetic field of mgnitude 1.00 mt. The ngulr momentum of the electron out the center of the circle is kg? m 2 /s. Determine () the rdius of the circulr pth nd () the speed of the electron. 20. Review. A 30.0-g metl ll hving net chrge Q 5 Q/C 5.00 mc is thrown out of window horizontlly north t speed v m/s. The window is t height h m ove the ground. A uniform, horizontl mgnetic field of mgnitude B T is perpendiculr to the plne of the ll s trjectory nd directed towrd the west. () Assuming the ll follows the sme trjectory s it would in the sence of the mgnetic field, find the mgnetic force cting on the ll just efore it hits the ground. () Bsed on the result of prt (), is it justified for three-significnt-digit precision to ssume the trjectory is unffected y the mgnetic field? Explin. 21. A cosmic-ry proton in interstellr spce hs n energy M of 10.0 MeV nd executes circulr orit hving rdius equl to tht of Mercury s orit round the un ( m). Wht is the mgnetic field in tht region of spce? 22. Assume the region to the right of certin plne contins uniform mgnetic field of mgnitude 1.00 mt nd the field is zero in the region to the left of the plne s shown in Figure P An electron, originlly trveling perpendiculr to the oundry plne, psses into the region of the field. () Determine the time intervl required for the electron to leve the field-filled region, noting tht the electron s pth is semicircle. () Assuming the mximum depth of penetrtion into the field is 2.00 cm, find the kinetic energy of the electron. e v B 0 B 1.00 mt Figure P A singly chrged ion of mss m is ccelerted from rest y potentil difference DV. It is then deflected y uniform mgnetic field (perpendiculr to the ion s velocity) into semicircle of rdius R. Now douly chrged ion of mss m9 is ccelerted through the sme potentil difference nd deflected y the sme mgnetic field into semicircle of rdius R9 5 2R. Wht is the rtio of the msses of the ions? ection 29.3 Applictions Involving Chrged Prticles Moving in Mgnetic Field 24. A cyclotron designed to ccelerte protons hs mgnetic field of mgnitude T over region of rdius M 1.20 m. Wht re () the cyclotron frequency nd () the mximum speed cquired y the protons? 25. Consider the mss spectrometer shown schemticlly W in Figure The mgnitude of the electric field etween the pltes of the velocity selector is V/m, nd the mgnetic field in oth the velocity selector nd the deflection chmer hs mgnitude of T. Clculte the rdius of the pth for singly chrged ion hving mss m kg. 26. ingly chrged urnium-238 ions re ccelerted Q/C through potentil difference of 2.00 kv nd enter uniform mgnetic field of mgnitude 1.20 T directed perpendiculr to their velocities. () Determine the rdius of their circulr pth. () Repet this clcultion for urnium-235 ions. (c) Wht If? How does the rtio of these pth rdii depend on the ccelerting voltge? (d) On the mgnitude of the mgnetic field? 27. A cyclotron (Fig ) designed to ccelerte protons hs n outer rdius of m. The protons re emitted nerly t rest from source t the center nd re ccelerted through 600 V ech time they cross the gp etween the dees. The dees re etween the poles of n electromgnet where the field is T. () Find the cyclotron frequency for the protons in

240 Prolems 897 this cyclotron. Find () the speed t which protons exit the cyclotron nd (c) their mximum kinetic energy. (d) How mny revolutions does proton mke in the cyclotron? (e) For wht time intervl does the proton ccelerte? 28. A prticle in the cyclotron shown in Figure Q/C gins energy q DV from the lternting power supply ech time it psses from one dee to the other. The time intervl for ech full orit is T 5 2p v 5 2pm qb so the prticle s verge rte of increse in energy is 2q DV T 5 q 2 B DV pm Notice tht this power input is constnt in time. On the other hnd, the rte of increse in the rdius r of its pth is not constnt. () how tht the rte of increse in the rdius r of the prticle s pth is given y dr dt 5 1 DV r pb () Descrie how the pth of the prticles in Figure is consistent with the result of prt (). (c) At wht rte is the rdil position of the protons in cyclotron incresing immeditely efore the protons leve the cyclotron? Assume the cyclotron hs n outer rdius of m, n ccelerting voltge of DV V, nd mgnetic field of mgnitude T. (d) By how much does the rdius of the protons pth increse during their lst full revolution? 29. A velocity selector consists of electric nd mgnetic fields descried y the expressions E 5 E k^ nd B W 5 B j^, with B mt. Find the vlue of E such tht 750-eV electron moving in the negtive x direction is undeflected. 30. In his experiments on cthode rys during which he Q/C discovered the electron, J. J. Thomson showed tht the sme em deflections resulted with tues hving cthodes mde of different mterils nd contining vrious gses efore evcution. () Are these oservtions importnt? Explin your nswer. () When he pplied vrious potentil differences to the deflection pltes nd turned on the mgnetic coils, lone or in comintion with the deflection pltes, Thomson oserved tht the fluorescent screen continued to show single smll glowing ptch. Argue whether his oservtion is importnt. (c) Do clcultions to show tht the chrge-tomss rtio Thomson otined ws huge compred with tht of ny mcroscopic oject or of ny ionized tom or molecule. How cn one mke sense of this comprison? (d) Could Thomson oserve ny deflection of the em due to grvittion? Do clcultion to rgue for your nswer. Note: To otin visily glowing ptch on the fluorescent screen, the potentil difference etween the slits nd the cthode must e 100 V or more. 31. The picture tue in n old lck-nd-white television uses mgnetic deflection coils rther thn electric deflection pltes. uppose n electron em is ccelerted through 50.0-kV potentil difference nd then through region of uniform mgnetic field 1.00 cm wide. The screen is locted 10.0 cm from the center of the coils nd is 50.0 cm wide. When the field is turned off, the electron em hits the center of the screen. Ignoring reltivistic corrections, wht field mgnitude is necessry to deflect the em to the side of the screen? ection 29.4 Mgnetic Force Acting on Current- Crrying Conductor 32. A stright wire crrying 3.00-A current is plced in Q/C uniform mgnetic field of mgnitude T directed perpendiculr to the wire. () Find the mgnitude of the mgnetic force on section of the wire hving length of 14.0 cm. () Explin why you cn t determine the direction of the mgnetic force from the informtion given in the prolem. 33. A conductor crrying current I A is directed long the positive x xis nd perpendiculr to uniform mgnetic field. A mgnetic force per unit length of N/m cts on the conductor in the negtive y direction. Determine () the mgnitude nd () the direction of the mgnetic field in the region through which the current psses. 34. A wire 2.80 m in length crries current of 5.00 A in W region where uniform mgnetic field hs mgnitude of T. Clculte the mgnitude of the mgnetic force on the wire ssuming the ngle etween the mgnetic field nd the current is () 60.08, () 90.08, nd (c) A wire crries stedy current of 2.40 A. A stright W section of the wire is m long nd lies long the x xis within uniform mgnetic field, B k^ T. If the current is in the positive x direction, wht is the mgnetic force on the section of wire? 36. Why is the following sitution impossile? Imgine copper wire with rdius 1.00 mm encircling the Erth t its mgnetic equtor, where the field direction is horizontl. A power supply delivers 100 MW to the wire to mintin current in it, in direction such tht the mgnetic force from the Erth s mgnetic field is upwrd. Due to this force, the wire is levitted immeditely ove the ground. 37. Review. A rod of mss kg nd rdius 6.00 cm AMT rests on two prllel rils (Fig. P29.37) tht re d 5 W 12.0 cm prt nd L cm long. The rod crries d I L Figure P29.37 Prolems 37 nd 38. B

241 898 Chpter 29 Mgnetic Fields current of I A in the direction shown nd rolls long the rils without slipping. A uniform mgnetic field of mgnitude T is directed perpendiculr to the rod nd the rils. If it strts from rest, wht is the speed of the rod s it leves the rils? 38. Review. A rod of mss m nd rdius R rests on two prllel rils (Fig. P29.37) tht re distnce d prt nd hve length L. The rod crries current I in the direction shown nd rolls long the rils without slipping. A uniform mgnetic field B is directed perpendiculr to the rod nd the rils. If it strts from rest, wht is the speed of the rod s it leves the rils? 39. A wire hving mss per unit length of g/cm M crries 2.00-A current horizontlly to the south. Wht re () the direction nd () the mgnitude of the minimum mgnetic field needed to lift this wire verticlly upwrd? 40. Consider the system pictured in Figure P A Q/C 15.0-cm horizontl wire of mss 15.0 g is plced etween two thin, verticl conductors, nd uniform mgnetic field cts perpendiculr to the pge. The wire is free to move verticlly without friction on the two verticl conductors. When 5.00-A current is directed s shown in the figure, the horizontl wire moves upwrd t constnt velocity in the presence of grvity. () Wht forces ct on the horizontl wire, nd () under wht condition is the wire le to move upwrd t constnt velocity? (c) Find the mgnitude nd direction of the minimum mgnetic field required to move the wire t constnt speed. (d) Wht hppens if the mgnetic field exceeds this minimum vlue? 5.00 A 15.0 cm 5.00 A 5.00 A Figure P A horizontl power line of length 58.0 m crries current of 2.20 ka northwrd s shown in Figure P The Erth s mgnetic field t this loction hs mgnitude of T. The field t this loction is directed towrd the north t n ngle elow the I power line. Find () the mgnitude nd () the direction of the mgnetic force on the power line. 42. A strong mgnet is plced under horizontl conducting ring of rdius r tht crries current I s shown in Figure P If the mgnetic field B mkes n ngle u with the verticl t the ring s loction, wht re () the mgnitude nd () the direction of the resultnt mgnetic force on the ring? u I N Figure P Assume the Erth s mgnetic field is 52.0 mt northwrd t elow the horizontl in Atlnt, Georgi. A tue in neon sign stretches etween two digonlly opposite corners of shop window which lies in north south verticl plne nd crries current 35.0 ma. The current enters the tue t the ottom south corner of the shop s window. It exits t the opposite corner, which is 1.40 m frther north nd m higher up. Between these two points, the glowing tue spells out DONUT. Determine the totl vector mgnetic force on the tue. Hint: You my use the first importnt generl sttement presented in the Finlize section of Exmple In Figure P29.44, the cue is 40.0 cm on ech edge. Q/C Four stright segments of wire, c, cd, nd d form closed loop tht crries current I A in the direction shown. A uniform mgnetic field of mgnitude B T is in the positive y direction. Determine the mgnetic force vector on (), () c, (c) cd, nd (d) d. (e) Explin how you could find the force exerted on the fourth of these segments from the forces on the other three, without further clcultion involving the mgnetic field. d y r B B I u B Figure P z c Figure P29.44 ection 29.5 Torque on Current Loop in Uniform Mgnetic Field 45. A typicl mgnitude of the externl mgnetic field in BIO crdic ctheter ltion procedure using remote x

242 Prolems 899 mgnetic nvigtion is B T. uppose tht the permnent mgnet in the ctheter used in the procedure is inside the left trium of the hert nd suject to this externl mgnetic field. The permnent mgnet hs mgnetic moment of 0.10 A? m 2. The orienttion of the permnent mgnet is 30 from the direction of the externl mgnetic field lines. () Wht is the mgnitude of the torque on the tip of the ctheter contining this permnent mgnet? () Wht is the potentil energy of the system consisting of the permnent mgnet in the ctheter nd the mgnetic field provided y the externl mgnets? 46. A 50.0-turn circulr coil of rdius 5.00 cm cn e oriented in ny direction in uniform mgnetic field hving mgnitude of T. If the coil crries current of 25.0 ma, find the mgnitude of the mximum possile torque exerted on the coil. 47. A mgnetized sewing needle hs mgnetic moment Q/C of 9.70 ma? m 2. At its loction, the Erth s mgnetic field is 55.0 mt northwrd t elow the horizontl. Identify the orienttions of the needle tht represent () the minimum potentil energy nd () the mximum potentil energy of the needle field system. (c) How much work must e done on the system to move the needle from the minimum to the mximum potentil energy orienttion? 48. A current of 17.0 ma is mintined in single circulr loop of 2.00 m circumference. A mgnetic field of W T is directed prllel to the plne of the loop. () Clculte the mgnetic moment of the loop. () Wht is the mgnitude of the torque exerted y the mgnetic field on the loop? 49. An eight-turn coil encloses n ellipticl re hving M mjor xis of 40.0 cm nd minor xis of 30.0 cm (Fig. P29.49). The coil lies in the plne of the pge nd hs 6.00-A current flowing clockwise round it. If the coil is in uniform mgnetic field of T directed towrd the left of the pge, wht is the mgnitude of the torque on the coil? Hint: The re of n ellipse is A 5 p, where nd re, respectively, the semimjor nd semiminor xes of the ellipse. B I 30.0 cm 40.0 cm Figure P The rotor in certin electric motor is flt, rectngulr coil with 80 turns of wire nd dimensions 2.50 cm y 4.00 cm. The rotor rottes in uniform mgnetic field of T. When the plne of the rotor is perpendiculr to the direction of the mgnetic field, the rotor crries current of 10.0 ma. In this orienttion, the mgnetic moment of the rotor is directed opposite the mgnetic field. The rotor then turns through one- hlf revolution. This process is repeted to cuse the rotor to turn stedily t n ngulr speed of rev/min. () Find the mximum torque cting on the rotor. () Find the pek power output of the motor. (c) Determine the mount of work performed y the mgnetic field on the rotor in every full revolution. (d) Wht is the verge power of the motor? 51. A rectngulr coil consists of N closely wrpped M turns nd hs dimensions m nd m. The coil is hinged long the y xis, nd its plne mkes n ngle u with the x xis (Fig. P29.51). () Wht is the mgnitude of the torque exerted on the coil y uniform mgnetic field B T directed in the positive x direction when the current is I A in the direction shown? () Wht is the expected direction of rottion of the coil? z y Figure P A rectngulr loop of wire hs dimensions m y GP m. The loop is pivoted t the x xis nd lies in the Q/C xy plne s shown in Figure P A uniform mgnetic field of mgnitude 1.50 T is directed t n ngle of with respect to the y xis with field lines prllel to the yz plne. The loop crries current of A in the direction shown. (Ignore grvittion.) We wish to evlute the torque on the current loop. () Wht is the direction of the mgnetic force exerted on wire segment? () Wht is the direction of the torque ssocited with this force out n xis through the origin? (c) Wht is the direction of the mgnetic force exerted on segment cd? (d) Wht is the direction of the torque ssocited with this force out n xis through the origin? (e) Cn the forces exmined in prts () nd (c) comine to cuse the loop to rotte round the x xis? (f) Cn they ffect the motion of the loop in ny wy? Explin. (g) Wht is the direction of the mgnetic force exerted on segment c? (h) Wht is the direction of the torque ssocited with this force out n xis through the origin? (i) Wht is the torque on segment d out n xis through the origin? ( j) From the point of view of Figure P29.52, once the loop is relesed from rest t z x d I c I B Figure P29.52 u x B 40.0 y

243 900 Chpter 29 Mgnetic Fields the position shown, will it rotte clockwise or counterclockwise round the x xis? (k) Compute the mgnitude of the mgnetic moment of the loop. (l) Wht is the ngle etween the mgnetic moment vector nd the mgnetic field? (m) Compute the torque on the loop using the results to prts (k) nd (l). 53. A wire is formed into circle hving dimeter of W 10.0 cm nd is plced in uniform mgnetic field of 3.00 mt. The wire crries current of 5.00 A. Find () the mximum torque on the wire nd () the rnge of potentil energies of the wire field system for different orienttions of the circle. ection 29.6 The Hll Effect 54. A Hll-effect proe opertes with 120-mA current. When the proe is plced in uniform mgnetic field of mgnitude T, it produces Hll voltge of mv. () When it is used to mesure n unknown mgnetic field, the Hll voltge is mv. Wht is the mgnitude of the unknown field? () The thickness of the proe in the direction of B is 2.00 mm. Find the density of the chrge crriers, ech of which hs chrge of mgnitude e. 55. In n experiment designed to mesure the Erth s M mgnetic field using the Hll effect, copper r cm thick is positioned long n est west direction. Assume n electrons/m 3 nd the plne of the r is rotted to e perpendiculr to the direction of B. If current of 8.00 A in the conductor results in Hll voltge of V, wht is the mgnitude of the Erth s mgnetic field t this loction? Additionl Prolems 56. Cron-14 nd cron-12 ions (ech with chrge of mgnitude e) re ccelerted in cyclotron. If the cyclotron hs mgnetic field of mgnitude 2.40 T, wht is the difference in cyclotron frequencies for the two ions? 57. In Niels Bohr s 1913 model of the hydrogen tom, the single electron is in circulr orit of rdius m nd its speed is m/s. () Wht is the mgnitude of the mgnetic moment due to the electron s motion? () If the electron moves in horizontl circle, counterclockwise s seen from ove, wht is the direction of this mgnetic moment vector? 58. Hert lung mchines nd rtificil kidney mchines BIO employ electromgnetic lood pumps. The lood is Q/C confined to n electriclly insulting tue, cylindricl in prctice ut represented here for simplicity s rectngle of interior width w nd height h. Figure P29.58 shows rectngulr section of lood within the tue. Two electrodes fit into the top nd the ottom of the tue. The potentil difference etween them estlishes n electric current through the lood, with current density J over the section of length L shown in Figure P A perpendiculr mgnetic field exists in the sme region. () Explin why this rrngement produces on the liquid force tht is directed long the length of the pipe. () how tht the section of liquid in the mgnetic field experiences pressure increse JLB. (c) After the lood leves the pump, is it chrged? (d) Is it crrying current? (e) Is it mgnetized? (The sme electromgnetic pump cn e used for ny fluid tht conducts electricity, such s liquid sodium in nucler rector.) h w J Figure P A prticle with positive chrge q C M moves with velocity v 5 12i^ 1 3j^ 2 k^ 2 m/s through region where oth uniform mgnetic field nd uniform electric field exist. () Clculte the totl force on the moving prticle (in unit-vector nottion), tking B 5 12 i^ 1 4j^ 1 k^ 2 T nd E 5 14i^ 2 j^ 2 2k^ 2 V/m. () Wht ngle does the force vector mke with the positive x xis? 60. Figure shows chrged prticle trveling in Q/C nonuniform mgnetic field forming mgnetic ottle. () Explin why the positively chrged prticle in the figure must e moving clockwise when viewed from the right of the figure. The prticle trvels long helix whose rdius decreses nd whose pitch decreses s the prticle moves into stronger mgnetic field. If the prticle is moving to the right long the x xis, its velocity in this direction will e reduced to zero nd it will e reflected from the right-hnd side of the ottle, cting s mgnetic mirror. The prticle ends up ouncing ck nd forth etween the ends of the ottle. () Explin qulittively why the xil velocity is reduced to zero s the prticle moves into the region of strong mgnetic field t the end of the ottle. (c) Explin why the tngentil velocity increses s the prticle pproches the end of the ottle. (d) Explin why the oriting prticle hs mgnetic dipole moment. 61. Review. The upper portion of the circuit in Figure AMT P29.61 is fixed. The horizontl wire t the ottom hs mss of 10.0 g nd is 5.00 cm long. This wire hngs in the grvittionl field of the Erth from identicl light springs connected to the upper portion of the circuit. The springs stretch cm under the weight of the L 5.00 cm 24.0 V B Figure P29.61 B out

244 Prolems 901 wire, nd the circuit hs totl resistnce of 12.0 V. When mgnetic field is turned on, directed out of the pge, the springs stretch n dditionl cm. Only the horizontl wire t the ottom of the circuit is in the mgnetic field. Wht is the mgnitude of the mgnetic field? 62. Within cylindricl region of spce of rdius 100 Mm, Q/C mgnetic field is uniform with mgnitude 25.0 mt nd oriented prllel to the xis of the cylinder. The mgnetic field is zero outside this cylinder. A cosmicry proton trveling t one-tenth the speed of light is heding directly towrd the center of the cylinder, moving perpendiculr to the cylinder s xis. () Find the rdius of curvture of the pth the proton follows when it enters the region of the field. () Explin whether the proton will rrive t the center of the cylinder. 63. Review. A proton is t rest t the plne oundry of region contining uniform mgnetic field B (Fig. P29.63). An lph prticle moving horizontlly mkes hed-on elstic collision with the proton. Immeditely fter the collision, oth prticles enter the mgnetic field, moving perpendiculr to the direction of the field. The rdius of the proton s trjectory is R. The mss of the lph prticle is four times tht of the proton, nd its chrge is twice tht of the proton. Find the rdius of the lph prticle s trjectory. Proton v Alph prticle B 0 B Figure P () A proton moving with velocity v 5 v i i^ experiences Q/C mgnetic force F 5 F i j^. Explin wht you cn nd cnnot infer out B from this informtion. () Wht If? In terms of F i, wht would e the force on proton in the sme field moving with velocity v 5 2v i i^? (c) Wht would e the force on n electron in the sme field moving with velocity v 5 2v i i^? 65. Review. A kg metl rod crrying current of AMT 10.0 A glides on two horizontl rils m prt. If the coefficient of kinetic friction etween the rod nd rils is 0.100, wht verticl mgnetic field is required to keep the rod moving t constnt speed? 66. Review. A metl rod of mss m crrying current I glides on two horizontl rils distnce d prt. If the coefficient of kinetic friction etween the rod nd rils is m, wht verticl mgnetic field is required to keep the rod moving t constnt speed? 67. A proton hving n initil velocity of 20.0i^ Mm/s enters uniform mgnetic field of mgnitude T with direction perpendiculr to the proton s velocity. It leves the field-filled region with velocity j Mm/s. Determine () the direction of the mgnetic field, () the rdius of curvture of the proton s pth while in the field, (c) the distnce the proton trveled in the field, nd (d) the time intervl during which the proton is in the field. 68. Model the electric motor in hndheld electric mixer s single flt, compct, circulr coil crrying electric current in region where mgnetic field is produced y n externl permnent mgnet. You need consider only one instnt in the opertion of the motor. (We will consider motors gin in Chpter 31.) Mke order-ofmgnitude estimtes of () the mgnetic field, () the torque on the coil, (c) the current in the coil, (d) the coil s re, nd (e) the numer of turns in the coil. The input power to the motor is electric, given y P 5 I DV, nd the useful output power is mechnicl, P 5 tv. 69. A nonconducting sphere hs mss 80.0 g nd rdius AMT 20.0 cm. A flt, compct coil of wire with five turns is wrpped tightly round it, with ech turn concentric with the sphere. The sphere is plced on n inclined plne tht slopes downwrd to the left (Fig. P29.69), mking n ngle u with the horizontl so tht the coil is prllel to the inclined plne. A uniform mgnetic field of T verticlly upwrd exists in the region of the sphere. () Wht current in the coil will enle the sphere to rest in equilirium on the inclined plne? () how tht the result does not depend on the vlue of u. u Figure P Why is the following sitution impossile? Figure P29.70 shows n experimentl technique for ltering the direction of trvel for chrged prticle. A prticle of chrge q mc nd mss m kg enters the ottom of the region of uniform mgnetic field t speed v m/s, with velocity vector h v u Figure P29.70 B B

245 902 Chpter 29 Mgnetic Fields perpendiculr to the field lines. The mgnetic force on the prticle cuses its direction of trvel to chnge so tht it leves the region of the mgnetic field t the top trveling t n ngle from its originl direction. The mgnetic field hs mgnitude B T nd is directed out of the pge. The length h of the mgnetic field region is m. An experimenter performs the technique nd mesures the ngle u t which the prticles exit the top of the field. he finds tht the ngles of devition re exctly s predicted. 71. Figure P29.71 shows schemtic representtion of n Q/C pprtus tht cn e used to mesure mgnetic fields. A rectngulr coil of wire contins N turns nd hs width w. The coil is ttched to one rm of lnce nd is suspended etween the poles of mgnet. The mgnetic field is uniform nd perpendiculr to the plne of the coil. The system is first lnced when the current in the coil is zero. When the switch is closed nd the coil crries current I, mss m must e dded to the right side to lnce the system. () Find n expression for the mgnitude of the mgnetic field. () Why is the result independent of the verticl dimensions of the coil? (c) uppose the coil hs 50 turns nd width of 5.00 cm. When the switch is closed, the coil crries current of A, nd mss of 20.0 g must e dded to the right side to lnce the system. Wht is the mgnitude of the mgnetic field? w Coil R Figure P A hert surgeon monitors the flow rte of lood BIO through n rtery using n electromgnetic flowmeter Q/C (Fig. P29.72). Electrodes A nd B mke contct with the outer surfce of the lood vessel, which hs dimeter of 3.00 mm. () For mgnetic field mgnitude of T, n emf of 160 mv ppers etween the electrodes. Clculte the speed of the lood. () Explin why electrode A hs to e positive s shown. (c) Does the sign of the emf depend on whether the moile ions in the lood re predominntly positively or negtively chrged? Explin. Artery Electrodes A e To voltmeter Blood flow m 73. A uniform mgnetic field of mgnitude T is directed long the positive x xis. A positron moving t speed of m/s enters the field long direction tht mkes n ngle of u with the x xis (Fig. P29.73). The motion of the prticle is expected to e helix s descried in ection Clculte () the pitch p nd () the rdius r of the trjectory s defined in Figure P z y u v B Figure P29.73 r e x 74. Review. () how tht mgnetic dipole in uniform Q/C mgnetic field, displced from its equilirium orienttion nd relesed, cn oscillte s torsionl pendulum (ection 15.5) in simple hrmonic motion. () Is this sttement true for ll ngulr displcements, for ll displcements less thn 1808, or only for smll ngulr displcements? Explin. (c) Assume the dipole is compss needle light r mgnet with mgnetic moment of mgnitude m. It hs moment of inerti I out its center, where it is mounted on frictionless, verticl xle, nd it is plced in horizontl mgnetic field of mgnitude B. Determine its frequency of oscilltion. (d) Explin how the compss needle cn e conveniently used s n indictor of the mgnitude of the externl mgnetic field. (e) If its frequency is Hz in the Erth s locl field, with horizontl component of 39.2 mt, wht is the mgnitude of field prllel to the needle in which its frequency of oscilltion is 4.90 Hz? 75. The ccompnying tle shows mesurements of the Hll voltge nd corresponding mgnetic field for proe used to mesure mgnetic fields. () Plot these dt nd deduce reltionship etween the two vriles. () If the mesurements were tken with current of A nd the smple is mde from mteril hving chrge- crrier density of crriers/m 3, wht is the thickness of the smple? N B Figure P29.72 DV H (mv) B (T) p

246 Prolems A metl rod hving mss per unit length l crries current I. The rod hngs from two wires in uniform verticl mgnetic field s shown in Figure P The wires mke n ngle u with the verticl when in equilirium. Determine the mgnitude of the mgnetic field. p y x 1.00 m x u Figure P29.78 B u I Figure P29.76 Chllenge Prolems 77. Consider n electron oriting proton nd mintined in fixed circulr pth of rdius R m y the Coulom force. Tret the oriting prticle s current loop. Clculte the resulting torque when the electron proton system is plced in mgnetic field of T directed perpendiculr to the mgnetic moment of the loop. 78. Protons hving kinetic energy of 5.00 MeV (1 ev J) re moving in the positive x direction nd enter mgnetic field B k^ T directed out of the plne of the pge nd extending from x 5 0 to x m s shown in Figure P () Ignoring reltivistic effects, find the ngle etween the initil velocity vector of the proton em nd the velocity vector fter the em emerges from the field. () Clculte the y component of the protons moment s they leve the mgnetic field. g 79. Review. A wire hving liner mss density of 1.00 g/cm is plced on horizontl surfce tht hs coefficient of kinetic friction of The wire crries current of 1.50 A towrd the est nd slides horizontlly to the north t constnt velocity. Wht re () the mgnitude nd () the direction of the smllest mgnetic field tht enles the wire to move in this fshion? 80. A proton moving in the plne of the pge hs kinetic energy of 6.00 MeV. A mgnetic field of mgnitude B T is directed into the pge. The proton enters the mgnetic field with its velocity vector t n ngle u to the liner oundry of the field s shown in Figure P () Find x, the distnce from the point of entry to where the proton will leve the field. () Determine u, the ngle etween the oundry nd the proton s velocity vector s it leves the field. p x u u Figure P29.80

247 c h p t e r 30 ources of the Mgnetic Field 30.1 The Biot vrt Lw 30.2 The Mgnetic Force Between Two Prllel Conductors 30.3 Ampère s Lw 30.4 The Mgnetic Field of olenoid 30.5 Guss s Lw in Mgnetism 30.6 Mgnetism in Mtter A crdic ctheteriztion lortory stnds redy to receive ptient suffering from tril firilltion. The lrge white ojects on either side of the operting tle re strong mgnets tht plce the ptient in mgnetic field. The electrophysiologist performing ctheter ltion procedure sits t computer in the room to the left. With guidnce from the mgnetic field, he or she uses joystick nd other controls to thred the mgneticlly sensitive tip of crdic ctheter through lood vessels nd into the chmers of the hert. ( Courtesy of tereotxis, Inc.) In Chpter 29, we discussed the mgnetic force exerted on chrged prticle moving in mgnetic field. To complete the description of the mgnetic interction, this chpter explores the origin of the mgnetic field, moving chrges. We egin y showing how to use the lw of Biot nd vrt to clculte the mgnetic field produced t some point in spce y smll current element. This formlism is then used to clculte the totl mgnetic field due to vrious current distriutions. Next, we show how to determine the force etween two current-crrying conductors, leding to the definition of the mpere. We lso introduce Ampère s lw, which is useful in clculting the mgnetic field of highly symmetric configurtion crrying stedy current. This chpter is lso concerned with the complex processes tht occur in mgnetic mterils. All mgnetic effects in mtter cn e explined on the sis of tomic mgnetic moments, which rise oth from the oritl motion of electrons nd from n intrinsic property of electrons known s spin The Biot vrt Lw hortly fter Oersted s discovery in 1819 tht compss needle is deflected y current-crrying conductor, Jen-Bptiste Biot ( ) nd Félix vrt ( ) performed quntittive experiments on the force exerted y n electric current on nery mgnet. From their experimentl results, Biot nd vrt rrived t mthemticl expression tht gives the mgnetic field t some point in spce 904

248 The Biot vrt Lw in terms of the current tht produces the field. Tht expression is sed on the fol lowing experimentl oservtions for the mgnetic field d B t point P ssocited with length element d s of wire crrying stedy current I (Fig. 30.1): The vector d B is perpendiculr oth to d s (which points in the direction of the current) nd to the unit vector r directed from d s towrd P. ^ The mgnitude of d B is inversely proportionl to r 2, where r is the distnce from d s to P. The mgnitude of d B is proportionl to the current I nd to the mgnitude ds of the length element d s. The mgnitude of d B is proportionl to sin u, where u is the ngle etween the vectors d s nd r^. The Biot vrt Lw The mgnetic field descried y the Biot vrt lw is the field due to given current-crrying conductor. Do not confuse this field with ny externl field tht my e pplied to the conductor from some other source. om Pitfll Prevention 30.1 m0 I d s 3 r^ 4p r2 ly (30.1) where m0 is constnt clled the permeility of free spce: m0 5 4p T # m/a ee db 5.c These oservtions re summrized in the mthemticl expression known tody s the Biot vrt lw: (30.2) WW Biot vrt lw WW Permeility of free spce ic s. w Notice tht the field d B in Eqution 30.1 is the field creted t point y the current in only smll length element d s of the conductor. To find the totl mg netic field B creted t some point y current of finite size, we must sum up contriutions from ll current elements I d s tht mke up the current. Tht is, we must evlute B y integrting Eqution 30.1: m0 I d s 3 r^ 4p 3 r2 (30.3) r ph B 5 ys The direction of the field is out of the pge t P. d Bout w. sw where the integrl is tken over the entire current distriution. This expression must e hndled with specil cre ecuse the integrnd is cross product nd therefore vector quntity. We shll see one cse of such n integrtion in Exmple Although the Biot vrt lw ws discussed for current-crrying wire, it is lso vlid for current consisting of chrges flowing through spce such s the prticle s represents the length of smll segment of em in n ccelertor. In tht cse, d spce in which the chrges flow. Interesting similrities nd differences exist etween Eqution 30.1 for the mgnetic field due to current element nd Eqution 23.9 for the electric field due to point chrge. The mgnitude of the mgnetic field vries s the inverse squre of the distnce from the source, s does the electric field due to point chrge. The directions of the two fields re quite different, however. The electric field creted y point chrge is rdil, ut the mgnetic field creted y curs nd the unit vector r^ rent element is perpendiculr to oth the length element d s descried y the cross product in Eqution Hence, if the conductor lies in the plne of the pge s shown in Figure 30.1, d B points out of the pge t P nd into the pge t P 9. Another difference etween electric nd mgnetic fields is relted to the source of the field. An electric field is estlished y n isolted electric chrge. The Biot vrt lw gives the mgnetic field of n isolted current element t some point, ut such n isolted current element cnnot exist the wy n isolted electric chrge cn. A current element must e prt of n extended current distriution ecuse complete circuit is needed for chrges to flow. Therefore, r w w P I r u r ds P d Bin The direction of the field is into the pge t P. Figure30.1 The mgnetic field d B t point due to the current I through length element d s is given y the Biot vrt lw.

249 906 Chpter 30 ources of the Mgnetic Field B d s Figure 30.2 (Quick Quiz 30.1) Where is the mgnetic field due to the current element the gretest? C I A the Biot vrt lw (Eq. 30.1) is only the first step in clcultion of mgnetic field; it must e followed y n integrtion over the current distriution s in Eqution Q uick Quiz 30.1 Consider the mgnetic field due to the current in the wire shown in Figure Rnk the points A, B, nd C in terms of mgnitude of the mgnetic field tht is due to the current in just the length element d s shown from gretest to lest. Exmple 30.1 Mgnetic Field urrounding Thin, tright Conductor Consider thin, stright wire of finite length crrying constnt current I nd plced long the x xis s shown in Figure Determine the mgnitude nd direction of the mgnetic field t point P due to this current. olution Conceptulize From the Biot vrt lw, we expect tht the mgnitude of the field is proportionl to the current in the wire nd decreses s the distnce from the wire to point P increses. We lso expect the field to depend on the ngles u 1 nd u 2 in Figure We plce the origin t O nd let point P e long the positive y xis, with k^ eing unit vector pointing out of the pge. Ctegorize We re sked to find the mgnetic field due to simple current distriution, so this exmple is typicl prolem for which the Biot vrt lw is pproprite. We must find the field contriution from smll element of current nd then integrte over the current distriution. Anlyze Let s strt y considering length element d s locted distnce r from P. The direction of the mgnetic field t point P due to the current in this element is out of the pge ecuse d s 3 r^ is out of the pge. In fct, ecuse ll the current elements I d s lie in the plne y d s dx P of the pge, they ll produce mgnetic field directed out of the pge t point P. Therefore, the direction of the mgnetic field t point P is out of the pge nd we need only find the mgnitude of the field. Evlute the cross product in the Biot vrt lw: ustitute into Eqution 30.1: From the geometry in Figure 30.3, express r in terms of u: Notice tht tn u 5 2x/ from the right tringle in Figure 30.3 (the negtive sign is necessry ecuse d s is locted t negtive vlue of x) nd solve for x: Find the differentil dx: rˆ r d s x u P u 1 y O Figure 30.3 (Exmple 30.1) () A thin, stright wire crrying current I. () The ngles u 1 nd u 2 used for determining the net field. d s 3 r^ 5 0d s 3 r^ 0 k^ 5 cdx sin p 2 2 u d k^ 5 1dx cos u 2 k^ (1) d B 5 1dB2 k^ 5 m 0I dx cos u k^ 4p r 2 (2) r 5 cos u ustitute Equtions (2) nd (3) into the expression for the z component of the field from Eqution (1): x 5 2 tn u (3) dx 5 2 sec 2 u du 5 2 du cos 2 u (4) db 5 2 m 0I 4p du cos 2 u cos2 u 2 cos u 5 2 m 0I 4p u2 I x x cos u du

250 The Biot vrt Lw 30.1 c o n t i n u e d Integrte Eqution (4) over ll length elements on the wire, where the sutending ngles rnge from u1 to u2 s defined in Figure 30.3: B5 2 m0 I u2 m0 I 1 sin u 1 2 sin u cos u du 5 4p u1 4p (30.4) Finlize We cn use this result to find the mgnitude of the mgnetic field of ny stright current- crrying wire if we know the geometry nd hence the ngles u1 nd u2. Consider the specil cse of n infinitely long, stright wire. If the wire in Figure 30.3 ecomes infinitely long, we see tht u1 5 p/2 nd u2 5 2p/2 for length elements rnging etween positions x 5 2` nd x 5 1`. Becuse (sin u1 2 sin u2) 5 [sin p/2 2 sin (2p/2)] 5 2, Eqution 30.4 ecomes m0 I 2p (30.5) om B5 ee ly.c Equtions 30.4 nd 30.5 oth show tht the mgnitude of the mgnetic field is proportionl to the current nd decreses with incresing distnce from the wire, s expected. Eqution 30.5 hs the sme mthemticl form s the expression for the mgnitude of the electric field due to long chrged wire (see Eq. 24.7). Exmple 30.2 Mgnetic Field Due to Curved Wire egment Clculte the mgnetic field t point O for the current-crrying wire segment shown in Figure The wire consists of two stright portions nd circulr rc of rdius, which sutends n ngle u. s. w A ic o l u ti o n A O Conceptulize The mgnetic field t O due to the current in the stright segments AA9 nd CC9 is zero ecuse d s is prllel to r^ long these pths, which mens tht d s 3 r^ 5 0 for these pths. Therefore, we expect the mgnetic field t O to e due only to the current in the curved portion of the wire. Ctegorize Becuse we cn ignore segments AA9 nd CC9, this exmple is ctegorized s n ppliction of the Biot vrt lw to the curved wire segment AC. Figure 30.4 ys r ds u C r ph I I C sw (Exmple 30.2) The length of the curved segment AC is s. Anlyze Ech length element d s long pth AC is t the sme distnce from O, nd the current in ech contriutes. field element d B directed into the pge t O. Furthermore, t every point on AC, d s is perpendiculr to r^ ; hence, 0d s 3 r^ 0 5 ds. w w From Eqution 30.1, find the mgnitude of the field t O due to the current in n element of length ds: w Integrte this expression over the curved pth AC, noting tht I nd re constnts: From the geometry, note tht s 5 u nd sustitute: db 5 B 5 B5 m 0 I ds 4p 2 m0 I 4p 2 m0 I 4p 2 3 ds 5 1 u 2 5 m0 I 4p 2 s m0 I u 4p (30.6) Finlize Eqution 30.6 gives the mgnitude of the mgnetic field t O. The direction of B is into the pge t O ecuse d s 3 r^ is into the pge for every length element. Wht if you were sked to find the mgnetic field t the center of circulr wire loop of rdius R tht crries current I? Cn this question e nswered t this point in our understnding of the source of mgnetic continued fields? W h t I f?

251 908 Chpter 30 ources of the Mgnetic Field 30.2 continued Answer Yes, it cn. The stright wires in Figure 30.4 do not contriute to the mgnetic field. The only contriution is from the curved segment. As the ngle u increses, the curved segment ecomes full circle when u 5 2p. Therefore, you cn find the mgnetic field t the center of wire loop y letting u 5 2p in Eqution 30.6: B 5 m 0I 4p 2p 5 m 0I 2 This result is limiting cse of more generl result discussed in Exmple Exmple 30.3 Mgnetic Field on the Axis of Circulr Current Loop Consider circulr wire loop of rdius locted in the yz plne nd crrying stedy current I s in Figure Clculte the mgnetic field t n xil point P distnce x from the center of the loop. olution u db Conceptulize Compre this prolem to Exmple 23.8 for d B O the electric field due to ring of chrge. Figure 30.5 shows r the mgnetic field contriution d z B t P due to single current element t the top of the ring. This field vector cn e u x I P resolved into components db x prllel to the xis of the ring db x x nd db perpendiculr to the xis. Think out the mgnetic field contriutions from current element t the ottom of the loop. Becuse of the symmetry of the sitution, Figure 30.5 (Exmple 30.3) Geometry for clculting the mgnetic field t point P lying on the xis of current loop. By symmetry, the totl field B is long this xis. the perpendiculr components of the field due to elements t the top nd ottom of the ring cncel. This cncelltion occurs for ll pirs of segments round the ring, so we cn ignore the perpendiculr component of the field nd focus solely on the prllel components, which simply dd. Ctegorize We re sked to find the mgnetic field due to simple current distriution, so this exmple is typicl prolem for which the Biot vrt lw is pproprite. Anlyze In this sitution, every length element d s is perpendiculr to the vector r^ t the loction of the element. Therefore, for ny element, 0d s 3 r^ 0 5 1ds sin ds. Furthermore, ll length elements round the loop re t the sme distnce r from P, where r x 2. Use Eqution 30.1 to find the mgnitude of d B due to the current in ny length element d s : Find the x component of the field element: Integrte over the entire loop: db 5 m 0I 4p db x 5 m 0I 4p From the geometry, evlute cos u: cos u 5 0 d s 3 r^ 0 r 2 B x 5 C db x 5 m 0I 4p d s y ˆr 5 m 0I 4p ds cos u x x 2 2 1/2 C ds cos u 2 1 x 2 ds x 2 2 ustitute this expression for cos u into the integrl nd note tht x nd re oth constnt: B x 5 m 0I 4p C ds 2 1 x c x 2 2 1/2d 5 m 0I 4p x 2 2 3/2 C ds Integrte round the loop: B x 5 m 0I 4p x p2 5 m 0 I 2 (30.7) 3/ x 2 3/2 2

252 30.2 The Mgnetic Force Between Two Prllel Conductors continued Finlize To find the mgnetic field t the center of the loop, set x 5 0 in Eqution At this specil point, B 5 m 0I 2 1t x 5 02 (30.8) which is consistent with the result of the Wht If? feture of Exmple The pttern of mgnetic field lines for circulr current loop is shown in Figure For clrity, the lines re drwn for only the plne tht contins the xis of the loop. The field-line pttern is xilly symmetric nd looks like the pttern round r mgnet, which is shown in Figure Wht If? Wht if we consider points on the x xis very fr from the loop? How does the mgnetic field ehve t these distnt points? Answer In this cse, in which x.., we cn neglect the term 2 in the denomintor of Eqution 30.7 nd otin B < m 0I 2 (for x.. ) (30.9) 3 2x The mgnitude of the mgnetic moment m of the loop is defined s the product of current nd loop re (see Eq ): m 5 I(p 2 ) for our circulr loop. We cn express Eqution 30.9 s B < m 0 m 2p x (30.10) 3 This result is similr in form to the expression for the electric field due to n electric dipole, E 5 k e (p/y 3 ) (see Exmple 23.6), where p 5 2q is the electric dipole moment s defined in Eqution The Mgnetic Force Between Two Prllel Conductors In Chpter 29, we descried the mgnetic force tht cts on current-crrying conductor plced in n externl mgnetic field. Becuse current in conductor sets up its own mgnetic field, it is esy to understnd tht two current-crrying conductors exert mgnetic forces on ech other. One wire estlishes the mgnetic field nd the other wire is modeled s collection of prticles in mgnetic field. uch forces etween wires cn e used s the sis for defining the mpere nd the coulom. Consider two long, stright, prllel wires seprted y distnce nd crrying currents I 1 nd I 2 in the sme direction s in Figure Let s determine the force exerted on one wire due to the mgnetic field set up y the other wire. Wire 2, which crries current I 2 nd is identified ritrrily s the source wire, cretes mgnetic field B 2 t the loction of wire 1, the test wire. The mgnitude of this mgnetic field is the sme t ll points on wire 1. The direction of B 2 is perpendiculr to wire 1 s shown in Figure According to Eqution 29.10, the mgnetic force on length, of wire 1 is F1 5 I 1 < 3 B2. Becuse < is perpendiculr to B 2 in this sitution, the mgnitude of F1 is F 1 5 I 1,B 2. Becuse the mgnitude of B 2 is given y Eqution 30.5, F 1 5 I 1,B 2 5 I 1, m 0I 2 2p 5 m 0I 1 I 2 2p, (30.11) The direction of F 1 is towrd wire 2 ecuse < 3 B 2 is in tht direction. When the field set up t wire 2 y wire 1 is clculted, the force F 2 cting on wire 2 is found to e equl in mgnitude nd opposite in direction to F 1, which is wht we expect ecuse Newton s third lw must e oeyed. When the currents re in opposite directions (tht is, when one of the currents is reversed in Fig. 30.7), the forces N I Figure 30.6 (Exmple 30.3) () Mgnetic field lines surrounding current loop. () Mgnetic field lines surrounding r mgnet. Notice the similrity etween this line pttern nd tht of current loop. 1 2 B 2 N The field B 2 due to the current in wire 2 exerts mgnetic force of mgnitude F 1 I 1 B 2 on wire 1. F 1 Figure 30.7 Two prllel wires tht ech crry stedy current exert mgnetic force on ech other. The force is ttrctive if the currents re prllel (s shown) nd repulsive if the currents re ntiprllel. I 1 I 2

253 910 Chpter 30 ources of the Mgnetic Field re reversed nd the wires repel ech other. Hence, prllel conductors crrying currents in the sme direction ttrct ech other, nd prllel conductors crrying currents in opposite directions repel ech other. Becuse the mgnitudes of the forces re the sme on oth wires, we denote the mgnitude of the mgnetic force etween the wires s simply F B. We cn rewrite this mgnitude in terms of the force per unit length: F B, 5 m 0 I 1 I 2 2p (30.12) Definition of the mpere Exmple 30.4 The force etween two prllel wires is used to define the mpere s follows: When the mgnitude of the force per unit length etween two long, prllel wires tht crry identicl currents nd re seprted y 1 m is N/m, the current in ech wire is defined to e 1 A. The vlue N/m is otined from Eqution with I 1 5 I A nd 5 1 m. Becuse this definition is sed on force, mechnicl mesurement cn e used to stndrdize the mpere. For instnce, the Ntionl Institute of tndrds nd Technology uses n instrument clled current lnce for primry current mesurements. The results re then used to stndrdize other, more conventionl instruments such s mmeters. The I unit of chrge, the coulom, is defined in terms of the mpere: When conductor crries stedy current of 1 A, the quntity of chrge tht flows through cross section of the conductor in 1 s is 1 C. In deriving Equtions nd 30.12, we ssumed oth wires re long compred with their seprtion distnce. In fct, only one wire needs to e long. The equtions ccurtely descrie the forces exerted on ech other y long wire nd stright, prllel wire of limited length,. Q uick Quiz 30.2 A loose spirl spring crrying no current is hung from ceiling. When switch is thrown so tht current exists in the spring, do the coils () move closer together, () move frther prt, or (c) not move t ll? uspending Wire AM Two infinitely long, prllel wires re lying on the ground distnce cm prt s shown in Figure A third wire, of length L m nd mss 400 g, crries current of I A nd is levitted ove the first two wires, t horizontl position midwy etween them. The infinitely long wires crry equl currents I 2 in the sme direction, ut in the direction opposite tht in the levitted wire. Wht current must the infinitely long wires crry so tht the three wires form n equilterl tringle? olution Conceptulize Becuse the current in the short wire is opposite those in the long wires, the short wire is repelled from oth of the others. Imgine the currents in the long wires in Figure 30.8 re incresed. The repulsive force ecomes stronger, nd the levitted wire rises to the point t which the wire is once gin levitted in equilirium t higher position. Figure 30.8 shows the desired sitution with the three wires forming n equilterl tringle. Ctegorize Becuse the levitted wire is suject to forces ut does not ccelerte, it is modeled s prticle in equilirium. I 2 I 1 F B,R I 1 L F g I 2 I 2 I 2 u F B,L Figure 30.8 (Exmple 30.4) () Two current-crrying wires lie on the ground nd suspend third wire in the ir y mgnetic forces. () End view. In the sitution descried in the exmple, the three wires form n equilterl tringle. The two mgnetic forces on the levitted wire re F B,L, the force due to the left-hnd wire on the ground, nd F B,R, the force due to the right-hnd wire. The grvittionl force F g on the levitted wire is lso shown.

254 30.3 Ampère s Lw continued Anlyze The horizontl components of the mgnetic forces on the levitted wire cncel. The verticl components re oth positive nd dd together. Choose the z xis to e upwrd through the top wire in Figure 30.8 nd in the plne of the pge. Find the totl mgnetic force in the upwrd direction on the levitted wire: Find the grvittionl force on the levitted wire: Apply the prticle in equilirium model y dding the forces nd setting the net force equl to zero: olve for the current in the wires on the ground: I 2 5 ustitute numericl vlues: I 2 5 B F 5 2 m 0I 1 I 2 2p, cos u k^ 5 m 0I 1 I 2, cos u k^ p F g 5 2mg k^ F 5 FB 1 Fg 5 m 0I 1 I 2, cos u k^ 2 mg k^ 5 0 p mg p m 0 I 1, cos u kg m/s 2 2p m2 14p T # m/a A m2 cos A Finlize The currents in ll wires re on the order of 10 2 A. uch lrge currents would require specilized equipment. Therefore, this sitution would e difficult to estlish in prctice. Is the equilirium of wire 1 stle or unstle? 30.3 Ampère s Lw Looking ck, we cn see tht the result of Exmple 30.1 is importnt ecuse current in the form of long, stright wire occurs often. Figure 30.9 is perspective view of the mgnetic field surrounding long, stright, current-crrying wire. Becuse of the wire s symmetry, the mgnetic field lines re circles concentric with the wire nd lie in plnes perpendiculr to the wire. The mgnitude of B is constnt on ny circle of rdius nd is given y Eqution A convenient rule for determining the direction of B is to grsp the wire with the right hnd, positioning the thum long the direction of the current. The four fingers wrp in the direction of the mgnetic field. Figure 30.9 lso shows tht the mgnetic field line hs no eginning nd no end. Rther, it forms closed loop. Tht is mjor difference etween mgnetic field lines nd electric field lines, which egin on positive chrges nd end on negtive chrges. We will explore this feture of mgnetic field lines further in ection Oersted s 1819 discovery out deflected compss needles demonstrtes tht current-crrying conductor produces mgnetic field. Figure (pge 912) shows how this effect cn e demonstrted in the clssroom. everl compss needles re plced in horizontl plne ner long, verticl wire. When no current is present in the wire, ll the needles point in the sme direction (tht of the horizontl component of the Erth s mgnetic field) s expected. When the wire crries strong, stedy current, the needles ll deflect in direction tngent to the circle s in Figure These oservtions demonstrte tht the direction of the mgnetic field produced y the current in the wire is consistent with the right-hnd rule descried in Figure When the current is reversed, the needles in Figure lso reverse. Now let s evlute the product B? d s for smll length element d s on the circulr pth defined y the compss needles nd sum the products for ll elements Figure 30.9 The right-hnd rule for determining the direction of the mgnetic field surrounding long, stright wire crrying current. Notice tht the mgnetic field lines form circles round the wire. I B

255 912 Chpter 30 ources of the Mgnetic Field itockphoto.com/hultonarchive Andre-Mrie Ampère French Physicist ( ) Ampère is credited with the discovery of electromgnetism, which is the reltionship etween electric currents nd mgnetic fields. Ampère s genius, prticulrly in mthemtics, ecme evident y the time he ws 12 yers old; his personl life, however, ws filled with trgedy. His fther, welthy city officil, ws guillotined during the French Revolution, nd his wife died young, in Ampère died t the ge of 61 of pneumoni. Pitfll Prevention 30.2 Avoiding Prolems with igns When using Ampère s lw, pply the following right-hnd rule. Point your thum in the direction of the current through the mperin loop. Your curled fingers then point in the direction tht you should integrte when trversing the loop to void hving to define the current s negtive. Ampère s lw When no current is present in the wire, ll compss needles point in the sme direction (towrd the Erth s north pole). I 0 Figure () nd () Compsses show the effects of the current in nery wire. (c) Circulr mgnetic field lines surrounding current-crrying conductor, displyed with iron filings. over the closed circulr pth. 1 Along this pth, the vectors d s nd B re prllel t ech point (see Fig ), so B? d s 5 B ds. Furthermore, the mgnitude of B is constnt on this circle nd is given y Eqution Therefore, the sum of the products B ds over the closed pth, which is equivlent to the line integrl of B? d s, is C B? d s 5 B C ds 5 m 0I 2pr 12pr2 5 m 0I where r ds 5 2pr is the circumference of the circulr pth of rdius r. Although this result ws clculted for the specil cse of circulr pth surrounding wire of infinite length, it holds for closed pth of ny shpe (n mperin loop) surrounding current tht exists in n unroken circuit. The generl cse, known s Ampère s lw, cn e stted s follows: The line integrl of B? d s round ny closed pth equls m 0 I, where I is the totl stedy current pssing through ny surfce ounded y the closed pth: C B? d s 5 m 0 I (30.13) Ampère s lw descries the cretion of mgnetic fields y ll continuous current configurtions, ut t our mthemticl level it is useful only for clculting the mgnetic field of current configurtions hving high degree of symmetry. Its use is similr to tht of Guss s lw in clculting electric fields for highly symmetric chrge distriutions. Q uick Quiz 30.3 Rnk the mgnitudes of r B? d s for the closed pths through d in Figure from gretest to lest. When the wire crries strong current, the compss needles deflect in direction tngent to the circle, which is the direction of the mgnetic field creted y the current. B d s 1 A c I d 5 A 2 A c Figure (Quick Quiz 30.3) Four closed pths round three current-crrying wires. Richrd Megn, Fundmentl Photogrphs, New York 1 You my wonder why we would choose to evlute this sclr product. The origin of Ampère s lw is in 19th-century science, in which mgnetic chrge (the supposed nlog to n isolted electric chrge) ws imgined to e moved round circulr field line. The work done on the chrge ws relted to B? d s, just s the work done moving n electric chrge in n electric field is relted to E? d s. Therefore, Ampère s lw, vlid nd useful principle, rose from n erroneous nd ndoned work clcultion!

256 30.3 Ampère s Lw 913 Q uick Quiz 30.4 Rnk the mgnitudes of r B? d s for the closed pths through d in Figure from gretest to lest. c d Figure (Quick Quiz 30.4) everl closed pths ner single current-crrying wire. Exmple 30.5 The Mgnetic Field Creted y Long Current-Crrying Wire A long, stright wire of rdius R crries stedy current I tht is uniformly distriuted through the cross section of the wire (Fig ). Clculte the mgnetic field distnce r from the center of the wire in the regions r $ R nd r, R. olution Conceptulize tudy Figure to understnd the structure of the wire nd the current in the wire. The current cretes mgnetic fields everywhere, oth inside nd outside the wire. Bsed on our discussions out long, stright wires, we expect the mgnetic field lines to e circles centered on the centrl xis of the wire. Ctegorize Becuse the wire hs high degree of symmetry, we ctegorize this exmple s n Ampère s lw prolem. For the r $ R cse, we should rrive t the sme result s ws otined in Exmple 30.1, where we pplied the Biot vrt lw to the sme sitution. Anlyze For the mgnetic field exterior to the wire, let us choose for our pth of integrtion circle 1 in Figure From symmetry, B must e constnt in mgnitude nd prllel to d s t every point on this circle. Note tht the totl current pssing through the plne of the circle is I nd pply Ampère s lw: olve for B: C B? d s 5 B C ds 5 B 12pr2 5 m 0 I B 5 m 0I 2pr 1 2 R r d s Figure (Exmple 30.5) A long, stright wire of rdius R crrying stedy current I uniformly distriuted cross the cross section of the wire. The mgnetic field t ny point cn e clculted from Ampère s lw using circulr pth of rdius r, concentric with the wire. (for r $ R) (30.14) Now consider the interior of the wire, where r, R. Here the current I 9 pssing through the plne of circle 2 is less thn the totl current I. et the rtio of the current I 9 enclosed y circle 2 to the entire current I equl to the rtio of the re pr 2 enclosed y circle 2 to the cross-sectionl re pr 2 of the wire: olve for I9: I r 5 r 2 R I 2 Apply Ampère s lw to circle 2: C B? d s 5 B 12pr2 5 m 0 I r 5 m 0 r 2 I r I 5 pr 2 pr 2 R 2 I I olve for B: B 5 m 0I 2pR 2 r (for r, R) (30.15) continued

257 914 Chpter 30 ources of the Mgnetic Field 30.5 continued Finlize The mgnetic field exterior to the wire is identicl in form to Eqution As is often the cse in highly symmetric situtions, it is much esier to use Ampère s lw thn the Biot vrt lw (Exmple 30.1). The mgnetic field interior to the wire is similr in form to the expression for the electric field inside uniformly chrged sphere (see Exmple 24.3). The mgnitude of the mgnetic field versus r for this configurtion is plotted in Figure Inside the Figure (Exmple 30.5) Mgnitude of the mgnetic field versus r for the wire shown in Figure The field is proportionl to r inside the wire nd vries s 1/r outside the wire. wire, B 0 s r 0. Furthermore, Equtions nd give the sme vlue of the mgnetic field t r 5 R, demonstrting tht the mgnetic field is continuous t the surfce of the wire. Exmple 30.6 The Mgnetic Field Creted y Toroid A device clled toroid (Fig ) is often used to crete n lmost uniform mgnetic field in some enclosed re. The device consists of conducting wire wrpped round ring ( torus) mde of nonconducting mteril. For toroid hving N closely spced turns of wire, clculte the mgnetic field in the region occupied y the torus, distnce r from the center. olution Conceptulize tudy Figure crefully to understnd how the wire is wrpped round the torus. The torus could e solid mteril or it could e ir, with stiff wire wrpped into the shpe shown in Figure to form n empty toroid. Imgine ech turn of the wire to e circulr loop s in Exmple The mgnetic field t the center of the loop is perpendiculr to the plne of the loop. Therefore, the mgnetic field lines of the collection of loops will form circles within the toroid such s suggested y loop 1 in Figure Ctegorize Becuse the toroid hs high degree of symmetry, we ctegorize this exmple s n Ampère s lw prolem. B B r R B 1/r Anlyze Consider the circulr mperin loop (loop 1) of rdius r in the plne of Figure By symmetry, the mgnitude of the field is constnt on this circle nd tngent to it, so B? d s 5 B ds. Furthermore, the wire psses through the loop N times, so the totl current through the loop is NI. Apply Ampère s lw to loop 1: olve for B: I I d s c r B r Loop 1 Loop 2 Figure (Exmple 30.6) A toroid consisting of mny turns of wire. If the turns re closely spced, the mgnetic field in the interior of the toroid is tngent to the dshed circle (loop 1) nd vries s 1/r. The dimension is the cross-sectionl rdius of the torus. The field outside the toroid is very smll nd cn e descried y using the mperin loop (loop 2) t the right side, perpendiculr to the pge. C B? d s 5 B C ds 5 B 12pr2 5 m 0 NI B 5 m 0NI 2pr (30.16) Finlize This result shows tht B vries s 1/r nd hence is nonuniform in the region occupied y the torus. If, however, r is very lrge compred with the cross-sectionl rdius of the torus, the field is pproximtely uniform inside the torus. For n idel toroid, in which the turns re closely spced, the externl mgnetic field is close to zero, ut it is not exctly zero. In Figure 30.15, imgine the rdius r of mperin loop 1 to e either smller thn or lrger thn c. In either cse, the loop encloses zero net current, so r B? d s 5 0. You might think this result proves tht B 5 0, ut it does not. Consider the mperin loop (loop 2) on the right side of the toroid in Figure The plne of this loop is perpendiculr to the pge, nd the toroid psses through the loop. As chrges enter the toroid s indicted y the current directions in Figure 30.15,

258 30.4 The Mgnetic Field of olenoid continued they work their wy counterclockwise round the toroid. Therefore, there is counterclockwise current round the toroid, so tht current psses through mperin loop 2! This current is smll, ut not zero. As result, the toroid cts s current loop nd produces wek externl field of the form shown in Figure The reson r B? d s 5 0 for mperin loop 1 of rdius r, or r. c is tht the field lines re perpendiculr to d s, not ecuse B The Mgnetic Field of olenoid A solenoid is long wire wound in the form of helix. With this configurtion, resonly uniform mgnetic field cn e produced in the spce surrounded y the turns of wire which we shll cll the interior of the solenoid when the solenoid crries current. When the turns re closely spced, ech cn e pproximted s circulr loop; the net mgnetic field is the vector sum of the fields resulting from ll the turns. Figure shows the mgnetic field lines surrounding loosely wound solenoid. The field lines in the interior re nerly prllel to one nother, re uniformly distriuted, nd re close together, indicting tht the field in this spce is strong nd lmost uniform. If the turns re closely spced nd the solenoid is of finite length, the externl mgnetic field lines re s shown in Figure This field line distriution is similr to tht surrounding r mgnet (Fig ). Hence, one end of the solenoid ehves like the north pole of mgnet nd the opposite end ehves like the south pole. As the length of the solenoid increses, the interior field ecomes more uniform nd the exterior field ecomes weker. An idel solenoid is pproched when the turns re closely spced nd the length is much greter thn the rdius of the turns. Figure (pge 916) shows longitudinl cross section of prt of such solenoid crrying current I. In this cse, the externl field is close to zero nd the interior field is uniform over gret volume. Consider the mperin loop (loop 1) perpendiculr to the pge in Figure (pge 916), surrounding the idel solenoid. This loop encloses smll The mgnetic field lines resemle those of r mgnet, mening tht the solenoid effectively hs north nd south poles. N Exterior Interior Figure The mgnetic field lines for loosely wound solenoid. Henry Lep nd Jim Lehmn Figure () Mgnetic field lines for tightly wound solenoid of finite length, crrying stedy current. The field in the interior spce is strong nd nerly uniform. () The mgnetic field pttern of r mgnet, displyed with smll iron filings on sheet of pper.

259 916 Chpter 30 ources of the Mgnetic Field Ampère s lw pplied to the rectngulr dshed pth cn e used to clculte the mgnitude of the interior field. B Ampère s lw pplied to the circulr pth whose plne is perpendiculr to the pge cn e used to show tht there is wek field outside the solenoid. Figure Cross-sectionl view of n idel solenoid, where the interior mgnetic field is uniform nd the exterior field is close to zero. Mgnetic field inside solenoid w Loop 2 Loop 1 current s the chrges in the wire move coil y coil long the length of the solenoid. Therefore, there is nonzero mgnetic field outside the solenoid. It is wek field, with circulr field lines, like those due to line of current s in Figure For n idel solenoid, this wek field is the only field externl to the solenoid. We cn use Ampère s lw to otin quntittive expression for the interior mgnetic field in n idel solenoid. Becuse the solenoid is idel, B in the interior spce is uniform nd prllel to the xis nd the mgnetic field lines in the exterior spce form circles round the solenoid. The plnes of these circles re perpendiculr to the pge. Consider the rectngulr pth (loop 2) of length, nd width w shown in Figure Let s pply Ampère s lw to this pth y evluting the integrl of B? d s over ech side of the rectngle. The contriution long side 3 is zero ecuse the externl mgnetic field lines re perpendiculr to the pth in this region. The contriutions from sides 2 nd 4 re oth zero, gin ecuse B is perpendiculr to d s long these pths, oth inside nd outside the solenoid. ide 1 gives contriution to the integrl ecuse long this pth B is uniform nd prllel to d s. The integrl over the closed rectngulr pth is therefore C B? d s 5 3 B? d s 5 B 3 ds 5 B, pth 1 The right side of Ampère s lw involves the totl current I through the re ounded y the pth of integrtion. In this cse, the totl current through the rectngulr pth equls the current through ech turn multiplied y the numer of turns. If N is the numer of turns in the length,, the totl current through the rectngle is NI. Therefore, Ampère s lw pplied to this pth gives pth 1 C B? d s 5 B, 5 m 0 NI B 5 m 0 N, I 5 m 0nI (30.17) where n 5 N/, is the numer of turns per unit length. We lso could otin this result y reconsidering the mgnetic field of toroid (see Exmple 30.6). If the rdius r of the torus in Figure contining N turns is much greter thn the toroid s cross-sectionl rdius, short section of the toroid pproximtes solenoid for which n 5 N/2pr. In this limit, Eqution grees with Eqution Eqution is vlid only for points ner the center (tht is, fr from the ends) of very long solenoid. As you might expect, the field ner ech end is smller thn the vlue given y Eqution As the length of solenoid increses, the mgnitude of the field t the end pproches hlf the mgnitude t the center (see Prolem 69). Q uick Quiz 30.5 Consider solenoid tht is very long compred with its rdius. Of the following choices, wht is the most effective wy to increse the mgnetic field in the interior of the solenoid? () doule its length, keeping the numer of turns per unit length constnt () reduce its rdius y hlf, keeping the numer of turns per unit length constnt (c) overwrp the entire solenoid with n dditionl lyer of current-crrying wire 30.5 Guss s Lw in Mgnetism The flux ssocited with mgnetic field is defined in mnner similr to tht used to define electric flux (see Eq. 24.3). Consider n element of re da on n

260 30.5 Guss s Lw in Mgnetism 917 ritrrily shped surfce s shown in Figure If the mgnetic field t this element is B, the mgnetic flux through the element is B? d A, where d A is vector tht is perpendiculr to the surfce nd hs mgnitude equl to the re da. Therefore, the totl mgnetic flux F B through the surfce is F B ; 3 B? d A (30.18) WWDefinition of mgnetic flux Consider the specil cse of plne of re A in uniform field B tht mkes n ngle u with d A. The mgnetic flux through the plne in this cse is F B 5 BA cos u (30.19) If the mgnetic field is prllel to the plne s in Figure 30.20, then u nd the flux through the plne is zero. If the field is perpendiculr to the plne s in Figure 30.20, then u 5 0 nd the flux through the plne is BA (the mximum vlue). The unit of mgnetic flux is T? m 2, which is defined s weer (W); 1 W 5 1 T? m 2. The flux through the plne is zero when the mgnetic field is prllel to the plne surfce. d A Exmple 30.7 B The flux through the plne is mximum when the mgnetic field is perpendiculr to the plne. Mgnetic Flux Through Rectngulr Loop A rectngulr loop of width nd length is locted ner long wire crrying current I (Fig ). The distnce etween the wire nd the closest side of the loop is c. The wire is prllel to the long side of the loop. Find the totl mgnetic flux through the loop due to the current in the wire. olution Conceptulize As we sw in ection 30.3, the mgnetic field lines due to the wire will e circles, mny of which will pss through the rectngulr loop. We know tht the mgnetic field is function of distnce r from long wire. Therefore, the mgnetic field vries over the re of the rectngulr loop. Ctegorize Becuse the mgnetic field vries over the re of the loop, we must integrte over this re to find the totl flux. Tht identifies this s n nlysis prolem. da B Figure (Exmple 30.7) The mgnetic field due to the wire crrying current I is not uniform over the rectngulr loop. I Figure Mgnetic flux through plne lying in mgnetic field. r c B dr u d A Figure The mgnetic flux through n re element da is B? d A 5 B da cos u, where d A is vector perpendiculr to the surfce. Anlyze Noting tht B is prllel to d A t ny point within the loop, find the mgnetic flux through the rectngulr re using Eqution nd incorporte Eqution for the mgnetic field: F B 5 3 B? d A 5 3 B da 5 3 m 0I 2pr da continued

261 918 Chpter 30 ources of the Mgnetic Field 30.7 continued Express the re element (the tn strip in Fig ) s da 5 dr nd sustitute: Integrte from r 5 c to r 5 1 c: F B 5 m 0I 3 2pr dr 5 m 0I dr 2p 3 r F B 5 m 0I 2p 3 c 1c dr r 5 m 0I 2p ln r ` 1c c Guss s lw in mgnetism 5 m 0I 2p 1 c ln 5 c m 0I 2p ln 1 1 c Finlize Notice how the flux depends on the size of the loop. Incresing either or increses the flux s expected. If c ecomes lrge such tht the loop is very fr from the wire, the flux pproches zero, lso s expected. If c goes to zero, the flux ecomes infinite. In principle, this infinite vlue occurs ecuse the field ecomes infinite t r 5 0 (ssuming n infinitesimlly thin wire). Tht will not hppen in relity ecuse the thickness of the wire prevents the left edge of the loop from reching r 5 0. N In Chpter 24, we found tht the electric flux through closed surfce surrounding net chrge is proportionl to tht chrge (Guss s lw). In other words, the numer of electric field lines leving the surfce depends only on the net chrge within it. This ehvior exists ecuse electric field lines originte nd terminte on electric chrges. The sitution is quite different for mgnetic fields, which re continuous nd form closed loops. In other words, s illustrted y the mgnetic field lines of current in Figure 30.9 nd of r mgnet in Figure 30.22, mgnetic field lines do not egin or end t ny point. For ny closed surfce such s the one outlined y the dshed line in Figure 30.22, the numer of lines entering the surfce equls the numer leving the surfce; therefore, the net mgnetic flux is zero. In contrst, for closed surfce surrounding one chrge of n electric dipole (Fig ), the net electric flux is not zero. Guss s lw in mgnetism sttes tht the net mgnetic flux through ny closed surfce is lwys zero: The net mgnetic flux through closed surfce surrounding one of the poles or ny other closed surfce is zero. C B? d A 5 0 (30.20) The electric flux through closed surfce surrounding one of the chrges is not zero. Figure The mgnetic field lines of r mgnet form closed loops. (The dshed line represents the intersection of closed surfce with the pge.) Figure The electric field lines surrounding n electric dipole egin on the positive chrge nd terminte on the negtive chrge.

262 30.6 Mgnetism in Mtter 919 This sttement represents tht isolted mgnetic poles (monopoles) hve never een detected nd perhps do not exist. Nonetheless, scientists continue the serch ecuse certin theories tht re otherwise successful in explining fundmentl physicl ehvior suggest the possile existence of mgnetic monopoles Mgnetism in Mtter The mgnetic field produced y current in coil of wire gives us hint s to wht cuses certin mterils to exhiit strong mgnetic properties. Erlier we found tht solenoid like the one shown in Figure hs north pole nd south pole. In generl, ny current loop hs mgnetic field nd therefore hs mgnetic dipole moment, including the tomic-level current loops descried in some models of the tom. The Mgnetic Moments of Atoms Let s egin our discussion with clssicl model of the tom in which electrons move in circulr orits round the much more mssive nucleus. In this model, n oriting electron constitutes tiny current loop (ecuse it is moving chrge), nd the mgnetic moment of the electron is ssocited with this oritl motion. Although this model hs mny deficiencies, some of its predictions re in good greement with the correct theory, which is expressed in terms of quntum physics. In our clssicl model, we ssume n electron is prticle in uniform circulr motion: it moves with constnt speed v in circulr orit of rdius r out the nucleus s in Figure The current I ssocited with this oriting electron is its chrge e divided y its period T. Using Eqution 4.15 from the prticle in uniform circulr motion model, T 5 2pr/v, gives I 5 e T 5 ev 2pr The mgnitude of the mgnetic moment ssocited with this current loop is given y m 5 IA, where A 5 pr 2 is the re enclosed y the orit. Therefore, m 5 IA 5 ev 2pr pr evr (30.21) Becuse the mgnitude of the oritl ngulr momentum of the electron is given y L 5 m e vr (Eq with f 5 908), the mgnetic moment cn e written s m 5 e L (30.22) 2m e This result demonstrtes tht the mgnetic moment of the electron is proportionl to its oritl ngulr momentum. Becuse the electron is negtively chrged, the vectors m nd L point in opposite directions. Both vectors re perpendiculr to the plne of the orit s indicted in Figure A fundmentl outcome of quntum physics is tht oritl ngulr momentum is quntized nd is equl to multiples of " 5 h/2p J? s, where h is Plnck s constnt (see Chpter 40). The smllest nonzero vlue of the electron s mgnetic moment resulting from its oritl motion is m 5 "2 e 2m e U (30.23) We shll see in Chpter 42 how expressions such s Eqution rise. Becuse ll sustnces contin electrons, you my wonder why most sustnces re not mgnetic. The min reson is tht, in most sustnces, the mgnetic The electron hs n ngulr momentum L in one direction nd mgnetic moment m in the opposite direction. I L m r e Figure An electron moving in the direction of the gry rrow in circulr orit of rdius r. Becuse the electron crries negtive chrge, the direction of the current due to its motion out the nucleus is opposite the direction of tht motion. WWOritl mgnetic moment

263 920 Chpter 30 ources of the Mgnetic Field Pitfll Prevention 30.3 The Electron Does Not pin The electron is not physiclly spinning. It hs n intrinsic ngulr momentum s if it were spinning, ut the notion of rottion for point prticle is meningless. Rottion pplies only to rigid oject, with n extent in spce, s in Chpter 10. pin ngulr momentum is ctully reltivistic effect. m spin Figure Clssicl model of spinning electron. We cn dopt this model to remind ourselves tht electrons hve n intrinsic ngulr momentum. The model should not e pushed too fr, however; it gives n incorrect mgnitude for the mgnetic moment, incorrect quntum numers, nd too mny degrees of freedom. Tle 30.1 Mgnetic Moments of ome Atoms nd Ions Mgnetic Atom or Ion Moment ( J/T) H 9.27 He 0 Ne 0 Ce Y moment of one electron in n tom is cnceled y tht of nother electron oriting in the opposite direction. The net result is tht, for most mterils, the mgnetic effect produced y the oritl motion of the electrons is either zero or very smll. In ddition to its oritl mgnetic moment, n electron (s well s protons, neutrons, nd other prticles) hs n intrinsic property clled spin tht lso contriutes to its mgnetic moment. Clssiclly, the electron might e viewed s spinning out its xis s shown in Figure 30.25, ut you should e very creful with the clssicl interprettion. The mgnitude of the ngulr momentum ssocited with spin is on the sme order of mgnitude s the mgnitude of the ngulr momentum L due to the oritl motion. The mgnitude of the spin ngulr momentum of n electron predicted y quntum theory is 5 "3 2 U The mgnetic moment chrcteristiclly ssocited with the spin of n electron hs the vlue m spin 5 e U (30.24) 2m e This comintion of constnts is clled the Bohr mgneton m B : m B 5 e U 2m e J/T (30.25) Therefore, tomic mgnetic moments cn e expressed s multiples of the Bohr mgneton. (Note tht 1 J/T 5 1 A? m 2.) In toms contining mny electrons, the electrons usully pir up with their spins opposite ech other; therefore, the spin mgnetic moments cncel. Atoms contining n odd numer of electrons, however, must hve t lest one unpired electron nd therefore some spin mgnetic moment. The totl mgnetic moment of n tom is the vector sum of the oritl nd spin mgnetic moments, nd few exmples re given in Tle Notice tht helium nd neon hve zero moments ecuse their individul spin nd oritl moments cncel. The nucleus of n tom lso hs mgnetic moment ssocited with its constituent protons nd neutrons. The mgnetic moment of proton or neutron, however, is much smller thn tht of n electron nd cn usully e neglected. We cn understnd this smller vlue y inspecting Eqution nd replcing the mss of the electron with the mss of proton or neutron. Becuse the msses of the proton nd neutron re much greter thn tht of the electron, their mgnetic moments re on the order of 10 3 times smller thn tht of the electron. Ferromgnetism A smll numer of crystlline sustnces exhiit strong mgnetic effects clled ferromgnetism. ome exmples of ferromgnetic sustnces re iron, colt, nickel, gdolinium, nd dysprosium. These sustnces contin permnent tomic mgnetic moments tht tend to lign prllel to ech other even in wek externl mgnetic field. Once the moments re ligned, the sustnce remins mgnetized fter the externl field is removed. This permnent lignment is due to strong coupling etween neighoring moments, coupling tht cn e understood only in quntum-mechnicl terms. All ferromgnetic mterils re mde up of microscopic regions clled domins, regions within which ll mgnetic moments re ligned. These domins hve volumes of out to m 3 nd contin to toms. The oundries etween the vrious domins hving different orienttions re clled domin wlls. In n unmgnetized smple, the mgnetic moments in the domins re rndomly

264 30.6 Mgnetism in Mtter 921 oriented so tht the net mgnetic moment is zero s in Figure When the smple is plced in n externl mgnetic field B, the size of those domins with mgnetic moments ligned with the field grows, which results in mgnetized smple s in Figure As the externl field ecomes very strong s in Figure 30.26c, the domins in which the mgnetic moments re not ligned with the field ecome very smll. When the externl field is removed, the smple my retin net mgnetiztion in the direction of the originl field. At ordinry tempertures, therml gittion is not sufficient to disrupt this preferred orienttion of mgnetic moments. When the temperture of ferromgnetic sustnce reches or exceeds criticl temperture clled the Curie temperture, the sustnce loses its residul mgnetiztion. Below the Curie temperture, the mgnetic moments re ligned nd the sustnce is ferromgnetic. Aove the Curie temperture, the therml gittion is gret enough to cuse rndom orienttion of the moments nd the sustnce ecomes prmgnetic. Curie tempertures for severl ferromgnetic sustnces re given in Tle Prmgnetism Prmgnetic sustnces hve wek mgnetism resulting from the presence of toms (or ions) tht hve permnent mgnetic moments. These moments interct only wekly with one nother nd re rndomly oriented in the sence of n externl mgnetic field. When prmgnetic sustnce is plced in n externl mgnetic field, its tomic moments tend to line up with the field. This lignment process, however, must compete with therml motion, which tends to rndomize the mgnetic moment orienttions. Dimgnetism When n externl mgnetic field is pplied to dimgnetic sustnce, wek mgnetic moment is induced in the direction opposite the pplied field, cusing dimgnetic sustnces to e wekly repelled y mgnet. Although dimgnetism is present in ll mtter, its effects re much smller thn those of prmgnetism or ferromgnetism nd re evident only when those other effects do not exist. We cn ttin some understnding of dimgnetism y considering clssicl model of two tomic electrons oriting the nucleus in opposite directions ut with the sme speed. The electrons remin in their circulr orits ecuse of the ttrctive electrosttic force exerted y the positively chrged nucleus. Becuse the mgnetic moments of the two electrons re equl in mgnitude nd opposite in direction, they cncel ech other nd the mgnetic moment of the tom is zero. When n externl mgnetic field is pplied, the electrons experience n dditionl mgnetic force q v 3 B. This dded mgnetic force comines with the electrosttic force to increse the oritl speed of the electron whose mgnetic moment is ntiprllel to the field nd to decrese the speed of the electron whose mgnetic moment is prllel to the field. As result, the two mgnetic moments of the electrons no longer cncel nd the sustnce cquires net mgnetic moment tht is opposite the pplied field. Tle 30.2 Curie Tempertures for everl Ferromgnetic ustnces ustnce T Curie (K) Iron Colt Nickel 631 Gdolinium 317 Fe 2 O In n unmgnetized sustnce, the tomic mgnetic dipoles re rndomly oriented. When n externl field B is pplied, the domins with components of mgnetic moment in the sme direction s B grow lrger, giving the smple net mgnetiztion. c B da B B As the field is mde even stronger, the domins with mgnetic moment vectors not ligned with the externl field ecome very smll. Figure Orienttion of mgnetic dipoles efore nd fter mgnetic field is pplied to ferromgnetic sustnce.

265 922 Chpter 30 ources of the Mgnetic Field In the Meissner effect, the smll mgnet t the top induces currents in the superconducting disk elow, which is cooled to 321F (77 K). The currents crete repulsive mgnetic force on the mgnet cusing it to levitte ove the superconducting disk. Figure An illustrtion of the Meissner effect, shown y this mgnet suspended ove cooled cermic superconductor disk, hs ecome our most visul imge of high-temperture superconductivity. uperconductivity is the loss of ll resistnce to electricl current nd is key to more-efficient energy use. ummry Definition Courtesy Argonne Ntionl Lortory Liquid oxygen, prmgnetic mteril, is ttrcted to the poles of mgnet.. Cengge Lerning/Leon Lewndowski The levittion force is exerted on the dimgnetic wter molecules in the frog s ody. (Left) Prmgnetism. (Right) Dimgnetism: frog is levitted in 16-T mgnetic field t the Nijmegen High Field Mgnet Lortory in the Netherlnds. As you recll from Chpter 27, superconductor is sustnce in which the electricl resistnce is zero elow some criticl temperture. Certin types of superconductors lso exhiit perfect dimgnetism in the superconducting stte. As result, n pplied mgnetic field is expelled y the superconductor so tht the field is zero in its interior. This phenomenon is known s the Meissner effect. If permnent mgnet is rought ner superconductor, the two ojects repel ech other. This repulsion is illustrted in Figure 30.27, which shows smll permnent mgnet levitted ove superconductor mintined t 77 K. The mgnetic flux F B through surfce is defined y the surfce integrl Concepts nd Principles The Biot vrt lw sys tht the mgnetic field d B t point P due to length element d s tht crries stedy current I is F B ; 3 B? d A (30.18) d B 5 m 0 4p I d s 3 r^ r 2 (30.1) where m 0 is the permeility of free spce, r is the distnce from the element to the point P, nd r^ is unit vector pointing from d s towrd point P. We find the totl field t P y integrting this expression over the entire current distriution. The mgnetic force per unit length etween two prllel wires seprted y distnce nd crrying currents I 1 nd I 2 hs mgnitude F B, 5 m 0I 1 I 2 2p Courtesy of Dr. Andre Geim, Mnchester University (30.12) The force is ttrctive if the currents re in the sme direction nd repulsive if they re in opposite directions.