In Chapter 5 we mentioned the four kinds of fundamental forces. To this point

Transcription

1 ELECTRIC CHARGE AND ELECTRIC FIELD 21? Wter mkes life possile: The cells of your ody could not function without wter in which to dissolve essentil iologicl molecules. Wht electricl properties of wter mke it such good solvent? LEARNING GOAL By studying this chpter, you will lern: The nture of electric chrge, nd how we know tht electric chrge is conserved. How ojects ecome electriclly chrged. How to use Coulom s lw to clculte the electric force etween chrges. The distinction etween electric force nd electric field. In Chpter 5 we mentioned the four kinds of fundmentl forces. To this point the only one of these forces tht we hve exmined in ny detil is grvity. Now we re redy to exmine the force of electromgnetism, which encompsses oth electricity nd mgnetism. Electromgnetic phenomen will occupy our ttention for most of the reminder of this ook. Electromgnetic interctions involve prticles tht hve property clled electric chrge, n ttriute tht is s fundmentl s mss. Just s ojects with mss re ccelerted y grvittionl forces, so electriclly chrged ojects re ccelerted y electric forces. The shock you feel when you scuff your shoes cross crpet nd then rech for metl doorkno is due to chrged prticles leping etween your finger nd the doorkno. Electric currents re simply strems of chrged prticles flowing within wires in response to electric forces. Even the forces tht hold toms together to form solid mtter, nd tht keep the toms of solid ojects from pssing through ech other, re fundmentlly due to electric interctions etween the chrged prticles within toms. We egin our study of electromgnetism in this chpter y exmining the nture of electric chrge. We ll find tht chrge is quntized nd oeys conservtion principle. When chrges re t rest in our frme of reference, they exert electrosttic forces on ech other. These forces re of tremendous importnce in chemistry nd iology nd hve mny technologicl pplictions. Electrosttic forces re governed y simple reltionship known s Coulom s lw nd re most conveniently descried y using the concept of electric field. In lter chpters we ll expnd our discussion to include electric chrges in motion. This will led us to n understnding of mgnetism nd, remrkly, of the nture of light. While the key ides of electromgnetism re conceptully simple, pplying them to prcticl prolems will mke use of mny of your mthemticl skills, especilly your knowledge of geometry nd integrl clculus. For this reson you my find this chpter nd those tht follow to e more mthemticlly demnding How to clculte the electric field due to collection of chrges. How to use the ide of electric field lines to visulize nd interpret electric fields. How to clculte the properties of electric dipoles. 687

2 688 CHAPTER 21 Electric Chrge nd Electric Field thn erlier chpters. The rewrd for your extr effort will e deeper understnding of principles tht re t the hert of modern physics nd technology Electric Chrge The ncient Greeks discovered s erly s 600 B.C. tht fter they rued mer with wool, the mer could ttrct other ojects. Tody we sy tht the mer hs cquired net electric chrge, or hs ecome chrged. The word electric is derived from the Greek word elektron, mening mer. When you scuff your shoes cross nylon crpet, you ecome electriclly chrged, nd you cn chrge com y pssing it through dry hir. Plstic rods nd fur (rel or fke) re prticulrly good for demonstrting electrosttics, the interctions etween electric chrges tht re t rest (or nerly so). After we chrge oth plstic rods in Fig y ruing them with the piece of fur, we find tht the rods repel ech other. When we ru glss rods with silk, the glss rods lso ecome chrged nd repel ech other (Fig. 21.1). But chrged plstic rod ttrcts chrged glss rod; furthermore, the plstic rod nd the fur ttrct ech other, nd the glss rod nd the silk ttrct ech other (Fig. 21.1c). These experiments nd mny others like them hve shown tht there re exctly two kinds of electric chrge: the kind on the plstic rod rued with fur nd the kind on the glss rod rued with silk. Benjmin Frnklin ( ) suggested clling these two kinds of chrge negtive nd positive, respectively, nd these nmes re still used. The plstic rod nd the silk hve negtive chrge; the glss rod nd the fur hve positive chrge. Two positive chrges or two negtive chrges repel ech other. A positive chrge nd negtive chrge ttrct ech other. CAUTION Electric ttrction nd repulsion The ttrction nd repulsion of two chrged ojects re sometimes summrized s Like chrges repel, nd opposite chrges ttrct. But keep in mind tht the phrse like chrges does not men tht the two chrges re exctly identicl, only tht oth chrges hve the sme lgeric sign (oth positive or oth negtive). Opposite chrges mens tht oth ojects hve n electric chrge, nd those chrges hve different signs (one positive nd the other negtive) Experiments in electrosttics. () Negtively chrged ojects repel ech other. () Positively chrged ojects repel ech other. (c) Positvely chrged ojects nd negtively chrged ojects ttrct ech other. () Interction etween plstic rods rued on fur Plin plstic rods neither ttrct nor repel ech other... () Interction etween glss rods rued on silk Plin glss rods neither ttrct nor repel ech other... (c) Interction etween ojects with opposite chrges The fur-rued plstic rod nd the silkrued glss rod ttrct ech other... Fur Plstic ilk Glss... ut fter eing rued with fur, the rods repel ech other.... ut fter eing rued with silk, the rods repel ech other.... nd the fur nd silk ech ttrcts the rod it rued.

3 21.1 Electric Chrge chemtic digrm of the opertion of lser printer. 2 Lser em writes on the drum, leving negtively chrged res where the imge will e. 1 Wire sprys ions onto drum, giving the drum positive chrge. 6 Lmp dischrges the drum, redying it to strt the process over. 5 Fuser rollers het pper so toner remins permnently ttched. Rotting imging drum Toner (positively chrged) 3 Roller pplies positively chrged toner to drum. Toner dheres only to negtively chrged res of the drum written y the lser. Pper (feeding to left) 4 Wires spry stronger negtive chrge on pper so toner will dhere to it. One ppliction of forces etween chrged odies is in lser printer (Fig. 21.2). The printer s light-sensitive imging drum is given positive chrge. As the drum rottes, lser em shines on selected res of the drum, leving those res with negtive chrge. Positively chrged prticles of toner dhere only to the res of the drum written y the lser. When piece of pper is plced in contct with the drum, the toner prticles stick to the pper nd form n imge. Electric Chrge nd the tructure of Mtter When you chrge rod y ruing it with fur or silk s in Fig. 21.1, there is no visile chnge in the ppernce of the rod. Wht, then, ctully hppens to the rod when you chrge it? To nswer this question, we must look more closely t the structure of toms, the uilding locks of ordinry mtter. The structure of toms cn e descried in terms of three prticles: the negtively chrged electron, the positively chrged proton, nd the unchrged neutron (Fig. 21.3). The proton nd neutron re comintions of other entities clled qurks, which hve chrges of 1 nd times the electron chrge. Isolted qurks hve not een oserved, nd there re theoreticl resons to elieve tht it is impossile in principle to oserve qurk in isoltion. The protons nd neutrons in n tom mke up smll, very dense core clled the nucleus, with dimensions of the order of m. urrounding the nucleus re the electrons, extending out to distnces of the order of m from the nucleus. If n tom were few kilometers cross, its nucleus would e the size of tennis ll. The negtively chrged electrons re held within the tom y the ttrctive electric forces exerted on them y the positively chrged nucleus. (The protons nd neutrons re held within stle tomic nuclei y n ttrctive interction, clled the strong nucler force, tht overcomes the electric repulsion of the protons. The strong nucler force hs short rnge, nd its effects do not extend fr eyond the nucleus.) The msses of the individul prticles, to the precision tht they re presently known, re Mss of electron = m e = * kg Mss of proton = m p = * kg Mss of neutron = m n = * kg The numers in prentheses re the uncertinties in the lst two digits. Note tht the msses of the proton nd neutron re nerly equl nd re roughly 2000 times 21.3 The structure of n tom. The prticulr tom depicted here is lithium (see Fig. 21.4). Proton: Atom ~ m Tiny compred with the Nucleus rest of the tom, the nucleus contins over 99.9% of the tom s mss. ~ m Positive chrge Mss kg Neutron: No chrge Mss kg Electron: Negtive chrge Mss kg The chrges of the electron nd proton re equl in mgnitude. Most of the tom s volume is occupied sprsely y electrons.

4 690 CHAPTER 21 Electric Chrge nd Electric Field 21.4 () A neutrl tom hs s mny electrons s it does protons. () A positive ion hs deficit of electrons. (c) A negtive ion hs n excess of electrons. (The electron shells re schemtic representtion of the ctul electron distriution, diffuse cloud mny times lrger thn the nucleus.) Protons (1) Neutrons Electrons (2) () Neutrl lithium tom (Li): 3 protons (31) 4 neutrons 3 electrons (32) Electrons equl protons: Zero net chrge () Positive lithium ion (Li 1 ): 3 protons (31) 4 neutrons 2 electrons (22) Fewer electrons thn protons: Positive net chrge (c) Negtive lithium ion (Li 2 ): 3 protons (31) 4 neutrons 4 electrons (42) More electrons thn protons: Negtive net chrge the mss of the electron. Over 99.9% of the mss of ny tom is concentrted in its nucleus. The negtive chrge of the electron hs (within experimentl error) exctly the sme mgnitude s the positive chrge of the proton. In neutrl tom the numer of electrons equls the numer of protons in the nucleus, nd the net electric chrge (the lgeric sum of ll the chrges) is exctly zero (Fig. 21.4). The numer of protons or electrons in neutrl tom of n element is clled the tomic numer of the element. If one or more electrons re removed from n tom, wht remins is clled positive ion (Fig. 21.4). A negtive ion is n tom tht hs gined one or more electrons (Fig. 21.4c). This gin or loss of electrons is clled ioniztion. When the totl numer of protons in mcroscopic ody equls the totl numer of electrons, the totl chrge is zero nd the ody s whole is electriclly neutrl. To give ody n excess negtive chrge, we my either dd negtive chrges to neutrl ody or remove positive chrges from tht ody. imilrly, we cn crete n excess positive chrge y either dding positive chrge or removing negtive chrge. In most cses, negtively chrged (nd highly moile) electrons re dded or removed, nd positively chrged ody is one tht hs lost some of its norml complement of electrons. When we spek of the chrge of ody, we lwys men its net chrge. The net chrge is lwys very smll frction (typiclly no more thn ) of the totl positive chrge or negtive chrge in the ody. Electric Chrge Is Conserved Implicit in the foregoing discussion re two very importnt principles. First is the principle of conservtion of chrge: The lgeric sum of ll the electric chrges in ny closed system is constnt. If we ru together plstic rod nd piece of fur, oth initilly unchrged, the rod cquires negtive chrge (since it tkes electrons from the fur) nd the fur cquires positive chrge of the sme mgnitude (since it hs lost s mny electrons s the rod hs gined). Hence the totl electric chrge on the two odies together does not chnge. In ny chrging process, chrge is not creted or destroyed; it is merely trnsferred from one ody to nother. Conservtion of chrge is thought to e universl conservtion lw. No experimentl evidence for ny violtion of this principle hs ever een oserved. Even in high-energy interctions in which prticles re creted nd destroyed, such s the cretion of electronpositron pirs, the totl chrge of ny closed system is exctly constnt.

5 21.2 Conductors, Insultors, nd Induced Chrges 691 The second importnt principle is: The mgnitude of chrge of the electron or proton is nturl unit of chrge. Every oservle mount of electric chrge is lwys n integer multiple of this sic unit. We sy tht chrge is quntized. A fmilir exmple of quntiztion is money. When you py csh for n item in store, you hve to do it in one-cent increments. Csh cn t e divided into mounts smller thn one cent, nd electric chrge cn t e divided into mounts smller thn the chrge of one electron or proton. (The qurk chrges, 1 nd of the electron chrge, re proly not oservle s isolted chrges.) Thus the chrge on ny mcroscopic ody is lwys either zero or n integer multiple (negtive or positive) of the electron chrge. Understnding the electric nture of mtter gives us insight into mny spects of the physicl world (Fig. 21.5). The chemicl onds tht hold toms together to form molecules re due to electric interctions etween the toms. They include the strong ionic onds tht hold sodium nd chlorine toms together to mke tle slt nd the reltively wek onds etween the strnds of DNA tht record your ody s genetic code. The norml force exerted on you y the chir in which you re sitting rises from electric forces etween chrged prticles in the toms of your set nd in the toms of your chir. The tension force in stretched string nd the dhesive force of glue re likewise due to the electric interctions of toms Most of the forces on this wter skier re electric. Electric interctions etween djcent molecules give rise to the force of the wter on the ski, the tension in the tow rope, nd the resistnce of the ir on the skier s ody. Electric interctions lso hold the toms of the skier s ody together. Only one wholly nonelectric force cts on the skier: the force of grvity. Test Your Understnding of ection 21.1 () trictly speking, does the plstic rod in Fig weigh more, less, or the sme fter ruing it with fur? () Wht out the glss rod fter ruing it with silk? Wht out (c) the fur nd (d) the silk? 21.2 Conductors, Insultors, nd Induced Chrges ome mterils permit electric chrge to move esily from one region of the mteril to nother, while others do not. For exmple, Fig shows copper wire supported y nylon thred. uppose you touch one end of the wire to chrged plstic rod nd ttch the other end to metl ll tht is initilly unchrged; you then remove the chrged rod nd the wire. When you ring nother chrged ody up close to the ll (Figs nd 21.6c), the ll is ttrcted or repelled, showing tht the ll hs ecome electriclly chrged. Electric chrge hs een trnsferred through the copper wire etween the ll nd the surfce of the plstic rod. The copper wire is clled conductor of electricity. If you repet the experiment using ruer nd or nylon thred in plce of the wire, you find tht no chrge is trnsferred to the ll. These mterils re clled insultors. Conductors permit the esy movement of chrge through them, while insultors do not. (The supporting nylon threds shown in Fig re insultors, which prevents chrge from leving the metl ll nd copper wire.) As n exmple, crpet fiers on dry dy re good insultors. As you wlk cross crpet, the ruing of your shoes ginst the fiers cuses chrge to uild up on you, nd this chrge remins on you ecuse it cn t flow through the insulting fiers. If you then touch conducting oject such s doorkno, rpid chrge trnsfer tkes plce etween your finger nd the doorkno, nd you feel shock. One wy to prevent this is to wind some of the crpet fiers round conducting cores so tht ny chrge tht uilds up on you cn e trnsferred hrmlessly to the crpet. Another solution is to cot the crpet fiers with n ntisttic lyer tht does not esily trnsfer electrons to or from your shoes; this prevents ny chrge from uilding up on you in the first plce. PhET: Blloons nd ttic Electricity PhET: John Trvoltge

6 692 CHAPTER 21 Electric Chrge nd Electric Field 21.6 Copper is good conductor of electricity; nylon is good insultor. () The copper wire conducts chrge etween the metl ll nd the chrged plstic rod to chrge the ll negtively. Afterwrd, the metl ll is () repelled y negtively chrged plstic rod nd (c) ttrcted to positively chrged glss rod. () Metl ll Copper wire Insulting nylon threds A negtively chrged plstic rod now repels the ll... Chrged plstic rod Chrged glss rod Chrged plstic rod The wire conducts chrge from the negtively chrged plstic rod to the metl ll. () (c)... nd positively chrged glss rod ttrcts the ll Chrging metl ll y induction. Most metls re good conductors, while most nonmetls re insultors. Within solid metl such s copper, one or more outer electrons in ech tom ecome detched nd cn move freely throughout the mteril, just s the molecules of gs cn move through the spces etween the grins in ucket of snd. The other electrons remin ound to the positively chrged nuclei, which themselves re ound in nerly fixed positions within the mteril. In n insultor there re no, or very few, free electrons, nd electric chrge cnnot move freely through the mteril. ome mterils clled semiconductors re intermedite in their properties etween good conductors nd good insultors. Chrging y Induction We cn chrge metl ll using copper wire nd n electriclly chrged plstic rod, s in Fig In this process, some of the excess electrons on the rod re trnsferred from it to the ll, leving the rod with smller negtive chrge. But there is different technique in which the plstic rod cn give nother ody chrge of opposite sign without losing ny of its own chrge. This process is clled chrging y induction. Figure 21.7 shows n exmple of chrging y induction. An unchrged metl ll is supported on n insulting stnd (Fig. 21.7). When you ring negtively chrged rod ner it, without ctully touching it (Fig. 21.7), the free electrons in the metl ll re repelled y the excess electrons on the rod, nd they shift towrd the right, wy from the rod. They cnnot escpe from the ll ecuse the supporting stnd nd the surrounding ir re insultors. o we get excess negtive chrge t the right surfce of the ll nd deficiency of negtive chrge (tht is, net positive chrge) t the left surfce. These excess chrges re clled induced chrges. Not ll of the free electrons move to the right surfce of the ll. As soon s ny induced chrge develops, it exerts forces towrd the left on the other free electrons. These electrons re repelled y the negtive induced chrge on the right nd ttrcted towrd the positive induced chrge on the left. The system reches n equilirium stte in which the force towrd the right on n electron, due to the chrged rod, is just lnced y the force towrd the left due to the induced chrge. If we remove the chrged rod, the free electrons shift ck to the left, nd the originl neutrl condition is restored. Wht hppens if, while the plstic rod is nery, you touch one end of conducting wire to the right surfce of the ll nd the other end to the erth (Fig. 21.7c)? The erth is conductor, nd it is so lrge tht it cn ct s prcticlly infinite source of extr electrons or sink of unwnted electrons. ome of the negtive chrge flows through the wire to the erth. Now suppose you disconnect the wire (Fig. 21.7d) nd then remove the rod (Fig. 21.7e); net positive chrge is left on the ll. The chrge on the negtively chrged rod hs not chnged during this process. The erth cquires negtive chrge tht is equl in mgnitude to the induced positive chrge remining on the ll. Metl ll Insulting stnd () Unchrged metl ll Electron deficiency Negtively chrged rod Electron uildup () Negtive chrge on rod repels electrons, creting zones of negtive nd positive induced chrge. Wire Ground (c) Wire lets electron uildup (induced negtive chrge) flow into ground. Negtive chrge in ground (d) Wire removed; ll now hs only n electrondeficient region of positive chrge. (e) Rod removed; electrons rerrnge themselves, ll hs overll electron deficiency (net positive chrge).

7 21.3 Coulom s Lw The chrges within the molecules of n insulting mteril cn shift slightly. As result, com with either sign of chrge ttrcts neutrl insultor. By Newton s third lw the neutrl insultor exerts n equl-mgnitude ttrctive force on the com. () A chrged com picking up unchrged pieces of plstic () How negtively chrged com ttrcts n insultor (c) How positively chrged com ttrcts n insultor Electrons in ech molecule of the neutrl insultor shift wy from the com. 2F F Negtively chrged com As result, the (1) chrges in ech molecule re closer to the com thn re the (2) chrges nd so feel stronger force from the com. Therefore the net force is ttrctive. This time, electrons in the molecules shift towrd the com... 2F F Positively chrged com... so tht the (2) chrges in ech molecule re closer to the com, nd feel stronger force from it, thn the () chrges. Agin, the net force is ttrctive. Electric Forces on Unchrged Ojects Finlly, we note tht chrged ody cn exert forces even on ojects tht re not chrged themselves. If you ru lloon on the rug nd then hold the lloon ginst the ceiling, it sticks, even though the ceiling hs no net electric chrge. After you electrify com y running it through your hir, you cn pick up unchrged its of pper or plstic with the com (Fig. 21.8). How is this possile? This interction is n induced-chrge effect. Even in n insultor, electric chrges cn shift ck nd forth little when there is chrge nery. This is shown in Fig. 21.8; the negtively chrged plstic com cuses slight shifting of chrge within the molecules of the neutrl insultor, n effect clled polriztion. The positive nd negtive chrges in the mteril re present in equl mounts, ut the positive chrges re closer to the plstic com nd so feel n ttrction tht is stronger thn the repulsion felt y the negtive chrges, giving net ttrctive force. (In ection 21.3 we will study how electric forces depend on distnce.) Note tht neutrl insultor is lso ttrcted to positively chrged com (Fig. 21.8c). Now the chrges in the insultor shift in the opposite direction; the negtive chrges in the insultor re closer to the com nd feel n ttrctive force tht is stronger thn the repulsion felt y the positive chrges in the insultor. Hence chrged oject of either sign exerts n ttrctive force on n unchrged insultor. Figure 21.9 shows n industril ppliction of this effect. Test Your Understnding of ection 21.2 You hve two lightweight metl spheres, ech hnging from n insulting nylon thred. One of the spheres hs net negtive chrge, while the other sphere hs no net chrge. () If the spheres re close together ut do not touch, will they (i) ttrct ech other, (ii) repel ech other, or (iii) exert no force on ech other? () You now llow the two spheres to touch. Once they hve touched, will the two spheres (i) ttrct ech other, (ii) repel ech other, or (iii) exert no force on ech other? 21.9 The electrosttic pinting process (compre Figs nd 21.7c). A metl oject to e pinted is connected to the erth ( ground ), nd the pint droplets re given n electric chrge s they exit the spryer nozzle. Induced chrges of the opposite sign pper in the oject s the droplets pproch, just s in Fig. 21.7, nd they ttrct the droplets to the surfce. This process minimizes overspry from clouds of stry pint prticles nd gives prticulrly smooth finish. pry of negtively chrged pint droplets Pint spryer Metl oject to e pinted Positive chrge is induced on surfce of metl. Ground 21.3 Coulom s Lw Chrles Augustin de Coulom ( ) studied the interction forces of chrged prticles in detil in He used torsion lnce (Fig ) similr to the one used 13 yers lter y Cvendish to study the much weker grvittionl interction, s we discussed in ection For point chrges, chrged ActivPhysics 11.1: Electric Force: Coulom's Lw ActivPhysics 11.2: Electric Force: uperposition Principle ActivPhysics 11.3: Electric Force: uperposition (Quntittive)

8 694 CHAPTER 21 Electric Chrge nd Electric Field Appliction Electric Forces, wet, nd Cystic Firosis One wy to test for the genetic disese cystic firosis (CF) is y mesuring the slt content of person s swet. wet is mixture of wter nd ions, including the sodium ( N ) nd chloride ( Cl - ) ions tht mke up ordinry slt (NCl). When swet is secreted y epithelil cells, some of the Cl - ions flow from the swet ck into these cells ( process clled resorption). The electric ttrction etween negtive nd positive chrges pulls N ions long with the Cl -. Wter molecules cnnot flow ck into the epithelil cells, so swet on the skin hs low slt content. However, in persons with CF the resorption of Cl - ions is locked. Hence the swet of persons with CF is unusully slty, with up to four times the norml concentrtion of Cl - nd N. odies tht re very smll in comprison with the distnce r etween them, Coulom found tht the electric force is proportionl to 1>r 2. Tht is, when the distnce r doules, the force decreses to one-qurter of its initil vlue; when the distnce is hlved, the force increses to four times its initil vlue. The electric force etween two point chrges lso depends on the quntity of chrge on ech ody, which we will denote y q or Q. To explore this dependence, Coulom divided chrge into two equl prts y plcing smll chrged sphericl conductor into contct with n identicl ut unchrged sphere; y symmetry, the chrge is shred eqully etween the two spheres. (Note the essentil role of the principle of conservtion of chrge in this procedure.) Thus he could otin one-hlf, one-qurter, nd so on, of ny initil chrge. He found tht the forces tht two point chrges q 1 nd q 2 exert on ech other re proportionl to ech chrge nd therefore re proportionl to the product q 1 q 2 of the two chrges. Thus Coulom estlished wht we now cll Coulom s lw: The mgnitude of the electric force etween two point chrges is directly proportionl to the product of the chrges nd inversely proportionl to the squre of the distnce etween them. In mthemticl terms, the mgnitude F of the force tht ech of two point chrges nd distnce r prt exerts on the other cn e expressed s q 1 q 2 F = k ƒq 1q 2 ƒ r 2 (21.1) where k is proportionlity constnt whose numericl vlue depends on the system of units used. The solute vlue rs re used in Eq. (21.1) ecuse the chrges q 1 nd q 2 cn e either positive or negtive, while the force mgnitude F is lwys positive. The directions of the forces the two chrges exert on ech other re lwys long the line joining them. When the chrges q 1 nd q 2 hve the sme sign, either oth positive or oth negtive, the forces re repulsive; when the chrges hve opposite signs, the forces re ttrctive (Fig ). The two forces oey Newton s third lw; they re lwys equl in mgnitude nd opposite in direction, even when the chrges re not equl in mgnitude () Mesuring the electric force etween point chrges. () The electric forces etween point chrges oey Newton s third lw: F 1 on 2 F 2 on 1. () A torsion lnce of the type used y Coulom to mesure the electric force Torsion fier Chrged pith lls cle The negtively chrged ll ttrcts the positively chrged one; the positive ll moves until the elstic forces in the torsion fier lnce the electrosttic ttrction. () Interctions etween point chrges F 2 on 1 q 1 F 1 on 2 52F 2 on 1 F 1 on 2 5 F 2 on 1 5 k q 1 F 2 on 1 r r Chrges of the sme sign repel. 0q 1 q 2 0 r 2 F 1 on 2 q 2 Chrges of opposite sign ttrct. q 2 F 1 on 2

9 21.3 Coulom s Lw 695 The proportionlity of the electric force to 1>r 2 hs een verified with gret precision. There is no reson to suspect tht the exponent is different from precisely 2. Thus the form of Eq. (21.1) is the sme s tht of the lw of grvittion. But electric nd grvittionl interctions re two distinct clsses of phenomen. Electric interctions depend on electric chrges nd cn e either ttrctive or repulsive, while grvittionl interctions depend on mss nd re lwys ttrctive (ecuse there is no such thing s negtive mss). Fundmentl Electric Constnts The vlue of the proportionlity constnt k in Coulom s lw depends on the system of units used. In our study of electricity nd mgnetism we will use I units exclusively. The I electric units include most of the fmilir units such s the volt, the mpere, the ohm, nd the wtt. (There is no British system of electric units.) The I unit of electric chrge is clled one coulom (1 C). In I units the constnt k in Eq. (21.1) is k = * 10 9 N # m 2 >C * 10 9 N # m 2 >C 2 The vlue of k is known to such lrge numer of significnt figures ecuse this vlue is closely relted to the speed of light in vcuum. (We will show this in Chpter 32 when we study electromgnetic rdition.) As we discussed in ection 1.3, this speed is defined to e exctly c = * 10 8 m>s. The numericl vlue of k is defined in terms of c to e precisely k = N # s 2 >C 2 2c 2 You should check this expression to confirm tht k hs the right units. In principle we cn mesure the electric force F etween two equl chrges q t mesured distnce r nd use Coulom s lw to determine the chrge. Thus we could regrd the vlue of k s n opertionl definition of the coulom. For resons of experimentl precision it is etter to define the coulom insted in terms of unit of electric current (chrge per unit time), the mpere, equl to 1 coulom per second. We will return to this definition in Chpter 28. In I units we usully write the constnt k in Eq. (21.1) s 1>4pP 0, where P 0 ( epsilon-nought or epsilon-zero ) is nother constnt. This ppers to complicte mtters, ut it ctully simplifies mny formuls tht we will encounter in lter chpters. From now on, we will usully write Coulom s lw s F = 1 ƒq 1 q 2 ƒ 4pP 0 r 2 (Coulom s lw: force etween two point chrges) (21.2) The constnts in Eq. (21.2) re pproximtely P 0 = * C 2 >N # m 2 nd In exmples nd prolems we will often use the pproximte vlue 1 4pP 0 = 9.0 * 10 9 N # m 2 >C 2 which is within out 0.1% of the correct vlue. As we mentioned in ection 21.1, the most fundmentl unit of chrge is the mgnitude of the chrge of n electron or proton, which is denoted y e. The most precise vlue ville s of the writing of this ook is e = * C 1 4pP 0 = k = * 10 9 N # m 2 >C 2 One coulom represents the negtive of the totl chrge of out 6 * electrons. For comprison, copper cue 1 cm on side contins out 2.4 * 10 24

10 696 CHAPTER 21 Electric Chrge nd Electric Field electrons. Aout electrons pss through the glowing filment of flshlight ul every second. In electrosttics prolems (tht is, prolems tht involve chrges t rest), it s very unusul to encounter chrges s lrge s 1 coulom. Two 1-C chrges seprted y 1 m would exert forces on ech other of mgnitude 9 * 10 9 N (out 1 million tons)! The totl chrge of ll the electrons in copper one-cent coin is even greter, out 1.4 * 10 5 C, which shows tht we cn t distur electric neutrlity very much without using enormous forces. More typicl vlues of chrge rnge from out 10-9 to out 10-6 C. The microcoulom 11 mc = 10-6 C2 nd the nnocoulom 11 nc = 10-9 C2 re often used s prcticl units of chrge Exmple 21.1 Electric force versus grvittionl force An prticle (the nucleus of helium tom) hs mss m = 6.64 * kg nd chrge q = 2e = 3.2 * C. Compre the mgnitude of the electric repulsion etween two ( lph ) prticles with tht of the grvittionl ttrction etween them. OLUTION IDENTIFY nd ET UP: This prolem involves Newton s lw for the grvittionl force F g etween prticles (see ection 13.1) nd Coulom s lw for the electric force F e etween point chrges. To compre these forces, we mke our trget vrile the rtio F e >F g. We use Eq. (21.2) for F e nd Eq. (13.1) for Our sketch for this prolem. F g EXECUTE: Figure shows our sketch. From Eqs. (21.2) nd (13.1), 1 q 2 F e = 4pP 0 r 2 F g = G m 2 r 2 These re oth inverse-squre forces, so the we tke the rtio: F e F g = 1 4pP 0 G q 2 m 2 = 3.1 * fctors cncel when -19 C * 10 9 N # m 2 >C * 10 = 6.67 * N # m 2 >kg * kg2 2 r 2 EVALUATE: This stonishingly lrge numer shows tht the grvittionl force in this sitution is completely negligile in comprison to the electric force. This is lwys true for interctions of tomic nd sunucler prticles. But within ojects the size of person or plnet, the positive nd negtive chrges re nerly equl in mgnitude, nd the net electric force is usully much smller thn the grvittionl force. uperposition of Forces Coulom s lw s we hve stted it descries only the interction of two point chrges. Experiments show tht when two chrges exert forces simultneously on third chrge, the totl force cting on tht chrge is the vector sum of the forces tht the two chrges would exert individully. This importnt property, clled the principle of superposition of forces, holds for ny numer of chrges. By using this principle, we cn pply Coulom s lw to ny collection of chrges. Two of the exmples t the end of this section use the superposition principle. trictly speking, Coulom s lw s we hve stted it should e used only for point chrges in vcuum. If mtter is present in the spce etween the chrges, the net force cting on ech chrge is ltered ecuse chrges re induced in the molecules of the intervening mteril. We will descrie this effect lter. As prcticl mtter, though, we cn use Coulom s lw unltered for point chrges in ir. At norml tmospheric pressure, the presence of ir chnges the electric force from its vcuum vlue y only out one prt in 2000.

11 21.3 Coulom s Lw 697 Prolem-olving trtegy 21.1 Coulom s Lw IDENTIFY the relevnt concepts: Coulom s lw descries the electric force etween chrged prticles. ET UP the prolem using the following steps: 1. ketch the loctions of the chrged prticles nd lel ech prticle with its chrge. 2. If the chrges do not ll lie on single line, set up n xy- coordinte system. 3. The prolem will sk you to find the electric force on one or more prticles. Identify which these re. EXECUTE the solution s follows: 1. For ech prticle tht exerts n electric force on given prticle of interest, use Eq. (21.2) to clculte the mgnitude of tht force. 2. Using those mgnitudes, sketch free-ody digrm showing the electric force vectors cting on ech prticle of interest. The force exerted y prticle 1 on prticle 2 points from prticle 2 towrd prticle 1 if the chrges hve opposite signs, ut points from prticle 2 directly wy from prticle 1 if the chrges hve the sme sign. 3. Use the principle of superposition to clculte the totl electric force vector sum on ech prticle of interest. (Review the vector lger in ections 1.7 through 1.9. The method of components is often helpful.) 4. Use consistent units; I units re completely consistent. With 1>4pP 9.0 * 10 9 N # m 2 >C 2 0, distnces must e in meters, chrges in couloms, nd forces in newtons. 5. ome exmples nd prolems in this nd lter chpters involve continuous distriutions of chrge long line, over surfce, or throughout volume. In these cses the vector sum in step 3 ecomes vector integrl. We divide the chrge distriution into infinitesiml pieces, use Coulom s lw for ech piece, nd integrte to find the vector sum. ometimes this cn e done without ctul integrtion. 6. Exploit ny symmetries in the chrge distriution to simplify your prolem solving. For exmple, two identicl chrges q exert zero net electric force on chrge Q midwy etween them, ecuse the forces on Q hve equl mgnitude nd opposite direction. EVALUATE your nswer: Check whether your numericl results re resonle. Confirm tht the direction of the net electric force grees with the principle tht chrges of the sme sign repel nd chrges of opposite sign ttrct. Exmple 21.2 Force etween two point chrges Two point chrges, q 1 = 25 nc nd q 2 = -75 nc, re seprted y distnce r = 3.0 cm (Fig ). Find the mgnitude nd direction of the electric force () tht q 1 exerts on q 2 nd () tht q 2 exerts on. q 1 OLUTION IDENTIFY nd ET UP: This prolem sks for the electric forces tht two chrges exert on ech other. We use Coulom s lw, Eq. (21.2), to clculte the mgnitudes of the forces. The signs of the chrges will determine the directions of the forces. EXECUTE: () After converting the units of r to meters nd the units of q 1 nd q 2 to couloms, Eq. (21.2) gives us 1 ƒ q 1 q 2 ƒ F 1on2 = 4pP 0 r 2 = 19.0 * 10 9 N # m 2 >C 2 2 ƒ 125 * 10-9 C21-75 * 10-9 C2 ƒ m2 2 = N The chrges hve opposite signs, so the force is ttrctive (to the left in Fig ); tht is, the force tht cts on q 2 is directed towrd q 1 long the line joining the two chrges Wht force does q 1 exert on q 2, nd wht force does exert on q 1? Grvittionl forces re negligile. () The two chrges q 1 q 2 r () Proceeding s in prt (), we hve 1 ƒ q 2 q 1 ƒ F 1on2 = = F 4pP 2on1 = N 0 The ttrctive force tht cts on (Fig c). is to the right, towrd EVALUATE: Newton s third lw pplies to the electric force. Even though the chrges hve different mgnitudes, the mgnitude of the force tht q 2 exerts on q 1 is the sme s the mgnitude of the force tht q 1 exerts on, nd these two forces re in opposite directions. q 2 () Free-ody digrm for chrge q 2 F 1 on 2 r 2 q 2 q 1 (c) Free-ody digrm for chrge q 1 q 1 F2 on 1 q 2 q 2 Exmple 21.3 Vector ddition of electric forces on line Two point chrges re locted on the x-xis of coordinte system: q 1 = 1.0 nc is t x = 2.0 cm, nd q 2 = -3.0 nc is t x = 4.0 cm. Wht is the totl electric force exerted y q 1 nd q 2 on chrge q 3 = 5.0 nc t x = 0? OLUTION IDENTIFY nd ET UP: Figure shows the sitution. To find the totl force on q 3, our trget vrile, we find the vector sum of the two electric forces on it. Continued

12 698 CHAPTER 21 Electric Chrge nd Electric Field EXECUTE: Figure is free-ody digrm for q 3 which is repelled y q 1 (which hs the sme sign) nd ttrcted to,q 2 (which hs the opposite sign): F is in the x-direction nd F 1 on 3 2 on 3 is in the x-direction. After unit conversions, we hve from Eq. (21.2) 1 ƒ q 1 q 3 ƒ F 1 on 3 = 4pP 0 In the sme wy you cn show tht F 2 on 3 = 84 mn. We thus hve F nd F 1 on m 2ın 2 on m 2ın. The net force on is q 3 2 r 13 = 19.0 * 10 9 N # m 2 >C * 10-9 C215.0 * 10-9 C m2 2 = 1.12 * 10-4 N = 112 mn F 3 F 1 on 3 F 2 on m 2ın 184 m 2ın 1-28 m 2ın Our sketches for this prolem. () Our digrm of the sitution () Free-ody digrm for q 3 EVALUATE: As check, note tht the mgnitude of q 2 is three times tht of q 1, ut q 2 is twice s fr from q 3 s q 1. Eqution (21.2) then sys tht F must e 3>2 2 2 on 3 = 3>4 = 0.75 s lrge s F 1 on 3. This grees with our clculted vlues: F 2 on 3 >F 1 on 3 = 184 mn2>1112 mn2 = Becuse F 2on3 is the weker force, the direction of the net force is tht of F 1 on 3 tht is, in the negtive x-direction. Exmple 21.4 Vector ddition of electric forces in plne Two equl positive chrges q 1 = q 2 = 2.0 mc re locted t EXECUTE: Figure shows the forces F nd F 1 on Q 2 on Q due to x = 0, y = 0.30 m nd x = 0, y = m, respectively. Wht re the mgnitude nd direction of the totl electric force tht q 1 nd q 2 exert on third chrge Q = 4.0 mc t x = 0.40 m, y = 0? the identicl chrges q 1 nd q 2, which re t equl distnces from Q. From Coulom s lw, oth forces hve mgnitude F 1 or 2 on Q = 19.0 * 10 9 N # m 2 >C 2 2 OLUTION IDENTIFY nd ET UP: As in Exmple 21.3, we must compute the force tht ech chrge exerts on Q nd then find the vector sum of those forces. Figure shows the sitution. ince the three chrges do not ll lie on line, the est wy to clculte the forces is to use components Our sketch for this prolem. * 14.0 * 10-6 C212.0 * 10-6 C m2 2 = 0.29 N The x-components of the two forces re equl: 1F 1 or 2 on Q 2 x = 1F 1 or 2 on Q 2cos = N m 0.50 m = 0.23 N From symmetry we see tht the y-components of the two forces re equl nd opposite. Hence their sum is zero nd the totl force F on Q hs only n x-component F x = 0.23 N 0.23 N = 0.46 N. The totl force on Q is in the x-direction, with mgnitude 0.46 N. EVALUATE: The totl force on Q points neither directly wy from q 1 nor directly wy from q 2. Rther, this direction is compromise tht points wy from the system of chrges q 1 nd q 2. Cn you see tht the totl force would not e in the x-direction if q 1 nd q 2 were not equl or if the geometricl rrngement of the chnges were not so symmetric? Test Your Understnding of ection 21.3 uppose tht chrge q 2 in Exmple 21.4 were -2.0 mc. In this cse, the totl electric force on Q would e (i) in the positive x-direction; (ii) in the negtive x-direction; (iii) in the positive y-direction; (iv) in the negtive y-direction; (v) zero; (vi) none of these Electric Field nd Electric Forces When two electriclly chrged prticles in empty spce interct, how does ech one know the other is there? We cn egin to nswer this question, nd t the sme time reformulte Coulom s lw in very useful wy, y using the concept of electric field.

13 21.4 Electric Field nd Electric Forces 699 Electric Field To introduce this concept, let s look t the mutul repulsion of two positively chrged odies A nd B (Fig ). uppose B hs chrge q nd let F 0, 0 e the electric force of A on B. One wy to think out this force is s n ction-t-distnce force tht is, s force tht cts cross empty spce without needing ny mtter (such s push rod or rope) to trnsmit it through the intervening spce. (Grvity cn lso e thought of s n ction-t--distnce force.) But more fruitful wy to visulize the repulsion etween A nd B is s two-stge process. We first envision tht ody A, s result of the chrge tht it crries, somehow modifies the properties of the spce round it. Then ody B, s result of the chrge tht it crries, senses how spce hs een modified t its position. The response of ody B is to experience the force F 0. To elorte how this two-stge process occurs, we first consider ody A y itself: We remove ody B nd lel its former position s point P (Fig ). We sy tht the chrged ody A produces or cuses n electric field t point P (nd t ll other points in the neighorhood). This electric field is present t P even if there is no chrge t P; it is consequence of the chrge on ody A only. If point chrge q is then plced t point, it experiences the force F We tke the point of view tht this force is exerted on Pq y the field t P (Fig c). Thus the electric field is the intermediry through which A communictes its presence to q 0. Becuse the point chrge q 0 would experience force t ny point in the neighorhood of A, the electric field tht A produces exists t ll points in the region round A. We cn likewise sy tht the point chrge q 0 produces n electric field in the spce round it nd tht this electric field exerts the force F 0 on ody A. For ech force (the force of A on q 0 nd the force of q 0 on A), one chrge sets up n electric field tht exerts force on the second chrge. We emphsize tht this is n interction etween two chrged odies. A single chrge produces n electric field in the surrounding spce, ut this electric field cnnot exert net force on the chrge tht creted it; s we discussed in ection 4.3, ody cnnot exert net force on itself. (If this wsn t true, you would e le to lift yourself to the ceiling y pulling up on your elt!) The electric force on chrged ody is exerted y the electric field creted y other chrged odies. To find out experimentlly whether there is n electric field t prticulr point, we plce smll chrged ody, which we cll test chrge, t the point (Fig c). If the test chrge experiences n electric force, then there is n electric field t tht point. This field is produced y chrges other thn q 0. Force is vector quntity, so electric field is lso vector quntity. (Note the use of vector signs s well s oldfce letters nd plus, minus, nd equls signs in the following discussion.) We define the electric field E t point s the electric force F 0 experienced y test chrge q 0 t the point, divided y the chrge q 0. Tht is, the electric field t certin point is equl to the electric force per unit chrge experienced y chrge t tht point: E F 0 q 0 (definition of electric field s electric force per unit chrge) (21.3) A chrged ody cretes n electric field in the spce round it. () A nd B exert electric forces on ech other. 2F 0 A () Remove ody B nd lel its former position s P. A (c) Body A sets up n electric field E t point P. A q 0 B P F0 Test chrge q 0 F 0 E 5 q 0 E is the force per unit chrge exerted y A on test chrge t P. Appliction hrks nd the ixth ense hrks hve the ility to locte prey (such s flounder nd other ottom-dwelling fish) tht re completely hidden eneth the snd t the ottom of the ocen. They do this y sensing the wek electric fields produced y muscle contrctions in their prey. hrks derive their sensitivity to electric fields ( sixth sense ) from jelly-filled cnls in their odies. These cnls end in pores on the shrk s skin (shown in this photogrph). An electric field s wek s 5 * 10-7 N> C cuses chrge flow within the cnls nd triggers signl in the shrk s nervous system. Becuse the shrk hs cnls with different orienttions, it cn mesure different components of the electric-field vector nd hence determine the direction of the field. In I units, in which the unit of force is 1 N nd the unit of chrge is 1 C, the unit of electric field mgnitude is 1 newton per coulom 11 N>C2. If the field E t certin point is known, rerrnging Eq. (21.3) gives the force F 0 experienced y point chrge q 0 plced t tht point. This force is just equl to the electric field E produced t tht point y chrges other thn q 0, multiplied y the chrge q 0 : (force exerted on point chrge q 0 F 0 q 0 E (21.4) y n electric field E )

14 700 CHAPTER 21 Electric Chrge nd Electric Field The force F 0 q 0 E exerted on point chrge plced in n electric field E. Q Q q 0 F 0 q 0 q 0 E (due to chrge Q) F 0 The force on positive test chrge q 0 points in the direction of the electric field. E (due to chrge Q) The force on negtive test chrge q 0 points opposite to the electric field. ActivPhysics 11.4: Electric Field: Point Chrge ActivPhysics 11.9: Motion of Chrge in n Electric Field: Introduction ActivPhysics 11.10: Motion in n Electric Field: Prolems The chrge q cn e either positive or negtive. If is positive, the force F 0 q 0 experienced y the chrge is the sme direction s E if is negtive, F nd E 0 ; q 0 0 re in opposite directions (Fig ). While the electric field concept my e new to you, the sic ide tht one ody sets up field in the spce round it nd second ody responds to tht field is one tht you ve ctully used efore. Compre Eq. (21.4) to the fmilir expression for the grvittionl force tht the erth exerts on mss m 0 : (21.5) In this expression, is the ccelertion due to grvity. If we divide oth sides of Eq. (21.5) y the mss m 0, we otin g g F g F g m 0 g g F g m 0 Thus cn e regrded s the grvittionl force per unit mss. By nlogy to Eq. (21.3), we cn interpret g s the grvittionl field. Thus we tret the grvittionl interction etween the erth nd the mss m 0 s two-stge process: The erth sets up grvittionl field g in the spce round it, nd this grvittionl field exerts force given y Eq. (21.5) on the mss m 0 (which we cn regrd s test mss). The grvittionl field g, or grvittionl force per unit mss, is useful concept ecuse it does not depend on the mss of the ody on which the grvittionl force is exerted; likewise, the electric field E, or electric force per unit chrge, is useful ecuse it does not depend on the chrge of the ody on which the electric force is exerted. CAUTION F 0 q 0 E 0 is for point test chrges only The electric force experienced y test chrge q 0 cn vry from point to point, so the electric field cn lso e different t different points. For this reson, Eq. (21.4) cn e used only to find the electric force on point chrge. If chrged ody is lrge enough in size, the electric field E my e noticely different in mgnitude nd direction t different points on the ody, nd clculting the net electric force on the ody cn ecome rther complicted. Electric Field of Point Chrge If the source distriution is point chrge q, it is esy to find the electric field tht it produces. We cll the loction of the chrge the source point, nd we cll the point P where we re determining the field the field point. It is lso useful to introduce unit vector rn tht points long the line from source point to field point (Fig ). This unit vector is equl to the displcement vector r from the source point to the field point, divided y the distnce r = ƒ r ƒ etween these two points; tht is, rn r>r. If we plce smll test chrge t the field point P, t q The electric field produced t point P y n isolted point chrge q t. Note tht in oth () nd (c), is produced y q [see Eq. (21.7)] ut cts on the chrge t point P [see Eq. (21.4)]. E q 0 E () q r^ r q 0 P Unit vector r ^ points from source point to field point P. () q r^ q 0 P At ech point P, the electric field set up y n isolted positive point chrge q points directly wy from the chrge in the sme direction s r. ^ E (c) q E q 0 P r^ At ech point P, the electric field set up y n isolted negtive point chrge q points directly towrd the chrge in the opposite direction from r. ^

15 21.4 Electric Field nd Electric Forces 701 distnce r from the source point, the mgnitude F 0 of the force is given y Coulom s lw, Eq. (21.2): From Eq. (21.3) the mgnitude E of the electric field t P is E = F 0 = 1 ƒqq 0 ƒ 4pP 0 r 2 1 ƒqƒ 4pP 0 r 2 (mgnitude of electric field of point chrge) (21.6) Using the unit vector rn, we cn write vector eqution tht gives oth the mgnitude nd direction of the electric field E : E 1 4pP 0 q r 2 r N (electric field of point chrge) (21.7) By definition, the electric field of point chrge lwys points wy from positive chrge (tht is, in the sme direction s rn; see Fig ) ut towrd negtive chrge (tht is, in the direction opposite rn; see Fig c) A point chrge q produces n electric field E t ll points in spce. The field We hve emphsized clculting the electric field E strength decreses with incresing distnce. t certin point. But since E cn vry from point to point, it is not single vector quntity ut rther () The field produced y positive point chrge points wy from the chrge. n infinite set of vector quntities, one ssocited with ech point in spce. This is n exmple of vector field. Figure shows numer of the field vectors produced y positive or negtive point chrge. If we use rectngulr 1x, y, z2 coordinte system, ech component of E t ny point is in generl function of the coordintes 1x, y, z2 of the point. We cn represent the functions s E E x 1x, y, z2, E y 1x, y, z2, nd E z 1x, y, z2. Vector fields re n importnt prt of the lnguge of physics, not just in electricity nd mgnetism. One everydy exmple of vector field is the velocity Y q of wind currents; the mgnitude nd direction of Y, nd hence its vector components, vry from point to point in the tmosphere. In some situtions the mgnitude nd direction of the field (nd hence its vector components) hve the sme vlues everywhere throughout certin region; we then sy tht the field is uniform in this region. An importnt exmple of this is the electric field inside conductor. If there is n electric field within conductor, the field exerts force on every chrge in the conductor, giving the free chrges net motion. By definition n electrosttic sitution is one in which the () The field produced y negtive point chrges hve no net motion. We conclude tht in electrosttics the electric field t chrge points towrd the chrge. every point within the mteril of conductor must e zero. (Note tht we re not sying tht the field is necessrily zero in hole inside conductor.) In summry, our description of electric interctions hs two prts. First, given E chrge distriution cts s source of electric field. econd, the electric field exerts force on ny chrge tht is present in the field. Our nlysis often hs two q corresponding steps: first, clculting the field cused y source chrge distriution; second, looking t the effect of the field in terms of force nd motion. The second step often involves Newton s lws s well s the principles of electric interctions. In the next section we show how to clculte fields cused y vrious source distriutions, ut first here re three exmples of clculting the field due to point chrge nd of finding the force on chrge due to given field E. Exmple 21.5 Electric-field mgnitude for point chrge Wht is the mgnitude of the electric field from point chrge q = 4.0 nc? E t field point 2.0 m OLUTION IDENTIFY nd ET UP: This prolem concerns the electric field due to point chrge. We re given the mgnitude of the chrge Continued

16 702 CHAPTER 21 Electric Chrge nd Electric Field nd the distnce from the chrge to the field point, so we use Eq. (21.6) to clculte the field mgnitude E. EXECUTE: From Eq. (21.6), 1 ƒ q ƒ E = 4pP 0 r 2 = 19.0 * 109 N # m 2 >C * 10-9 C 12.0 m2 2 = 9.0 N>C EVALUATE: Our result E = 9.0 N>C mens tht if we plced 1.0-C chrge t point 2.0 m from q, it would experience 9.0-N force. The force on 2.0-C chrge t tht point would e 12.0 C219.0 N>C2 = 18 N, nd so on. Exmple 21.6 Electric-field vector for point chrge A point chrge q = -8.0 nc is locted t the origin. Find the electric-field vector t the field point x = 1.2 m, y = -1.6 m. the unit vector rn r>r tht points from to P. gin O) to the field point P, nd we must otin n expression for EXECUTE: The distnce from to P is OLUTION IDENTIFY nd ET UP: We must find the electric-field vector E r = 2x 2 y 2 = m m2 2 = 2.0 m due to point chrge. Figure shows the sitution. We use Eq. The unit vector rn is then (21.7); to do this, we must find the distnce r from the source point xın y n (the position of the chrge q, which in this exmple is t the ori- rn r r r 11.2 m2ın m2 n Our sketch for this prolem. 0.60ın 0.80 n 2.0 m Then, from Eq. (21.7), E 1 q 4pP 0 r 2 rn 19.0 * 10 9 N # m 2 >C * 10-9 C ın 12.0 m n N>C2ın 114 N>C2 n EVALUATE: ince q is negtive, points from the field point to the chrge (the source point), in the direction opposite to rn (compre Fig c). We leve the clcultion of the mgnitude nd direction of E to you (see Exercise 21.36). E Exmple 21.7 Electron in uniform field When the terminls of ttery re connected to two prllel conducting pltes with smll gp etween them, the resulting chrges on the pltes produce nerly uniform electric field E etween the pltes. (In the next section we ll see why this is.) If the pltes re 1.0 cm prt nd re connected to 100-volt ttery s shown in Fig , the field is verticlly upwrd nd hs mgnitude A uniform electric field etween two prllel conducting pltes connected to 100-volt ttery. (The seprtion of the pltes is exggerted in this figure reltive to the dimensions of the pltes.) 100 V The thin rrows represent the uniform electric field. x O E F 52eE 1.0 cm y E = 1.00 * 10 4 N>C. () If n electron (chrge -e = * 10-9 C, mss m = 9.11 * kg) is relesed from rest t the upper plte, wht is its ccelertion? () Wht speed nd kinetic energy does it cquire while trveling 1.0 cm to the lower plte? (c) How long does it tke to trvel this distnce? OLUTION IDENTIFY nd ET UP: This exmple involves the reltionship etween electric field nd electric force. It lso involves the reltionship etween force nd ccelertion, the definition of kinetic energy, nd the kinemtic reltionships mong ccelertion, distnce, velocity, nd time. Figure shows our coordinte system. We re given the electric field, so we use Eq. (21.4) to find the force on the electron nd Newton s second lw to find its ccelertion. Becuse the field is uniform, the force is constnt nd we cn use the constnt-ccelertion formuls from Chpter 2 to find the electron s velocity nd trvel time. We find the kinetic energy using K = 1 2 mv2.

17 21.5 Electric-Field Clcultions 703 EXECUTE: () Although E is upwrd (in the y-direction), is The velocity is downwrd, so v y = -5.9 * 10 6 m>s. The electron s downwrd (ecuse the electron s chrge is negtive) nd so F y is negtive. Becuse F y is constnt, the electron s ccelertion is constnt: kinetic energy is K = 1 2 mv2 = * kg215.9 * 10 6 m>s2 2 y = F y m = -ee m = * C * 10 4 N>C2 = 1.6 * J 9.11 * kg (c) From Eq. (2.8) for constnt ccelertion, v y = v 0y y t, = * m>s 2 t = v y - v 0y y () The electron strts from rest, so its motion is in the y-direction only (the direction of the ccelertion). We cn find the electron s speed t ny position y using the constnt-ccelertion Eq. (2.13), vy 2 = v0y 2 2 y 1y - y 0 2. We hve v 0y = 0 nd y so t y = -1.0 cm = -1.0 * = 0, m we hve ƒ v y ƒ = 22 y y = * m>s * 10-2 m2 = 5.9 * 10 6 m>s F = * 106 m>s2-10 m>s * m>s 2 = 3.4 * 10-9 s EVALUATE: Our results show tht in prolems concerning sutomic prticles such s electrons, mny quntities including ccelertion, speed, kinetic energy, nd time will hve very different vlues from those typicl of everydy ojects such s sells nd utomoiles. Test Your Understnding of ection 21.4 () A negtive point chrge moves long stright-line pth directly towrd sttionry positive point chrge. Which spect(s) of the electric force on the negtive point chrge will remin constnt s it moves? (i) mgnitude; (ii) direction; (iii) oth mgnitude nd direction; (iv) neither mgnitude nor direction. () A negtive point chrge moves long circulr orit round positive point chrge. Which spect(s) of the electric force on the negtive point chrge will remin constnt s it moves? (i) mgnitude; (ii) direction; (iii) oth mgnitude nd direction; (iv) neither mgnitude nor direction Electric-Field Clcultions Eqution (21.7) gives the electric field cused y single point chrge. But in most relistic situtions tht involve electric fields nd forces, we encounter chrge tht is distriuted over spce. The chrged plstic nd glss rods in Fig hve electric chrge distriuted over their surfces, s does the imging drum of lser printer (Fig. 21.2). In this section we ll lern to clculte electric fields cused y vrious distriutions of electric chrge. Clcultions of this kind re of tremendous importnce for technologicl pplictions of electric forces. To determine the trjectories of tomic nuclei in n ccelertor for cncer rdiotherpy or of chrged prticles in semiconductor electronic device, you hve to know the detiled nture of the electric field cting on the chrges. ActivPhysics 11.5: Electric Field Due to Dipole ActivPhysics 11.6: Electric Field: Prolems The uperposition of Electric Fields To find the field cused y chrge distriution, we imgine the distriution to e mde up of mny point chrges q 1, q 2, q 3,.... (This is ctully quite relistic description, since we hve seen tht chrge is crried y electrons nd protons tht re so smll s to e lmost pointlike.) At ny given point P, ech point chrge produces its own electric field E 1, E 2, E 3,..., so test chrge q 0 plced t P experiences force F from chrge force F 2 q 0 E 1 q 0 E 1 q 1, 2 from chrge q 2, nd so on. From the principle of superposition of forces discussed in ection 21.3, the totl force F 0 tht the chrge distriution exerts on q 0 is the vector sum of these individul forces: F 0 F 1 F 2 F 3 Á q 0 E 1 q 0 E 2 q 0 E 3 Á The comined effect of ll the chrges in the distriution is descried y the totl electric field E t point P. From the definition of electric field, Eq. (21.3), this is E F 0 q 0 E 1 E 2 E 3 Á

18 704 CHAPTER 21 Electric Chrge nd Electric Field Illustrting the principle of superposition of electric fields. q 1 Electric field t P due to q 2 E 2 q 2 E P E 1 Electric field t P due to q 1 The totl electric field E t point P is the vector sum of E 1 nd E 2. The totl electric field t P is the vector sum of the fields t P due to ech point chrge in the chrge distriution (Fig ). This is the principle of superposition of electric fields. When chrge is distriuted long line, over surfce, or through volume, few dditionl terms re useful. For line chrge distriution (such s long, thin, chrged plstic rod), we use l (the Greek letter lmd) to represent the liner chrge density (chrge per unit length, mesured in C>m). When chrge is distriuted over surfce (such s the surfce of the imging drum of lser printer), we use s (sigm) to represent the surfce chrge density (chrge per unit re, mesured in C>m 2 ). And when chrge is distriuted through volume, we use r (rho) to represent the volume chrge density (chrge per unit volume, C>m 3 ). ome of the clcultions in the following exmples my look firly intricte. After you ve worked through the exmples one step t time, the process will seem less formidle. We will use mny of the clcultionl techniques in these exmples in Chpter 28 to clculte the mgnetic fields cused y chrges in motion. Prolem-olving trtegy 21.2 Electric-Field Clcultions IDENTIFY the relevnt concepts: Use the principle of superposition to clculte the electric field due to discrete or continuous chrge distriution. ET UP the prolem using the following steps: 1. Mke drwing showing the loctions of the chrges nd your choice of coordinte xes. 2. On your drwing, indicte the position of the field point P (the point t which you wnt to clculte the electric field E ). EXECUTE the solution s follows: 1. Use consistent units. Distnces must e in meters nd chrge must e in couloms. If you re given centimeters or nnocouloms, don t forget to convert. 2. Distinguish etween the source point nd the field point P. The field produced y point chrge lwys points from to P if the chrge is positive, nd from P to if the chrge is negtive. 3. Use vector ddition when pplying the principle of superposition; review the tretment of vector ddition in Chpter 1 if necessry. 4. implify your clcultions y exploiting ny symmetries in the chrge distriution. 5. If the chrge distriution is continuous, define smll element of chrge tht cn e considered s point, find its electric field t P, nd find wy to dd the fields of ll the chrge elements y doing n integrl. Usully it is esiest to do this for ech component of E seprtely, so you my need to evlute more thn one integrl. Ensure tht the limits on your integrls re correct; especilly when the sitution hs symmetry, don t count chrge twice. EVALUATE your nswer: Check tht the direction of is resonle. If your result for the electric-field mgnitude E is function of position (sy, the coordinte x), check your result in ny limits for which you know wht the mgnitude should e. When possile, check your nswer y clculting it in different wy. E Exmple 21.8 Field of n electric dipole Point chrges q 1 12 nc nd q 2-12 nc re m prt (Fig ). (uch pirs of point chrges with equl mgnitude nd opposite sign re clled electric dipoles.) Compute the electric field cused y q 1, the field cused y q 2, nd the totl field () t point ; () t point ; nd (c) t point c. OLUTION IDENTIFY nd ET UP: We must find the totl electric field t vrious points due to two point chrges. We use the principle of superposition: E E 1 E 2. Figure shows the coordinte system nd the loctions of the field points,, nd c. EXECUTE: At ech field point, E depends on E nd E 1 2 there; we first clculte the mgnitudes E 1 nd E 2 t ech field point. At the mgnitude of the field cused y is E 1 q 1 1 ƒ q 1 ƒ E 1 = 4pP 0 r 2 = 3.0 * 10 4 N>C We clculte the other field mgnitudes in similr wy. The results re E 1 = 3.0 * 10 4 N>C E 1c = 6.39 * 10 3 N>C E 2 = 6.8 * 10 4 N>C = 19.0 * 10 9 N # m 2 >C * 10-9 C m2 2 E 2c = E 1c = 6.39 * 10 3 N>C The directions of the corresponding fields re in ll cses wy from the positive chrge nd towrd the negtive chrge. q 1 E 1 = 6.8 * 10 4 N>C E 2 = 0.55 * 10 4 N>C q 2

19 21.5 Electric-Field Clcultions Electric field t three points,,, nd c, set up y chrges nd q 2, which form n electric dipole. E q cm y c E 1 E 2 E c 13.0 cm 13.0 cm q 1 E q cm cm E 2 () At, nd re oth directed to the right, so E 1 E E 1 ın E 2 ın 19.8 * 10 4 N>C2ın E 2 () At, E 1 is directed to the left nd is directed to the right, so E -E 1 ın E 2 ın * 10 4 N>C2ın x (c) Figure shows the directions of E nd E 1 2 t c. Both vectors hve the sme x-component: E 1cx = E 2cx = E 1c cos = * 10 3 N>C2A 5 13 B = 2.46 * 10 3 N>C E 1y E 2y From symmetry, nd re equl nd opposite, so their sum is zero. Hence E c * 10 3 N>C2ın 14.9 * 10 3 N>C2ın EVALUATE: We cn lso find E c using Eq. (21.7) for the field of point chrge. The displcement vector r 1 from q 1 to point c is r 1 r cos ın r sin n. Hence the unit vector tht points from q to point c is rn 1 1 r1>r cos ın sin n. By symmetry, the unit vector tht points from q 2 to point c hs the opposite x- component ut the sme y-component: rn 2 -cos ın sin n. We cn now use Eq. (21.7) to write the fields E nd E 1c 2c t c in vector form, then find their sum. ince q 2 = -q 1 nd the distnce r to c is the sme for oth chrges, E c E 1c E 2c 1 q 1 4pP 0 r 2 rn 1 1 q 2 4pP 0 r 2 rn 2 1 4pP 0 r 2 1q 1rN 1 q 2 rn 2 2 q 1 4pP 0 r 2 1rN 1 rn q 1 12 cosın2 4pP 2 0 r * 10 9 N # m 2 >C * 10-9 C m2 2 A 5 13 B ın 14.9 * 10 3 N>C2ın This is the sme s we clculted in prt (c). Exmple 21.9 Field of ring of chrge Chrge Q is uniformly distriuted round conducting ring of rdius (Fig ). Find the electric field t point P on the ring xis t distnce x from its center. OLUTION IDENTIFY nd ET UP: This is prolem in the superposition of electric fields. Ech it of chrge round the ring produces n electric field t n ritrry point on the x-xis; our trget vrile is the totl field t this point due to ll such its of chrge Clculting the electric field on the xis of ring of chrge. In this figure, the chrge is ssumed to e positive. Q y O dq ds r x 2 2 x de y P de x de x EXECUTE: We divide the ring into infinitesiml segments ds s shown in Fig In terms of the liner chrge density l = Q>2p, the chrge in segment of length ds is dq = lds. Consider two identicl segments, one s shown in the figure t y = nd nother hlfwy round the ring t y = -. From Exmple 21.4, we see tht the net force they exert on point test chrge t P, nd thus their net field de df, re directed long the x-xis. The sme is true for ny such pir of segments round the ring, so the net field t P is long the x-xis: E E x ın. To clculte E x, note tht the squre of the distnce r from single ring segment to the point P is r 2 = x 2 2. Hence the mgnitude of this segment s contriution de to the electric field t P is 1 dq de = 4pP 0 x 2 2 The x-component of this field is de x = decos. We know dq lds nd Fig shows tht cos = x>r = x>1x >2, so 1 dq x de x = decos = 4pP 0 x 2 2 2x lx = 4pP 0 1x 2 2 ds 3>2 2 Continued

20 706 CHAPTER 21 Electric Chrge nd Electric Field To find E x we integrte this expression over the entire ring tht is, EVALUATE: Eqution (21.8) shows tht E 0 t the center of the for s from 0 to 2p (the circumference of the ring). The integrnd ring 1x 02. This mkes sense; chrges on opposite sides of the hs the sme vlue for ll points on the ring, so it cn e tken outside the integrl. Hence we get the vector sum of ech such pir of forces is zero. When the field ring push in opposite directions on test chrge t the center, nd 2p point P is much frther from the ring thn the ring s rdius, we 1 lx hve x W nd the denomintor in Eq. (21.8) ecomes pproximtely equl to x 3. In this limit the electric field t P E x = de x = L 4pP 0 1x >2 ds 3 is = 1 lx 4pP 0 1x >212p2 E E x ın 1 4pP 0 Qx 1x >2 ın 0 (21.8) E 1 4pP 0 Q x 2 ın Tht is, when the ring is so fr wy tht its rdius is negligile in comprison to the distnce x, its field is the sme s tht of point chrge. Exmple Field of chrged line segment Positive chrge Q is distriuted uniformly long the y-xis etween y = - nd y =. Find the electric field t point P on the x-xis t distnce x from the origin. OLUTION IDENTIFY nd ET UP: Figure shows the sitution. As in Exmple 21.9, we must find the electric field due to continuous distriution of chrge. Our trget vrile is n expression for the electric field t P s function of x. The x-xis is perpendiculr isector of the segment, so we cn use symmetry rgument. EXECUTE: We divide the line chrge of length 2 into infinitesiml segments of length dy. The liner chrge density is l = Q>2, nd the chrge in segment is dq = ldy = 1Q>22dy. The distnce r from segment t height y to the field point P is r = 1x 2 y 2 2 1>2, so the mgnitude of the field t P due to the segment t height y is Figure shows tht the x- nd y-components of this field re de x = de cos nd de y = -de sin, where cos xrnd > sin = yr. > Hence To find the totl field t P, we must sum the fields from ll segments long the line tht is, we must integrte from y = - to y =. You should work out the detils of the integrtion ( tle of integrls will help). The results re E x = E y = de = 1 Q 4pP 0 2 L 1 Q 4pP 0 2 L or, in vector form, 1 dq 4pP 0 r 2 = 1 Q 4pP 0 2 de x = de y = Q 4pP Q 4pP 0 2 xdy 1x 2 y 2 2 3>2 ydy 1x 2 y 2 2 3>2 xdy 1x 2 y 2 2 3>2 = ydy 1x 2 y 2 2 3>2 = 0 E 1 Q 4pP 0 x2x 2 ın 2 dy 1x 2 y 2 2 Q 4pP 0 1 x2x 2 2 (21.9) Our sketch for this prolem. points wy from the line of chrge if l is positive nd towrd the line of chrge if l is negtive. EVALUATE: Using symmetry rgument s in Exmple 21.9, we could hve guessed tht E y would e zero; if we plce positive test chrge t P, the upper hlf of the line of chrge pushes downwrd on it, nd the lower hlf pushes up with equl mgnitude. ymmetry lso tells us tht the upper nd lower hlves of the segment contriute eqully to the totl field t P. If the segment is very short (or the field point is very fr from the segment) so tht x W, we cn neglect in the denomintor of Eq. (21.9). Then the field ecomes tht of point chrge, just s in Exmple 21.9: E 1 Q 4pP 0 x 2 ın To see wht hppens if the segment is very long (or the field point is very close to it) so tht W x, we first rewrite Eq. (21.9) slightly: E 1 l (21.10) 2pP 0 x21x 2 > ın E In the limit W x we cn neglect x 2 > 2 in the denomintor of Eq. (21.10), so E l 2pP 0 x ın

21 21.5 Electric-Field Clcultions 707 This is the field of n infinitely long line of chrge. At ny point P t perpendiculr distnce r from the line in ny direction, E hs mgnitude l E = (infinite line of chrge) 2pP 0 r Note tht this field is proportionl to 1>r rther thn to 1>r 2 s for point chrge. There s relly no such thing in nture s n infinite line of chrge. But when the field point is close enough to the line, there s very little difference etween the result for n infinite line nd the rel-life finite cse. For exmple, if the distnce r of the field point from the center of the line is 1% of the length of the line, the vlue of E differs from the infinite-length vlue y less thn 0.02%. Exmple Field of uniformly chrged disk A nonconducting disk of rdius R hs uniform positive surfce To find the totl field due to ll the rings, we integrte de x over r chrge density s. Find the electric field t point long the xis of from r = 0 to r = R (not from R to R): the disk distnce x from its center. Assume tht x is positive. R 1 12psr dr2x R sx 2r dr E x = OLUTION L 0 4pP 0 1x 2 r 2 = 3>2 2 4P 0 L0 1x 2 r 2 2 3>2 IDENTIFY nd ET UP: Figure shows the sitution. We represent the chrge distriution s collection of concentric rings of x 2 r 2 (which yields dt = 2rdr); you cn work out the detils. You cn evlute this integrl y mking the sustitution t = chrge dq. In Exmple 21.9 we otined Eq. (21.8) for the field on The result is the xis of single uniformly chrged ring, so ll we need do here is integrte the contriutions of our rings. EXECUTE: A typicl ring hs chrge dq, inner rdius r, nd outer rdius r dr. Its re is pproximtely equl to its width dr times its circumference 2pr, or da = 2pr dr. The chrge per unit re is s = dq>da, so the chrge of the ring is dq = sda = 2psr dr. We use dq in plce of Q in Eq. (21.8), the expression for the field due to ring tht we found in Exmple 21.9, nd replce the ring rdius with r. Then the field component de x t point P due to this ring is 1 2psrx dr de x = 4pP 0 1x 2 r 2 2 3> Our sketch for this prolem. E x = sx 2P 0 c - = s 2P 0 c1-1 2x 2 R 2 1 x d 1 21R 2 >x d (21.11) EVALUATE: If the disk is very lrge (or if we re very close to it), so tht R W x, the term 1> 21R 2 >x in Eq. (21.11) is very much less thn 1. Then Eq. (21.11) ecomes E = s (21.12) 2P 0 Our finl result does not contin the distnce x from the plne. Hence the electric field produced y n infinite plne sheet of chrge is independent of the distnce from the sheet. The field direction is everywhere perpendiculr to the sheet, wy from it. There is no such thing s n infinite sheet of chrge, ut if the dimensions of the sheet re much lrger thn the distnce x of the field point P from the sheet, the field is very nerly given y Eq. (21.12). If P is to the left of the plne 1x 6 02, the result is the sme except tht the direction of E is to the left insted of the right. If the surfce chrge density is negtive, the directions of the fields on oth sides of the plne re towrd it rther thn wy from it. Exmple Field of two oppositely chrged infinite sheets Two infinite plne sheets with uniform surfce chrge densities s nd -s re plced prllel to ech other with seprtion d (Fig ). Find the electric field etween the sheets, ove the upper sheet, nd elow the lower sheet Finding the electric field due to two oppositely chrged infinite sheets. The sheets re seen edge-on; only portion of the infinite sheets cn e shown! y OLUTION IDENTIFY nd ET UP: Eqution (21.12) gives the electric field due to single infinite plne sheet of chrge. To find the field due to two such sheets, we comine the fields using the principle of superposition (Fig ). heet 2 heet 1 2s 1s d E 1 E 2 E E 1 E 2 0 E 1 E 2 E 1 E 2 E E 1 E 2 E E 1 E 2 0 x Continued

22 708 CHAPTER 21 Electric Chrge nd Electric Field EXECUTE: From Eq. (21.12), oth E nd E 1 2 hve the sme mgnitude t ll points, independent of distnce from either sheet: E 1 = E 2 = s 2P 0 From Exmple 21.11, E is everywhere directed wy from sheet 1, nd E 1 2 is everywhere directed towrd sheet 2. Between the sheets, E nd E 1 2 reinforce ech other; ove the upper sheet nd elow the lower sheet, they cncel ech other. Thus the totl field is 0 ove the upper sheet E E 1 E s 2 d n etween the sheets P 0 0 elow the lower sheet EVALUATE: Becuse we considered the sheets to e infinite, our result does not depend on the seprtion d. Our result shows tht the field etween oppositely chrged pltes is essentilly uniform if the plte seprtion is much smller thn the dimensions of the pltes. We ctully used this result in Exmple 21.7 (ection 21.4). CAUTION Electric fields re not flows You my hve thought tht the field E of sheet 1 would e unle to penetrte sheet 2, nd tht field E 1 2 cused y sheet 2 would e unle to penetrte sheet 1. You might conclude this if you think of the electric field s some kind of physicl sustnce tht flows into or out of chrges. But in fct there is no such sustnce, nd the electric fields E nd E 1 2 depend only on the individul chrge distriutions tht crete them. The totl field t every point is just the vector sum of nd E 2. E 1 Test Your Understnding of ection 21.5 uppose tht the line of chrge in Fig (Exmple 21.11) hd chrge Q distriuted uniformly etween y = 0 nd y = nd hd chrge -Q distriuted uniformly etween y = 0 nd y = -. In this sitution, the electric field t P would e (i) in the positive x-direction; (ii) in the negtive x-direction; (iii) in the positive y-direction; (iv) in the negtive y-direction; (v) zero; (vi) none of these Electric Field Lines The direction of the electric field t ny point is tngent to the field line through tht point. Electric field line Field t point P P E P PhET: Chrges nd Fields PhET: Electric Field of Drems PhET: Electric Field Hockey Field t point R E R R The concept of n electric field cn e little elusive ecuse you cn t see n electric field directly. Electric field lines cn e ig help for visulizing electric fields nd mking them seem more rel. An electric field line is n imginry line or curve drwn through region of spce so tht its tngent t ny point is in the direction of the electric-field vector t tht point. Figure shows the sic ide. (We used similr concept in our discussion of fluid flow in ection A stremline is line or curve whose tngent t ny point is in the direction of the velocity of the fluid t tht point. However, the similrity etween electric field lines nd fluid stremlines is mthemticl one only; there is nothing flowing in n electric field.) The English scientist Michel Frdy ( ) first introduced the concept of field lines. He clled them lines of force, ut the term field lines is preferle. Electric field lines show the direction of t ech point, nd their spcing gives generl ide of the mgnitude of E E t ech point. Where E is strong, we drw lines unched closely together; where E is weker, they re frther prt. At ny prticulr point, the electric field hs unique direction, so only one field line cn pss through ech point of the field. In other words, field lines never intersect. Figure shows some of the electric field lines in plne contining () single positive chrge; () two equl-mgnitude chrges, one positive nd one negtive ( dipole); nd (c) two equl positive chrges. Digrms such s these re sometimes clled field mps; they re cross sections of the ctul threedimensionl ptterns. The direction of the totl electric field t every point in ech digrm is long the tngent to the electric field line pssing through the point. Arrowheds indicte the direction of the E -field vector long ech field line. The ctul field vectors hve een drwn t severl points in ech pttern. Notice tht in generl, the mgnitude of the electric field is different t different points on given field line; field line is not curve of constnt electric-field mgnitude! Figure shows tht field lines re directed wy from positive chrges (since close to positive point chrge, E points wy from the chrge) nd

23 21.7 Electric Dipoles Electric field lines for three different chrge distriutions. In generl, the mgnitude of E is different t different points long given field line. () A single positive chrge () Two equl nd opposite chrges ( dipole) (c) Two equl positive chrges E E E E E E E E Field lines lwys point wy from (1) chrges nd towrd (2) chrges. At ech point in spce, the electric field vector is tngent to the field line pssing through tht point. Field lines re close together where the field is strong, frther prt where it is weker. towrd negtive chrges (since close to negtive point chrge, points towrd the chrge). In regions where the field mgnitude is lrge, such s etween the positive nd negtive chrges in Fig , the field lines re drwn close together. In regions where the field mgnitude is smll, such s etween the two positive chrges in Fig c, the lines re widely seprted. In uniform field, the field lines re stright, prllel, nd uniformly spced, s in Fig Figure is view from ove of demonstrtion setup for visulizing electric field lines. In the rrngement shown here, the tips of two positively chrged wires re inserted in continer of insulting liquid, nd some grss seeds re floted on the liquid. The grss seeds re electriclly neutrl insultors, ut the electric field of the two chrged wires cuses polriztion of the grss seeds; there is slight shifting of the positive nd negtive chrges within the molecules of ech seed, like tht shown in Fig The positively chrged end of ech grss seed is pulled in the direction of E nd the negtively chrged end is pulled opposite E. Hence the long xis of ech grss seed tends to orient prllel to the electric field, in the direction of the field line tht psses through the position of the seed (Fig ). CAUTION Electric field lines re not the sme s trjectories It s common misconception tht if chrged prticle of chrge q is in motion where there is n electric field, the prticle must move long n electric field line. Becuse E t ny point is tngent to the field line tht psses through tht point, it is indeed true tht the force F qe on the prticle, nd hence the prticle s ccelertion, re tngent to the field line. But we lerned in Chpter 3 tht when prticle moves on curved pth, its ccelertion cnnot e tngent to the pth. o in generl, the trjectory of chrged prticle is not the sme s field line. Test Your Understnding of ection 21.6 uppose the electric field lines in region of spce re stright lines. If chrged prticle is relesed from rest in tht region, will the trjectory of the prticle e long field line? 21.7 Electric Dipoles An electric dipole is pir of point chrges with equl mgnitude nd opposite sign ( positive chrge q nd negtive chrge -q) seprted y distnce d. We introduced electric dipoles in Exmple 21.8 (ection 21.5); the concept is worth exploring further ecuse mny physicl systems, from molecules to TV ntenns, cn e descried s electric dipoles. We will lso use this concept extensively in our discussion of dielectrics in Chpter 24. E () Electric field lines produced y two equl point chrges. The pttern is formed y grss seeds floting on liquid ove two chrged wires. Compre this pttern with Fig c. () The electric field cuses polriztion of the grss seeds, which in turn cuses the seeds to lign with the field. () () Field line E Grss seed

24 710 CHAPTER 21 Electric Chrge nd Electric Field () A wter molecule is n exmple of n electric dipole. () Ech test tue contins solution of different sustnce in wter. The lrge electric dipole moment of wter mkes it n excellent solvent. () A wter molecule, showing positive chrge s red nd negtive chrge s lue H O () Vrious sustnces dissolved in wter H The electric dipole moment p is directed from the negtive end to the positive end of the molecule The net force on this electric dipole is zero, ut there is torque directed into the pge tht tends to rotte the dipole clockwise. E F 2 5 2qE p d 2q p F 1 5 qe 1q d sin f f Figure shows molecule of wter 1H 2 O2, which in mny wys ehves like n electric dipole. The wter molecule s whole is electriclly neutrl, ut the chemicl onds within the molecule cuse displcement of chrge; the result is net negtive chrge on the oxygen end of the molecule nd net positive chrge on the hydrogen end, forming n electric dipole. The effect is equivlent to shifting one electron only out 4 * m (out the rdius of hydrogen tom), ut the consequences of this shift re profound. Wter is n excellent solvent for ionic sustnces such s tle slt (sodium chloride, NCl) precisely ecuse the wter molecule is n electric dipole (Fig ). When dissolved in wter, slt dissocites into positive sodium ion 1N 2 nd negtive chlorine ion 1Cl - 2, which tend to e ttrcted to the negtive nd positive ends, respectively, of wter molecules; this holds the ions in solution. If wter molecules were not electric dipoles, wter would e poor solvent, nd lmost ll of the chemistry tht occurs in queous solutions would e impossile. This includes ll of the iochemicl rections tht occur in ll of the life on erth. In very rel sense, your existence s living eing depends on electric dipoles! We exmine two questions out electric dipoles. First, wht forces nd torques does n electric dipole experience when plced in n externl electric field (tht is, field set up y chrges outside the dipole)? econd, wht electric field does n electric dipole itself produce? Force nd Torque on n Electric Dipole To strt with the first question, let s plce n electric dipole in uniform externl electric field E, s shown in Fig The forces F nd F - on the two chrges oth hve mgnitude qe, ut their directions re opposite, nd they dd to zero. The net force on n electric dipole in uniform externl electric field is zero. However, the two forces don t ct long the sme line, so their torques don t dd to zero. We clculte torques with respect to the center of the dipole. Let the ngle etween the electric field E nd the dipole xis e f; then the lever rm for oth F nd F is The torque of F nd the torque of F - 1d>22sin f. - oth hve the sme mgnitude of 1qE21d>22 sin f, nd oth torques tend to rotte the dipole clockwise (tht is, T is directed into the pge in Fig ). Hence the mgnitude of the net torque is twice the mgnitude of either individul torque: t = 1qE21d sin f2? (21.13) where d sin f is the perpendiculr distnce etween the lines of ction of the two forces. The product of the chrge q nd the seprtion d is the mgnitude of quntity clled the electric dipole moment, denoted y p: p = qd (mgnitude of electric dipole moment) (21.14) The units of p re chrge times distnce 1C # m2. For exmple, the mgnitude of the electric dipole moment of wter molecule is p = 6.13 * C # m. CAUTION The symol p hs multiple menings Be creful not to confuse dipole moment with momentum or pressure. There ren t s mny letters in the lphet s there re physicl quntities, so some letters re used severl times. The context usully mkes it cler wht we men, ut e creful. We further define the electric dipole moment to e vector quntity p. The mgnitude of p is given y Eq. (21.14), nd its direction is long the dipole xis from the negtive chrge to the positive chrge s shown in Fig In terms of p, Eq. (21.13) for the mgnitude t of the torque exerted y the field ecomes

25 21.7 Electric Dipoles 711 t = pe sin f (mgnitude of the torque on n electric dipole) (21.15) ince the ngle f in Fig is the ngle etween the directions of the vectors p nd E, this is reminiscent of the expression for the mgnitude of the vector product discussed in ection (You my wnt to review tht discussion.) Hence we cn write the torque on the dipole in vector form s T p : E (torque on n electric dipole, in vector form) (21.16) You cn use the right-hnd rule for the vector product to verify tht in the sitution shown in Fig , T is directed into the pge. The torque is gretest when p nd E re perpendiculr nd is zero when they re prllel or ntiprllel. The torque lwys tends to turn p to line it up with E. The position f = 0, with p prllel to E, is position of stle equilirium, nd the position f = p, with p nd E ntiprllel, is position of unstle equilirium. The polriztion of grss seed in the pprtus of Fig gives it n electric dipole moment; the torque exerted y E then cuses the seed to lign with E nd hence with the field lines. Potentil Energy of n Electric Dipole When dipole chnges direction in n electric field, the electric-field torque does work on it, with corresponding chnge in potentil energy. The work dw done y torque t during n infinitesiml displcement df is given y Eq. (10.19): dw = t df. Becuse the torque is in the direction of decresing f, we must write the torque s t = -pe sin f, nd dw = t df = -pesin fdf In finite displcement from to the totl work done on the dipole is f 1 f 2 PhET: Microwves f 2 W = 1-pEsin f2 df L = pecos f 2 - pecos f 1 The work is the negtive of the chnge of potentil energy, just s in Chpter 7: W = U 1 - U 2. o suitle definition of potentil energy U for this system is U1f2 = -pecos f In this expression we recognize the sclr product p # E = pecos f, lso write f 1 (21.17) so we cn U = -p # E (potentil energy for dipole in n electric field) (21.18) The potentil energy hs its minimum (most negtive) vlue U = -pe t the stle equilirium position, where f = 0 nd p is prllel to E. The potentil energy is mximum when f = p nd p is ntiprllel to E ; then U = pe. At f = p>2, where p is perpendiculr to E, U is zero. We could define U differently so tht it is zero t some other orienttion of p, ut our definition is simplest. Eqution (21.18) gives us nother wy to look t the effect shown in Fig The electric field E gives ech grss seed n electric dipole moment, nd the grss seed then ligns itself with E to minimize the potentil energy. Exmple Force nd torque on n electric dipole Figure shows n electric dipole in uniform electric field of mgnitude 5.0 * 10 5 N>C tht is directed prllel to the plne of the figure. The chrges re 1.6 * C; oth lie in the plne nd re seprted y nm = * 10-9 m. Find () the net force exerted y the field on the dipole; () the mgnitude nd Continued

26 712 CHAPTER 21 Electric Chrge nd Electric Field () An electric dipole. () Directions of the electric dipole moment, electric field, nd torque ( T points out of the pge). () E 1q q direction of the electric dipole moment; (c) the mgnitude nd direction of the torque; (d) the potentil energy of the system in the position shown. OLUTION IDENTIFY nd ET UP: This prolem uses the ides of this section out n electric dipole plced in n electric field. We use the reltionship F qe for ech point chrge to find the force on the dipole s whole. Eqution (21.14) gives the dipole moment, Eq. (21.16) gives the torque on the dipole, nd Eq. (21.18) gives the potentil energy of the system. () p t E 145 EXECUTE: () The field is uniform, so the forces on the two chrges re equl nd opposite. Hence the totl force on the dipole is zero. () The mgnitude p of the electric dipole moment p is p = qd = 11.6 * C * 10-9 m2 = 2.0 * C # m The direction of p is from the negtive to the positive chrge, 145 clockwise from the electric-field direction (Fig ). (c) The mgnitude of the torque is t = pesin f = 12.0 * C215.0 * 10 5 N>C21sin145 2 = 5.7 * N # m From the right-hnd rule for vector products (see ection 1.10), the direction of the torque T p : E is out of the pge. This corresponds to counterclockwise torque tht tends to lign p with E. (d) The potentil energy U = -pecos f = * C # m215.0 * 10 5 N>C21cos = 8.2 * J EVALUATE: The chrge mgnitude, the distnce etween the chrges, the dipole moment, nd the potentil energy re ll very smll, ut re ll typicl of molecules. In this discussion we hve ssumed tht is uniform, so there is no net force on the dipole. If E is not uniform, the forces t the ends my not cncel completely, nd the net force my not e zero. Thus ody with zero net chrge ut n electric dipole moment cn experience net force in nonuniform electric field. As we mentioned in ection 21.1, n unchrged ody cn e polrized y n electric field, giving rise to seprtion of chrge nd n electric dipole moment. This is how unchrged odies cn experience electrosttic forces (see Fig. 21.8). Field of n Electric Dipole Now let s think of n electric dipole s source of electric field. Wht does the field look like? The generl shpe of things is shown y the field mp of Fig At ech point in the pttern the totl E field is the vector sum of the fields from the two individul chrges, s in Exmple 21.8 (ection 21.5). Try drwing digrms showing this vector sum for severl points. To get quntittive informtion out the field of n electric dipole, we hve to do some clculting, s illustrted in the next exmple. Notice the use of the principle of superposition of electric fields to dd up the contriutions to the field of the individul chrges. Also notice tht we need to use pproximtion techniques even for the reltively simple cse of field due to two chrges. Field clcultions often ecome very complicted, nd computer nlysis is typiclly used to determine the field due to n ritrry chrge distriution. E Exmple Field of n electric dipole, revisited An electric dipole is centered t the origin, with in the direction of the y-xis (Fig ). Derive n pproximte expression for the electric field t point P on the y-xis for which y is much lrger thn d. To do this, use the inomil expnsion 11 x2 n 1 nx n1n - 12x 2 >2 Á (vlid for the cse ƒ x ƒ 6 1). p OLUTION IDENTIFY nd ET UP: We use the principle of superposition: The totl electric field is the vector sum of the field produced y the positive chrge nd the field produced y the negtive chrge. At the field point P shown in Fig , the field E of the positive chrge hs positive (upwrd) y-component nd the field of E -

27 21.7 Electric Dipoles Finding the electric field of n electric dipole t point on its xis. y 1 d/2 the negtive chrge hs negtive (downwrd) y-component. We dd these components to find the totl field nd then pply the pproximtion tht y is much greter thn d. EXECUTE: The totl y-component E y of electric field from the two chrges is q 1 E y = c 4pP 0 1y - d> y d>22 2 d = y 2 d/2 y E 1 P E 2 1q d p O x 2q q 4pP 0 y 2 c1 - d -2 2y - 1 d -2 2y d We used this sme pproch in Exmple 21.8 (ection 21.5). Now the pproximtion: When we re fr from the dipole compred to its size, so y W d, we hve d>2y V 1. With n = -2 nd with d>2y replcing x in the inomil expnsion, we keep only the first two terms (the terms we discrd re much smller). We then hve Hence 1 - d -2 2y 1 d y nd 1 d -2 2y 1 - d y E y is given pproximtely y E y q 4pP 0 y 2 c1 d y d y d = qd 2pP 0 y 3 = p 2pP 0 y 3 EVALUATE: An lterntive route to this result is to put the frctions in the first expression for E y over common denomintor, dd, nd then pproximte the denomintor 1y - d>22 2 1y d>22 2 s y 4. We leve the detils to you (see Exercise 21.60). For points P off the coordinte xes, the expressions re more complicted, ut t ll points fr wy from the dipole (in ny direction) the field drops off s 1>r 3. We cn compre this with the 1>r 2 ehvior of point chrge, the 1>r ehvior of long line chrge, nd the independence of r for lrge sheet of chrge. There re chrge distriutions for which the field drops off even more quickly. At lrge distnces, the field of n electric qudrupole, which consists of two equl dipoles with opposite orienttion, seprted y smll distnce, drops off s 1>r 4. Test Your Understnding of ection 21.7 An electric dipole is plced in region of uniform electric field E, with the electric dipole moment p, pointing in the direction opposite to E. Is the dipole (i) in stle equilirium, (ii) in unstle equilirium, or (iii) neither? (Hint: You mny wnt to review ection 7.5.)

28 CHAPTER 21 UMMARY Electric chrge, conductors, nd insultors: The fundmentl quntity in electrosttics is electric chrge. There re two kinds of chrge, positive nd negtive. Chrges of the sme sign repel ech other; chrges of opposite sign ttrct. Chrge is conserved; the totl chrge in n isolted system is constnt. All ordinry mtter is mde of protons, neutrons, nd electrons. The positive protons nd electriclly neutrl neutrons in the nucleus of n tom re ound together y the nucler force; the negtive electrons surround the nucleus t distnces much greter thn the nucler size. Electric interctions re chiefly responsile for the structure of toms, molecules, nd solids. Conductors re mterils in which chrge moves esily; in insultors, chrge does not move esily. Most metls re good conductors; most nonmetls re insultors. Coulom s lw: For chrges q 1 nd q 2 seprted y distnce r, the mgnitude of the electric force on either F = (21.2) 1 ƒq 1 q 2 ƒ 4pP 0 r 2 chrge is proportionl to the product q 1 q 2 nd inversely proportionl to r 2. The force on ech chrge is long the 1 line joining the two chrges repulsive if q 1 nd q 2 hve = * 10 9 N # m 2 >C 2 4pP the sme sign, ttrctive if they hve opposite signs. In I 0 units the unit of electric chrge is the coulom, revited C. (ee Exmples 21.1 nd 21.2.) When two or more chrges ech exert force on chrge, the totl force on tht chrge is the vector sum of the forces exerted y the individul chrges. (ee Exmples 21.3 nd 21.4.) F 2 on 1 q 1 r r q 1 F 2 on 1 F 1 on 2 q 2 q 2 F 1 on 2 Electric field: Electric field E, vector quntity, is the force per unit chrge exerted on test chrge t ny point. The electric field produced y point chrge is directed rdilly wy from or towrd the chrge. (ee Exmples ) E F 0 q 0 E 1 4pP 0 q r 2 rn (21.3) (21.7) q E E uperposition of electric fields: The electric field of ny comintion of chrges is the vector sum of the fields cused y the individul chrges. To clculte the electric field cused y continuous distriution of chrge, divide the distriution into smll elements, clculte the field cused y ech element, nd then crry out the vector sum, usully y integrting. Chrge distriutions re descried y liner chrge density l, surfce chrge density s, nd volume chrge density r. (ee Exmples ) Q y O dq ds r x 2 2 P x dey dex x de r Electric field lines: Field lines provide grphicl representtion of electric fields. At ny point on field line, the tngent to the line is in the direction of E t tht point. The numer of lines per unit re (perpendiculr to their direction) is proportionl to the mgnitude of E t the point. E E E Electric dipoles: An electric dipole is pir of electric chrges of equl mgnitude q ut opposite sign, seprted y distnce d. The electric dipole moment p hs mgnitude p = qd. The direction of p is from negtive towrd positive chrge. An electric dipole in n electric field E experiences torque T equl to the vector product of p nd E. The mgnitude of the torque depends on the ngle f etween p nd E. The potentil energy U for n electric dipole in n electric field lso depends on the reltive orienttion of p nd E. (ee Exmples nd ) t = pe sin f T p : E U = -p # E (21.15) (21.16) (21.18) E F 2 5 2qE p d 2q F 1 5 qe 1q d sin f f 714

29 Discussion Questions 715 BRIDGING PROBLEM Clculting Electric Field: Hlf Ring of Chrge Positive chrge Q is uniformly distriuted round semicircle of rdius s shown in Fig Find the mgnitude nd direction of the resulting electric field t point P, the center of curvture of the semicircle. OLUTION GUIDE ee MsteringPhysics study re for Video Tutor solution. IDENTIFY nd ET UP 1. The trget vriles re the components of the electric field t P. 2. Divide the semicircle into infinitesiml segments, ech of which is short circulr rc of rdius nd ngle d u. Wht is the length of such segment? How much chrge is on segment? 3. Consider n infinitesiml segment locted t n ngulr position u on the semicircle, mesured from the lower right corner of the semicircle t x, y 0. (Thus u p> 2 t x 0, y nd u = p t x = -, y = 0.) Wht re the x- nd y-components of the electric field t P ( de x nd de y ) produced y just this segment? EXECUTE 4. Integrte your expressions for de x nd de y from u 0 to u p. The results will e the x-component nd y-component of the electric field t P. 5. Use your results from step 4 to find the mgnitude nd direction of the field t P P y Q x EVALUATE 6. Does your result for the electric-field mgnitude hve the correct units? 7. Explin how you could hve found the x-component of the electric field using symmetry rgument. 8. Wht would e the electric field t P if the semicircle were extended to full circle centered t P? Prolems For instructor-ssigned homework, go to : Prolems of incresing difficulty. CP: Cumultive prolems incorporting mteril from erlier chpters. CALC: Prolems requiring clculus. BIO: Biosciences prolems. DICUION QUETION Q21.1 If you peel two strips of trnsprent tpe off the sme roll nd immeditely let them hng ner ech other, they will repel ech other. If you then stick the sticky side of one to the shiny side of the other nd rip them prt, they will ttrct ech other. Give plusile explntion, involving trnsfer of electrons etween the strips of tpe, for this sequence of events. Q21.2 Two metl spheres re hnging from nylon threds. When you ring the spheres close to ech other, they tend to ttrct. Bsed on this informtion lone, discuss ll the possile wys tht the spheres could e chrged. Is it possile tht fter the spheres touch, they will cling together? Explin. Q21.3 The electric force etween two chrged prticles ecomes weker with incresing distnce. uppose insted tht the electric force were independent of distnce. In this cse, would chrged com still cuse neutrl insultor to ecome polrized s in Fig. 21.8? Why or why not? Would the neutrl insultor still e ttrcted to the com? Agin, why or why not? Q21.4 Your clothing tends to cling together fter going through the dryer. Why? Would you expect more or less clinging if ll your clothing were mde of the sme mteril (sy, cotton) thn if you dried different kinds of clothing together? Agin, why? (You my wnt to experiment with your next lod of lundry.) Q21.5 An unchrged metl sphere hngs from nylon thred. When positively chrged glss rod is rought close to the metl sphere, the sphere is drwn towrd the rod. But if the sphere touches the rod, it suddenly flies wy from the rod. Explin why the sphere is first ttrcted nd then repelled. Q21.6 The free electrons in metl re grvittionlly ttrcted towrd the erth. Why, then, don t they ll settle to the ottom of the conductor, like sediment settling to the ottom of river? Q21.7. Figure Q21.7 shows some of the electric field lines due to three point chrges rrnged long the verticl xis. All three chrges hve the sme mgnitude. () Wht re the signs of the three chrges? Explin your resoning. () At wht point(s) is the mgnitude of the electric field the smllest? Explin your resoning. Explin how the fields produced y ech individul point chrge comine to give smll net field t this point or points. Q21.8 Good electricl conductors, such s metls, re typiclly good conductors of Figure Q21.7 het; electricl insultors, such s wood, re typiclly poor conductors of het. Explin why there should e reltionship etween electricl conduction nd het conduction in these mterils.

30 716 CHAPTER 21 Electric Chrge nd Electric Field Q21.9. uppose the chrge shown in Fig is fixed in position. A smll, positively chrged prticle is then plced t some point in the figure nd relesed. Will the trjectory of the prticle follow n electric field line? Why or why not? uppose insted tht the prticle is plced t some point in Fig nd relesed (the positive nd negtive chrges shown in the figure re fixed in position). Will its trjectory follow n electric field line? Agin, why or why not? Explin ny differences etween your nswers for the two different situtions. Q21.10 Two identicl metl ojects re mounted on insulting stnds. Descrie how you could plce chrges of opposite sign ut exctly equl mgnitude on the two ojects. Q21.11 You cn use plstic food wrp to cover continer y stretching the mteril cross the top nd pressing the overhnging mteril ginst the sides. Wht mkes it stick? (Hint: The nswer involves the electric force.) Does the food wrp stick to itself with equl tencity? Why or why not? Does it work with metllic continers? Agin, why or why not? Q21.12 If you wlk cross nylon rug nd then touch lrge metl oject such s doorkno, you my get sprk nd shock. Why does this tend to hppen more on dry dys thn on humid dys? (Hint: ee Fig ) Why re you less likely to get shock if you touch smll metl oject, such s pper clip? Q21.13 You hve negtively chrged oject. How cn you use it to plce net negtive chrge on n insulted metl sphere? To plce net positive chrge on the sphere? Q21.14 When two point chrges of equl mss nd chrge re relesed on frictionless tle, ech hs n initil ccelertion 0. If insted you keep one fixed nd relese the other one, wht will e its initil ccelertion: 0,2 0, or 0 >2? Explin. Q21.15 A point chrge of mss m nd chrge Q nd nother point chrge of mss m ut chrge 2Q re relesed on frictionless tle. If the chrge Q hs n initil ccelertion 0, wht will e the ccelertion of 2Q: 0,2 0,4 0, 0 >2, or 0 >4? Explin. Q21.16 A proton is plced in uniform electric field nd then relesed. Then n electron is plced t this sme point nd relesed. Do these two prticles experience the sme force? The sme ccelertion? Do they move in the sme direction when relesed? Q21.17 In Exmple 21.1 (ection 21.3) we sw tht the electric force etween two prticles is of the order of times s strong s the grvittionl force. o why do we redily feel the grvity of the erth ut no electricl force from it? Q21.18 Wht similrities do electricl forces hve with grvittionl forces? Wht re the most significnt differences? Q21.19 Two irregulr ojects A nd B crry chrges of opposite sign. Figure Q21.19 shows the electric field lines ner ech of these ojects. () Which oject is positive, A or B? How do you know? () Where is the electric field stronger, close to A or close to B? How do you know? Q21.20 Atomic nuclei re mde of protons nd neutrons. This shows tht there must e nother kind of interction in Figure Q ddition to grvittionl nd electric forces. Explin. Q21.21 ufficiently strong electric fields cn cuse toms to ecome positively ionized tht is, to lose one or more electrons. Explin how this cn hppen. Wht determines how strong the field must e to mke this hppen? A E B Q21.22 The electric fields t point P due to Figure Q21.22 the positive chrges q 1 nd q 2 re shown in Fig. Q Does the fct tht they cross E 2 E 1 ech other violte the sttement in ection 21.6 tht electric field lines never cross? Explin. P Q21.23 The ir temperture nd the velocity of the ir hve different vlues t different plces in the erth s tmosphere. Is the q 1 ir velocity vector field? Why or why not? Is the ir temperture vector field? Agin, why or why not? EXERCIE ection 21.3 Coulom s Lw Excess electrons re plced on smll led sphere with mss 8.00 g so tht its net chrge is * 10-9 C. () Find the numer of excess electrons on the sphere. () How mny excess electrons re there per led tom? The tomic numer of led is 82, nd its tomic mss is 207 g>mol Lightning occurs when there is flow of electric chrge (principlly electrons) etween the ground nd thundercloud. The mximum rte of chrge flow in lightning olt is out 20,000 C>s; this lsts for 100 ms or less. How much chrge flows etween the ground nd the cloud in this time? How mny electrons flow during this time? BIO Estimte how mny electrons there re in your ody. Mke ny ssumptions you feel re necessry, ut clerly stte wht they re. (Hint: Most of the toms in your ody hve equl numers of electrons, protons, nd neutrons.) Wht is the comined chrge of ll these electrons? Prticles in Gold Ring. You hve pure (24-krt) gold ring with mss 17.7 g. Gold hs n tomic mss of 197 g>mol nd n tomic numer of 79. () How mny protons re in the ring, nd wht is their totl positive chrge? () If the ring crries no net chrge, how mny electrons re in it? BIO ignl Propgtion in Neurons. Neurons re components of the nervous system of the ody tht trnsmit signls s electricl impulses trvel long their length. These impulses propgte when chrge suddenly rushes into nd then out of prt of the neuron clled n xon. Mesurements hve shown tht, during the inflow prt of this cycle, pproximtely 5.6 * N (sodium ions) per meter, ech with chrge e, enter the xon. How mny couloms of chrge enter 1.5-cm length of the xon during this process? Two smll spheres spced 20.0 cm prt hve equl chrge. How mny excess electrons must e present on ech sphere if the mgnitude of the force of repulsion etween them is 4.57 * N? An verge humn weighs out 650 N. If two such generic humns ech crried 1.0 coulom of excess chrge, one positive nd one negtive, how fr prt would they hve to e for the electric ttrction etween them to equl their 650-N weight? Two smll luminum spheres, ech hving mss kg, re seprted y 80.0 cm. () How mny electrons does ech sphere contin? (The tomic mss of luminum is g>mol, nd its tomic numer is 13.) () How mny electrons would hve to e removed from one sphere nd dded to the other to cuse n ttrctive force etween the spheres of mgnitude 1.00 * 10 4 N (roughly 1 ton)? Assume tht the spheres my e treted s point chrges. (c) Wht frction of ll the electrons in ech sphere does this represent? q 2

31 Exercises Two smll plstic spheres re given positive electricl chrges. When they re 15.0 cm prt, the repulsive force etween them hs mgnitude N. Wht is the chrge on ech sphere () if the two chrges re equl nd () if one sphere hs four times the chrge of the other? Wht If We Were Not Neutrl? A 75-kg person holds out his rms so tht his hnds re 1.7 m prt. Typiclly, person s hnd mkes up out 1.0% of his or her ody weight. For round numers, we shll ssume tht ll the weight of ech hnd is due to the clcium in the ones, nd we shll tret the hnds s point chrges. One mole of C contins g, nd ech tom hs 20 protons nd 20 electrons. uppose tht only 1.0% of the positive chrges in ech hnd were unlnced y negtive chrge. () How mny C toms does ech hnd contin? () How mny couloms of unlnced chrge does ech hnd contin? (c) Wht force would the person s rms hve to exert on his hnds to prevent them from flying off? Does it seem likely tht his rms re cple of exerting such force? Two very smll 8.55-g spheres, 15.0 cm prt from center to center, re chrged y dding equl numers of electrons to ech of them. Disregrding ll other forces, how mny electrons would you hve to dd to ech sphere so tht the two spheres will ccelerte t 25.0g when relesed? Which wy will they ccelerte? Just How trong Is the Electric Force? uppose you hd two smll oxes, ech contining 1.0 g of protons. () If one were plced on the moon y n stronut nd the other were left on the erth, nd if they were connected y very light (nd very long!) string, wht would e the tension in the string? Express your nswer in newtons nd in pounds. Do you need to tke into ccount the grvittionl forces of the erth nd moon on the protons? Why? () Wht grvittionl force would ech ox of protons exert on the other ox? In n experiment in spce, one proton is held fixed nd nother proton is relesed from rest distnce of 2.50 mm wy. () Wht is the initil ccelertion of the proton fter it is relesed? () ketch qulittive (no numers!) ccelertiontime nd velocitytime grphs of the relesed proton s motion A negtive chrge of mc exerts n upwrd N force on n unknown chrge m directly elow it. () Wht is the unknown chrge (mgnitude nd sign)? () Wht re the mgnitude nd direction of the force tht the unknown chrge exerts on the mC chrge? Three point chrges re rrnged on line. Chrge q 3 = 5.00 nc nd is t the origin. Chrge q 2 = nc nd is t x = 4.00 cm. Chrge q 1 is t x = 2.00 cm. Wht is q 1 (mgnitude nd sign) if the net force on q 3 is zero? In Exmple 21.4, suppose the point chrge on the y -xis t y = m hs negtive chrge -2.0 mc, nd the other chrges remin the sme. Find the mgnitude nd direction of the net force on Q. How does your nswer differ from tht in Exmple 21.4? Explin the differences In Exmple 21.3, clculte the net force on chrge q In Exmple 21.4, wht is the net force (mgnitude nd direction) on chrge q 1 exerted y the other two chrges? Three point chrges re rrnged long the x -xis. Chrge q 1 = 3.00 mc is t the origin, nd chrge q 2 = mc is t x = m. Chrge q 3 = mc. Where is q 3 locted if the net force on q 1 is 7.00 N in the - x-direction? Repet Exercise for q 3 = 8.00 mc Two point chrges re locted on the y -xis s follows: chrge q 1 = nc t y = m, nd chrge q 2 = 3.20 nc t the origin 1y = 02. Wht is the totl force (mgnitude nd direction) exerted y these two chrges on third chrge q 3 = 5.00 nc locted t y = m? Two point chrges re plced on the x -xis s follows: Chrge q 1 = 4.00 nc is locted t x = m, nd chrge q 2 = 5.00 nc is t x = m. Wht re the mgnitude nd direction of the totl force exerted y these two chrges on negtive point chrge q 3 = nc tht is plced t the origin? BIO Bse Piring in DNA, I. The two sides of the DNA doule helix re connected y pirs of ses (denine, thymine, cytosine, nd gunine). Becuse of the geometric shpe of these molecules, denine onds with thymine nd cytosine onds with gunine. Figure E21.23 shows the thyminedenine ond. Ech chrge shown is e, nd the H N distnce is nm. () Clculte the net force tht thymine exerts on denine. Is it ttrctive or repulsive? To keep the clcultions firly simple, yet resonle, consider only the forces due to the O H N nd the N H N comintions, ssuming tht these two comintions re prllel to ech other. Rememer, however, tht in the O H N set, the O - exerts force on oth the H nd the N -, nd likewise long the N H N set. () Clculte the force on the electron in the hydrogen tom, which is nm from the proton. Then compre the strength of the onding force of the electron in hydrogen with the onding force of the deninethymine molecules. Figure E21.23 Thymine H H C H C N C H C 1 C O N 2 O 2 H nm H nm H Adenine N 2 N C C C N 2 C BIO Bse Piring in DNA, II. Refer to Exercise Figure E21.24 shows the onding of the cytosine nd gunine molecules. The O H nd H N distnces re ech nm. In this cse, ssume tht the onding is due only to the forces long the O H O, N H N, nd O H N comintions, nd ssume lso tht these three comintions re prllel to ech other. Clculte the net force tht cytosine exerts on gunine due to the preceding three comintions. Is this force ttrctive or repulsive? Figure E21.24 Cytosine H C H N C C 1 H C N 2 O 2 O 2 H 1 ection 21.4 Electric Field nd Electric Forces CP A proton is plced in uniform electric field of 2.75 * 10 3 N>C. Clculte: () the mgnitude of the electric force felt y the proton; () the proton s ccelertion; (c) the proton s speed fter 1.00 ms in the field, ssuming it strts from rest. H nm nm nm H 1 H 1 N N 2 C N 2 H C O 2 1 C C N N C H Gunine N C N H

32 718 CHAPTER 21 Electric Chrge nd Electric Field A prticle hs chrge nc. () Find the mgnitude midwy etween the pltes. () If the electron just misses the upper nd direction of the electric field due to this prticle t point plte s it emerges from the field, find the mgnitude of the electric m directly ove it. () At wht distnce from this prticle field. () uppose tht in Fig. E21.33 the electron is replced y does its electric field hve mgnitude of 12.0 N>C? proton with the sme initil speed v 0. Would the proton hit one of CP A proton is trveling horizontlly to the right t the pltes? If the proton would not hit one of the pltes, wht 4.50 * 10 6 m>s. () Find the mgnitude nd direction of the would e the mgnitude nd direction of its verticl displcement wekest electric field tht cn ring the proton uniformly to rest s it exits the region etween the pltes? (c) Compre the pths over distnce of 3.20 cm. () How much time does it tke the trveled y the electron nd the proton nd explin the differences. proton to stop fter entering the field? (c) Wht minimum field (d) Discuss whether it is resonle to ignore the effects of grvity (mgnitude nd direction) would e needed to stop n electron for ech prticle. under the conditions of prt ()? Point chrge q 1 = nc is t the origin nd point CP An electron is relesed from rest in uniform electric chrge q 2 = 3.00 nc is on the x-xis t x = 3.00 cm. Point P is field. The electron ccelertes verticlly upwrd, trveling 4.50 m on the y-xis t y = 4.00 cm. () Clculte the electric fields E in the first 3.00 ms fter it is relesed. () Wht re the mgnitude nd E 1 2 t point P due to the chrges q 1 nd q 2. Express your results nd direction of the electric field? () Are we justified in ignoring in terms of unit vectors (see Exmple 21.6). () Use the results of the effects of grvity? Justify your nswer quntittively. prt () to otin the resultnt field t P, expressed in unit vector () Wht must the chrge (sign nd mgnitude) of form g prticle e for it to remin sttionry when plced in CP In Exercise 21.33, wht is the speed of the electron downwrd-directed electric field of mgnitude 650 N>C? () Wht s it emerges from the field? is the mgnitude of n electric field in which the electric force on () Clculte the mgnitude nd direction (reltive to the proton is equl in mgnitude to its weight? x-xis) of the electric field in Exmple () A -2.5-nC point A point chrge is plced t chrge is plced t point P in Fig Find the mgnitude nd Figure E21.30 ech corner of squre with side direction of (i) the force tht the -8.0-nC chrge t the origin 1q length. The chrges ll hve the 1q exerts on this chrge nd (ii) the force tht this chrge exerts on the sme mgnitude q. Two of the -8.0-nC chrge t the origin. chrges re positive nd two re negtive, s shown in Fig. E Wht m from proton, s shown in Fig If two electrons re ech 1.50 * Figure E21.37 e is the direction of the net electric E21.37, find the mgnitude nd direction field t the center of the squre due to of the net electric force they will exert on the four chrges, nd wht is its mgnitude in terms of q nd? CP A uniform electric field 65.0 the proton q 2q Two point chrges re seprted y 25.0 cm (Fig. E21.31). Find the net electric field these sitely chrged plne prllel pltes. A pro- exists in the region etween two oppo- p chrges produce t () point A nd () point B. (c) Wht would e ton is relesed from rest t the surfce of the positively chrged the mgnitude nd direction of the electric force this comintion plte nd strikes the surfce of the opposite plte, 1.60 cm distnt of chrges would produce on proton t A? Figure E21.31 B 10.0 cm nc A nc 25.0 cm 10.0 cm Electric Field of the Erth. The erth hs net electric chrge tht cuses field t points ner its surfce equl to 150 N>C nd directed in towrd the center of the erth. () Wht mgnitude nd sign of chrge would 60-kg humn hve to cquire to overcome his or her weight y the force exerted y the erth s electric field? () Wht would e the force of repulsion etween two people ech with the chrge clculted in prt () nd seprted y distnce of 100 m? Is use of the erth s electric field fesile mens of flight? Why or why not? CP An electron is projected Figure E21.33 with n initil speed v 0 = 1.60 * cm m>s into the uniform field etween the prllel pltes in Fig. v 0 E Assume tht the field etween 1.00 cm E the pltes is uniform nd directed verticlly downwrd, nd tht the field outside the pltes is zero. The electron enters the field t point from the first, in time intervl of 1.50 * 10-6 s. () Find the mgnitude of the electric field. () Find the speed of the proton when it strikes the negtively chrged plte A point chrge is t the origin. With this point chrge s the source point, wht is the unit vector rn in the direction of () the field point t x = 0, y = m; () the field point t x = 12.0 cm, y = 12.0 cm; (c) the field point t x = m, y = 2.60 m? Express your results in terms of the unit vectors ın nd n A 8.75-mC point Figure E21.40 chrge is glued down on horizontl frictionless tle. It is tied E to mC point chrge y light, nonconducting 2.50-cm 2.50 cm wire. A uniform electric field of mgnitude 1.85 * 10 8 N>C is directed prllel to the wire, s mc 8.75 mc shown in Fig. E () Find the tension in the wire. () Wht would the tension e if oth chrges were negtive? () An electron is moving est in uniform electric field of 1.50 N>C directed to the west. At point A, the velocity of the electron is 4.50 * 10 5 m>s towrd the est. Wht is the speed of the electron when it reches point B, m est of point A? () A proton is moving in the uniform electric field of prt (). At point A, the velocity of the proton is 1.90 * 10 4 m>s, est. Wht is the speed of the proton t point B?

33 Exercises 719 ection 21.5 Electric-Field Clcultions Two point chrges Q nd q Figure E21.42 (where q is positive) produce the net electric field shown t point P in Fig. E The Q field points prllel to the line connecting the two chrges. () Wht cn you conclude out the sign nd mgnitude of Q? Explin your resoning. () If the lower chrge were d E negtive insted, would it e possile for P the field to hve the direction shown in the d figure? Explin your resoning Two positive point chrges q re 1q plced on the x-xis, one t x = nd one t x = -. () Find the mgnitude nd direction of the electric field t x = 0. () Derive n expression for the electric field t points on the x-xis. Use your result to grph the x-component of the electric field s function of x, for vlues of x etween -4 nd The two chrges q1 Figure E21.44 nd q 2 shown in Fig. E21.44 hve equl mgnitudes. Wht is the direction of the net electric B field due to these two chrges t points A (midwy etween the chrges), B, nd C A if () oth chrges re negtive, () oth chrges re positive, (c) q 1 is positive nd q 2 is negtive. q 1 q 2 C A 2.00-nC point chrge is t the origin, nd second nC point chrge is on the x-xis t x = m. () Find the electric field (mgnitude nd direction) t ech of the following points on the x-xis: (i) x = m; (ii) x = 1.20 m; (iii) x = m. () Find the net electric force tht the two chrges would exert on n electron plced t ech point in prt () Repet Exercise 21.44, ut now let q 1 = nc Three negtive point chrges lie long line s shown in Fig. E Find the mgnitude nd direction of the electric field this comintion of chrges produces Figure E mc t point P, which lies 6.00 cm from the 8.00 cm mC chrge mesured perpendiculr 6.00 cm to the line connecting the three chrges. P BIO Electric Field of Axons. A mc nerve signl is trnsmitted through neuron when n excess of N ions suddenly 8.00 cm enters the xon, long cylindricl prt of the neuron. Axons re pproximtely mc 10.0 mm in dimeter, nd mesurements show tht out 5.6 * N ions per meter (ech of chrge e) enter during this process. Although the xon is long cylinder, the chrge does not ll enter everywhere t the sme time. A plusile model would e series of point chrges moving long the xon. Let us look t 0.10-mm length of the xon nd model it s point chrge. () If the chrge tht enters ech meter of the xon gets distriuted uniformly long it, how mny couloms of chrge enter 0.10-mm length of the xon? () Wht electric field (mgnitude nd direction) does the sudden influx of chrge produce t the surfce of the ody if the xon is 5.00 cm elow the skin? (c) Certin shrks cn respond to electric fields s wek s 1.0 mn>c. How fr from this segment of xon could shrk e nd still detect its electric field? In rectngulr coordinte system positive point chrge q = 6.00 * 10-9 C is plced t the point x = m, y = 0, nd n identicl point chrge is plced t x = m, y = 0. Find the x- nd y-components, the mgnitude, nd the direction of the electric field t the following points: () the origin; () x = m, y = 0; (c) x = m, y = m; (d) x = 0, y = m A point chrge q 1 = nc is t the point x = m, y = m, nd second point chrge q 2 = 6.00 nc is t the point x = m, y = 0. Clculte the mgnitude nd direction of the net electric field t the origin due to these two point chrges Repet Exercise for the cse where the point chrge t x = m, y = 0 is positive nd the other is negtive, ech with mgnitude 6.00 * 10-9 C A very long, stright wire hs chrge per unit length 1.50 * C>m. At wht distnce from the wire is the electricfield mgnitude equl to 2.50 N>C? A ring-shped conductor with rdius = 2.50 cm hs totl positive chrge Q = nc uniformly distriuted round it, s shown in Fig The center of the ring is t the origin of coordintes O. () Wht is the electric field (mgnitude nd direction) t point P, which is on the x-xis t x = 40.0 cm? () A point chrge q = mc is plced t the point P descried in prt (). Wht re the mgnitude nd direction of the force exerted y the chrge q on the ring? A stright, nonconducting plstic wire 8.50 cm long crries chrge density of 175 nc>m distriuted uniformly long its length. It is lying on horizontl tletop. () Find the mgnitude nd direction of the electric field this wire produces t point 6.00 cm directly ove its midpoint. () If the wire is now ent into circle lying flt on the tle, find the mgnitude nd direction of the electric field it produces t point 6.00 cm directly ove its center A chrge of nc is spred uniformly over the surfce of one fce of nonconducting disk of rdius 1.25 cm. () Find the mgnitude nd direction of the electric field this disk produces t point P on the xis of the disk distnce of 2.00 cm from its center. () uppose tht the chrge were ll pushed wy from the center nd distriuted uniformly on the outer rim of the disk. Find the mgnitude nd direction of the electric field t point P. (c) If the chrge is ll rought to the center of the disk, find the mgnitude nd direction of the electric field t point P. (d) Why is the field in prt () stronger thn the field in prt ()? Why is the field in prt (c) the strongest of the three fields? ection 21.7 Electric Dipoles The mmoni molecule 1NH 3 2 hs dipole moment of 5.0 * C # m. Ammoni molecules in the gs phse re plced in uniform electric field E with mgnitude 1.6 * 10 6 N>C. () Wht is the chnge in electric potentil energy when the dipole moment of molecule chnges its orienttion with respect to E from prllel to perpendiculr? () At wht solute temperture T 3 is the verge trnsltionl kinetic energy 2 kt of molecule equl to the chnge in potentil energy clculted in prt ()? (Note: Aove this temperture, therml gittion prevents the dipoles from ligning with the electric field.)

34 720 CHAPTER 21 Electric Chrge nd Electric Field Point chrges nd q 2 = 4.5 nc re seprted y 3.1 mm, forming n electric dipole. () Find the electric dipole moment (mgnitude nd direction). () The chrges re in uniform electric field whose direction mkes n ngle of 36.9 with the line connecting the chrges. Wht is the mgnitude of this field if the torque exerted on the dipole hs mgnitude 7.2 * 10-9 N # m? The dipole moment of the wter molecule 1H 2 O2 is 6.17 * C # m. Consider wter molecule locted t the origin whose dipole moment p points in the x-direction. A chlorine ion 1C1-2, of chrge * C, is locted t x = 3.00 * 10-9 m. Find the mgnitude nd direction of the electric force tht the wter molecule exerts on the chlorine ion. Is this force ttrctive or repulsive? Assume tht x is much lrger thn the seprtion d etween the chrges in the dipole, so tht the pproximte expression for the electric field long the dipole xis derived in Exmple cn e used Torque on Dipole. An electric dipole with dipole moment p is in uniform electric field E. () Find the orienttions of the dipole for which the torque on the dipole is zero. () Which of the orienttions in prt () is stle, nd which is unstle? (Hint: Consider smll displcement wy from the equilirium position nd see wht hppens.) (c) how tht for the stle orienttion in prt (), the dipole s own electric field tends to oppose the externl field Consider the electric dipole of Exmple () Derive n expression for the mgnitude of the electric field produced y the dipole t point on the x-xis in Fig Wht is the direction of this electric field? () How does the electric field t points on the x-xis depend on x when x is very lrge? Three chrges re t the corners of n isosceles tringle s shown in Fig. E mc Figure E21.61 The 5.00-mC chrges form dipole. () Find the force 2.00 cm (mgnitude nd direction) the mC chrge exerts on the 3.00 cm dipole. () For n xis perpendiculr to the line connecting the mc 5.00-mC chrges t the midpoint of this line, find the torque 2.00 cm (mgnitude nd direction) exerted mc on the dipole y the mC chrge A dipole consisting of chrges e, 220 nm prt, is plced etween two very lrge (essentilly infinite) sheets crrying equl ut opposite chrge densities of 125 mc>m 2. () Wht is the mximum potentil energy this dipole cn hve due to the sheets, nd how should it e oriented reltive to the sheets to ttin this vlue? () Wht is the mximum torque the sheets cn exert on the dipole, nd how should it e oriented reltive to the sheets to ttin this vlue? (c) Wht net force do the two sheets exert on the dipole? PROBLEM q 1 = -4.5 nc Two chrges, one of nd the other of Four identicl chrges Q re plced t the corners of squre of side L. () In free-ody digrm, show ll of the forces tht ct on one of the chrges. () Find the mgnitude nd direction of the totl force exerted on one chrge y the other three chrges mc mc, re plced on the x-xis, one t the origin nd the other t x = m, s shown in Fig. P Find the position on the x-xis where the net force on smll chrge q would e zero. Figure P mc mc x (m) m Three point chrges re rrnged long the x-xis. Chrge q 1 = nc is locted t x = m, nd chrge q 2 = 2.50 nc is t x = m. A positive point chrge q 3 is locted t the origin. () Wht must the vlue of q 3 e for the net force on this point chrge to hve mgnitude 4.00 mn? () Wht is the direction of the net force on q 3? (c) Where long the x-xis cn q 3 e plced nd the net force on it e zero, other thn the trivil nswers of x = q nd x = -q? A chrge q 1 = 5.00 nc is plced t the origin of n xy-coordinte system, nd chrge q 2 = nc is plced on the positive x-xis t x = 4.00 cm. () If third chrge q 3 = 6.00 nc is now plced t the point x = 4.00 cm, y = 3.00 cm, find the x- nd y-components of the totl force exerted on this chrge y the other two. () Find the mgnitude nd direction of this force CP Two positive point chrges Q re held fixed on the x-xis t x = nd x = -. A third positive point chrge q, with mss m, is plced on the x-xis wy from the origin t coordinte x such tht ƒ x ƒ V. The chrge q, which is free to move long the x-xis, is then relesed. () Find the frequency of oscilltion of the chrge q. (Hint: Review the definition of simple hrmonic motion in ection Use the inomil expnsion 11 z2 n = 1 nz n1n - 12z 2 >2 Á, vlid for the cse ƒ z ƒ 6 1. ) () uppose insted tht the chrge q were plced on the y-xis t coordinte y such tht ƒ y ƒ V, nd then relesed. If this chrge is free to move nywhere in the xy-plne, wht will hppen to it? Explin your nswer CP Two identicl spheres Figure P21.68 with mss m re hung from silk threds of length L, s shown in Fig. P Ech sphere hs the sme chrge, so q 1 = q 2 = q. The rdius of ech sphere is very smll compred to L L the distnce etween the spheres, so u u they my e treted s point chrges. how tht if the ngle u is smll, the equilirium seprtion d etween the spheres is d = 1q 2 L>2pP 0 mg2 1>3. (Hint: If u is smll, then tn u mss m mss m sin u. ) chrge q 1 chrge q CP Two smll spheres with mss m = 15.0 g re hung y silk threds of length L = 1.20 m from common point (Fig. P21.68). When the spheres re given equl quntities of negtive chrge, so tht q 1 = q 2 = q, ech thred hngs t u = 25.0 from the verticl. () Drw digrm showing the forces on ech sphere. Tret the spheres s point chrges. () Find the mgnitude of q. (c) Both threds re now shortened to length L = m, while the chrges q 1 nd q 2 remin unchnged. Wht new ngle will ech thred mke with the verticl? (Hint: This prt of the prolem cn e solved numericlly

35 Prolems 721 y using tril vlues for u nd djusting the vlues of u until selfconsistent nswer is otined.) q 2 re held in plce 4.50 cm prt Two point chrges q 1 nd Figure P q CP Two identicl spheres re ech ttched to silk Another point chrge Q = mc 1 threds of length L = m nd hung from common point of mss 5.00 g is initilly locted (Fig. P21.68). Ech sphere hs mss m = 8.00 g. The rdius of 3.00 cm from ech of these chrges ech sphere is very smll compred to the distnce etween the (Fig. P21.76) nd relesed from rest. spheres, so they my e treted s point chrges. One sphere is You oserve tht the initil ccelertion of is 324 m>s cm Q given positive chrge q 1, nd the other different positive chrge Q upwrd, prllel q 2 ; this cuses the spheres to seprte so tht when the spheres re to the line connecting the two point in equilirium, ech thred mkes n ngle u = 20.0 with the chrges. Find q 1 nd q 2. verticl. () Drw free-ody digrm for ech sphere when in Three identicl point chrges q equilirium, nd lel ll the forces tht ct on ech sphere. q re plced t ech of three corners of 2 () Determine the mgnitude of the electrosttic force tht cts on squre of side L. Find the mgnitude ech sphere, nd determine the tension in ech thred. (c) Bsed nd direction of the net force on point chrge -3q plced () t on the informtion you hve een given, wht cn you sy out the center of the squre nd () t the vcnt corner of the squre. the mgnitudes of q 1 nd q 2? Explin your nswers. (d) A smll In ech cse, drw free-ody digrm showing the forces exerted wire is now connected etween the spheres, llowing chrge to on the -3q chrge y ech of the other three chrges. e trnsferred from one sphere to the other until the two spheres Three point chrges re plced on the y -xis: chrge q hve equl chrges; the wire is then removed. Ech thred now t y =, chrge -2q t the origin, nd chrge q t y = -. mkes n ngle of 30.0 with the verticl. Determine the uch n rrngement is clled n electric qudrupole. () Find the originl chrges. (Hint: The totl chrge on the pir of spheres is mgnitude nd direction of the electric field t points on the positive x-xis. () Use the inomil expnsion to find n pproximte conserved.) odium chloride ( NCl, ordinry tle slt) is mde up expression for the electric field vlid for x W. Contrst this of positive sodium ions 1N 2 nd negtive chloride ions 1C1-2. ehvior to tht of the electric field of point chrge nd tht of () If point chrge with the sme chrge nd mss s ll the the electric field of dipole. N ions in mol of NCl is 2.00 cm from point chrge CP trength of the Electric Force. Imgine two with the sme chrge nd mss s ll the C1 - ions, wht is the 1.0-g gs of protons, one t the erth s north pole nd the other t mgnitude of the ttrctive force etween these two point the south pole. () How mny protons re in ech g? () Clculte the grvittionl ttrction nd the electricl repulsion tht chrges? () If the positive point chrge in prt () is held in plce nd the negtive point chrge is relesed from rest, wht is ech g exerts on the other. (c) Are the forces in prt () lrge its initil ccelertion? (ee Appendix D for tomic msses.) enough for you to feel if you were holding one of the gs? (c) Does it seem resonle tht the ions in NCl could e seprted in this wy? Why or why not? (In fct, when sodium chlosions of tomic nuclei re of the order of m(1fm2. () If Electric Force Within the Nucleus. Typicl dimenride dissolves in wter, it reks up into N nd C1 - ions. two protons in nucleus re 2.0 fm prt, find the mgnitude of However, in this sitution there re dditionl electric forces the electric force ech one exerts on the other. Express the nswer exerted y the wter molecules on the ions.) in newtons nd in pounds. Would this force e lrge enough for A nC point chrge is on the x-xis t x = 1.20 m. person to feel? () ince the protons repel ech other so strongly, A second point chrge Q is on the x-xis t m. Wht must why don t they shoot out of the nucleus? e the sign nd mgnitude of Q for the resultnt electric field t the If Atoms Were Not Neutrl... Becuse the chrges origin to e () 45.0 N> C in the x-direction, () 45.0 N> C in the on the electron nd proton hve the sme solute vlue, toms re -x-direction? electriclly neutrl. uppose this were not precisely true, nd the CP A smll 12.3-g plstic ll is tied Figure P21.73 solute vlue of the chrge of the electron were less thn the to very light 28.6-cm string tht is ttched to chrge of the proton y %. () Estimte wht the net the verticl wll of room (Fig. P21.73). A uniform horizontl electric field exists in this ny ssumptions you feel re justified, ut stte clerly wht they chrge of this textook would e under these circumstnces. Mke room. When the ll hs een given n excess re. (Hint: Most of the toms in this textook hve equl numers chrge of mc, you oserve tht it of electrons, protons, nd neutrons.) () Wht would e the mgnitude of the electric force etween two textooks plced 5.0 m remins suspended, with the string mking n ngle of 17.4 with the wll. Find the mgnitude nd direction of the electric field in the the ccelertion of ech ook would e if the ooks were 5.0 m 17.4 prt? Would this force e ttrctive or repulsive? Estimte wht room. prt nd there were no nonelectric forces on them. (c) Figure P CP At t = 0 very smll oject with mss mg nd chrge 9.00 mc is trveling t 125 m> s Discuss how the fct tht ordinry mtter is stle shows in the -x-direction. The chrge is moving in uniform electric field tht is in the y-direction nd tht hs mgnitude E = 895 N>C. tht the solute vlues of the u The grvittionl force on the prticle cn e neglected. How fr is chrges on the electron nd L L the prticle from the origin t t = 7.00 ms? proton must e identicl to Two prticles hving chrges q 1 = nc nd very high level of ccurcy. q 2 = 8.00 nc re seprted y distnce of 1.20 m. At wht point CP Two tiny spheres of mss 6.80 mg crry long the line connecting the two chrges is the totl electric field E due to the two chrges equl to zero? chrges of equl mgnitude, 3.00 cm 3.00 cm

36 722 CHAPTER 21 Electric Chrge nd Electric Field 72.0 nc, ut opposite sign. They re tied to the sme ceiling hook y light strings of length m. When horizontl uniform electric field E tht is directed to the left is turned on, the spheres hng t rest with the ngle u etween the strings equl to 50.0 o (Fig. P21.82). () Which ll (the one on the right or the one on the left) hs positive chrge? () Wht is the mgnitude E of the field? CP Consider model of hydrogen tom in which n electron is in circulr orit of rdius r = 5.29 * m round sttionry proton. Wht is the speed of the electron in its orit? CP A smll sphere with mss 9.00 mg nd chrge mc is moving in circulr orit round sttionry sphere tht hs chrge 7.50 mc. If the speed of the smll sphere is 5.90 * 10 3 m>s, wht is the rdius of its orit? Tret the spheres s point chrges nd ignore grvity Two smll copper spheres ech hve rdius 1.00 mm. () How mny toms does ech sphere contin? () Assume tht ech copper tom contins 29 protons nd 29 electrons. We know tht electrons nd protons hve chrges of exctly the sme mgnitude, ut let s explore the effect of smll differences (see lso Prolem 21.81). If the chrge of proton is e nd the mgnitude of the chrge of n electron is 0.100% smller, wht is the net chrge of ech sphere nd wht force would one sphere exert on the other if they were seprted y 1.00 m? CP Opertion of n Inkjet Printer. In n inkjet printer, letters re uilt up y squirting drops of ink t the pper from rpidly moving nozzle. The ink drops, which hve mss of 1.4 * 10-8 g ech, leve the nozzle nd trvel towrd the pper t 20 m>s, pssing through chrging unit tht gives ech drop positive chrge q y removing some electrons from it. The drops then pss etween prllel deflecting pltes 2.0 cm long where there is uniform verticl electric field with mgnitude 8.0 * 10 4 N>C. If drop is to e deflected 0.30 mm y the time it reches the end of the deflection pltes, wht mgnitude of chrge must e given to the drop? CP A proton is projected into uniform electric field tht points verticlly upwrd nd hs mgnitude E. The initil velocity of the proton hs mgnitude v 0 nd is directed t n ngle elow the horizontl. () Find the mximum distnce h mx tht the proton descends verticlly elow its initil elevtion. You cn ignore grvittionl forces. () After wht horizontl distnce d does the proton return to its originl elevtion? (c) ketch the trjectory of the proton. (d) Find the numericl vlues of h mx nd d if E = 500 N>C, v 0 = 4.00 * 10 5 m>s, nd = A negtive point chrge q 1 = nc is on the x -xis t x = 0.60 m. A second point chrge q 2 is on the x-xis t x = m. Wht must the sign nd mgnitude of q 2 e for the net electric field t the origin to e () 50.0 N>C in the x-direction nd () 50.0 N> C in the - x-direction? CALC Positive chrge Q is distriuted uniformly long Figure P21.89 the x-xis from x = 0 to x =. y A positive point chrge q is locted on the positive x-xis t x = r, distnce r to the q Q right of the end of Q (Fig. x O P21.89). () Clculte the x- nd r y-components of the electric field produced y the chrge distriution Q t points on the positive x-xis where x 7. () Clculte the force (mgnitude nd direction) tht the chrge distriution Q exerts on q. (c) how tht if r W, the mgnitude of the force in prt () is pproximtely Qq>4pP 0 r 2. Explin why this result is otined CALC Positive chrge Figure P21.90 Q is distriuted uniformly long y the positive y-xis etween y = 0 nd y =. A negtive point chrge -q lies on the positive x-xis, distnce x from the Q origin (Fig. P21.90). () Clculte the x- nd y-components of x O the electric field produced y the 2q chrge distriution Q t points on the positive x-xis. () Clculte the x- nd y-components of the force tht the chrge distriution Q exerts on q. (c) how tht if x W, F nd F y Qq>8pP 0 x 3 x -Qq>4pP 0 x 2. Explin why this result is otined A chrged line like tht shown in Fig extends from y = 2.50 cm to y = cm. The totl chrge distriuted uniformly long the line is nc. () Find the electric field (mgnitude nd direction) on the x-xis t x = 10.0 cm. () Is the mgnitude of the electric field you clculted in prt () lrger or smller thn the electric field 10.0 cm from point chrge tht hs the sme totl chrge s this finite line of chrge? In terms of the pproximtion used to derive E = Q>4pP 0 x 2 for point chrge from Eq. (21.9), explin why this is so. (c) At wht distnce x does the result for the finite line of chrge differ y 1.0% from tht for the point chrge? CP A Prllel Universe. Imgine prllel universe in which the electric force hs the sme properties s in our universe ut there is no grvity. In this prllel universe, the sun crries chrge Q, the erth crries chrge -Q, nd the electric ttrction etween them keeps the erth in orit. The erth in the prllel universe hs the sme mss, the sme oritl rdius, nd the sme oritl period s in our universe. Clculte the vlue of Q. (Consult Appendix F s needed.) A uniformly chrged disk like the disk in Fig hs rdius 2.50 cm nd crries totl chrge of 7.0 * C. () Find the electric field (mgnitude nd direction) on the x-xis t x = 20.0 cm. () how tht for x W R, Eq. (21.11) ecomes E = Q>4pP 0 x 2, where Q is the totl chrge on the disk. (c) Is the mgnitude of the electric field you clculted in prt () lrger or smller thn the electric field 20.0 cm from point chrge tht hs the sme totl chrge s this disk? In terms of the pproximtion used in prt () to derive E = Q>4pP 0 x 2 for point chrge from Eq. (21.11), explin why this is so. (d) Wht is the percent difference etween the electric fields produced y the finite disk nd y point chrge with the sme chrge t x = 20.0 cm nd t x = 10.0 cm? BIO Electrophoresis. Figure P21.94 Electrophoresis is process used y iologists to seprte different iologicl molecules (such s proteins) from ech other ccording to their rtio of chrge to size. The mterils to e seprted re in viscous solution tht produces drg force F D proportionl to the size nd speed of the molecule. We cn express this reltionship s F D = KRy, where R is the rdius of the molecule (modeled s eing sphericl), v is its speed, nd K is constnt tht depends on the viscosity of the

37 Chllenge Prolems 723 solution. The solution is plced in n externl electric field E so tht the electric force on prticle of chrge q is F = qe. () how tht when the electric field is djusted so tht the two forces (electric nd viscous drg) just lnce, the rtio of q to R is Kv>E. () how tht if we leve the electric field on for time T, the distnce x tht the molecule moves during tht time is x = 1ET>k21q>R2. (c) uppose you hve smple contining three different iologicl molecules for which the moleculr rtio q>r for mteril 2 is twice tht of mteril 1 nd the rtio for mteril 3 is three times tht of mteril 1. how tht the distnces migrted y these molecules fter the sme mount of time re x 2 = 2x 1 nd x 3 = 3x 1. In other words, mteril 2 trvels twice s fr s mteril 1, nd mteril 3 trvels three times s fr s mteril 1. Therefore, we hve seprted these molecules ccording to their rtio of chrge to size. In prctice, this process cn e crried out in specil gel or pper, long which the iologicl molecules migrte. (Fig. P21.94). The process cn e rther slow, requiring severl hours for seprtions of just centimeter or so CALC Positive chrge Q is distriuted uniformly long the x-xis from x = 0 to x =. Negtive chrge -Q is distriuted uniformly long the -x-xis from x = 0 to x = -. () A positive point chrge q lies on the positive y-xis, distnce y from the origin. Find the force (mgnitude nd direction) tht the positive nd negtive chrge distriutions together exert on q. how tht this force is proportionl to y -3 for y W. () uppose insted tht the positive point chrge q lies on the positive x-xis, distnce x 7 from the origin. Find the force (mgnitude nd direction) tht the chrge distriution exerts on q. how tht this force is proportionl to x -3 for x W CP A smll sphere with mss m crries positive chrge q nd is ttched to one end of silk fier of length L. The other end of the fier is ttched to lrge verticl insulting sheet tht hs positive surfce chrge density s. how tht when the sphere is in equilirium, the fier mkes n ngle equl to rctn 1qs>2mgP 0 2 with the verticl sheet CALC Negtive chrge -Q is distriuted uniformly round qurter-circle of rdius tht lies in the first qudrnt, with the center of curvture t the origin. Find the x- nd y-components of the net electric field t the origin CALC A semicircle of rdius is in the first nd Figure P21.98 second qudrnts, with the y center of curvture t the origin. Positive chrge Q is distriuted uniformly round the 1Q 2Q left hlf of the semicircle, nd negtive chrge -Q is distriuted uniformly round the right x hlf of the semicircle (Fig. P21.98). Wht re the mgnitude nd direction of the net Figure P21.99 electric field t the origin produced y this distriution of 1.20 m chrge? Two 1.20-m nonconducting wires meet t right ngle. One segment crries 2.50 mc of chrge distriuted uniformly long its length, 1.20 m P nd the other crries mc distriuted uniformly long it, s shown in Fig. P () Find the mgnitude nd direction of the electric field these wires produce t point P, which is 60.0 cm from ech wire. () If n electron is relesed t P, wht re the mgnitude nd direction of the net force tht these wires exert on it? Two very lrge prllel sheets re 5.00 cm prt. heet A crries uniform surfce chrge density of -9.50mC>m 2, nd sheet B, which is to the right of A, crries uniform chrge density of mc>m 2. Assume the sheets re lrge enough to e treted s infinite. Find the mgnitude nd direction of the net electric field these sheets produce t point () 4.00 cm to the right of sheet A; () 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B Repet Prolem for the cse where sheet B is positive Two very lrge horizontl sheets re 4.25 cm prt nd crry equl ut opposite uniform surfce chrge densities of mgnitude s. You wnt to use these sheets to hold sttionry in the region etween them n oil droplet of mss 324 mg tht crries n excess of five electrons. Assuming tht the drop is in vcuum, () which wy should the electric field etween the pltes point, nd () wht should s e? An infinite sheet with positive chrge per unit re s lies in the xy-plne. A second infinite sheet with negtive chrge per unit re -s lies in the yz-plne. Find the net electric field t ll points tht do not lie in either of these plnes. Express your nswer in terms of the unit vectors ın, n, nd k N. CP A thin disk with circulr hole t its center, clled Figure P n nnulus, hs inner rdius R 1 x nd outer rdius R 2 (Fig. P21.104). The disk hs uniform R 2 positive surfce chrge density s R on its surfce. () Determine the 1 z totl electric chrge on the nnulus. () The nnulus lies in the s y O yz-plne, with its center t the origin. For n ritrry point on the x-xis (the xis of the nnulus), find the mgnitude nd direction of the electric field E. Consider points oth ove nd elow the nnulus in Fig. P (c) how tht t points on the x-xis tht re sufficiently close to the origin, the mgnitude of the electric field is pproximtely proportionl to the distnce etween the center of the nnulus nd the point. How close is sufficiently close? (d) A point prticle with mss m nd negtive chrge -q is free to move long the x-xis (ut cnnot move off the xis). The prticle is originlly plced t rest t x = 0.01R 1 nd relesed. Find the frequency of oscilltion of the prticle. (Hint: Review ection The nnulus is held sttionry.) CHALLENGE PROBLEM Three chrges re Figure P plced s shown in Fig. q P The mgnitude of q is 3 1 F 2.00 mc, ut its sign nd the vlue of the chrge q 2 re not 4.00 cm 3.00 cm known. Chrge q 3 is mc, 4.00 nd the net force F on q 3 is q entirely in the negtive x-direc- tion. () Considering the different possile signs of q 1, there re four cm q 2 possile force digrms representing the forces F nd F 1 2 tht q 1 nd q 2 exert on q 3. ketch these four possile force configurtions.

38 724 CHAPTER 21 Electric Chrge nd Electric Field CALC Two thin rods of length lie long the -xis, ƒ ƒ E, () Using the sketches from prt () nd the direction of deduce L x F, the signs of the chrges q 1 nd q 2. (c) Clculte the mgnitude of one etween x = >2 nd x = >2 L nd the other etween q 2. (d) Determine F, the mgnitude of the net force on q 3. x = ->2 nd x = ->2 - L. Ech rod hs positive chrge Q Two chrges re distriuted uniformly long its length. () Clculte the electric Figure P plced s shown in Fig. field produced y the second rod t points long the positive P The mgnitude of q 1 is P x-xis. () how tht the mgnitude of the force tht one rod exerts 3.00 mc, ut its sign nd the on the other is vlue of the chrge q 2 re not 5.0 cm 12.0 cm Q 2 2 E ( L) known. The direction of the net F = ln c electric field E 2 t point P is 4pP 0 L 1 2L) d q entirely in the negtive y-direc cm q 2 (c) how tht if W L, the mgnitude of this force reduces to tion. () Considering the differ- F = Q 2 >4pP 0 2. (Hint: Use the expnsion ln11 z2 = z - ent possile signs of q 1 nd q 2, there re four possile digrms tht could represent the electric fields E nd E 1 2 produced y q 1 z 2 >2 z 3 >3 - Á, vlid for z V 1. Crry ll expnsions to t lest order L 2 > 2.) Interpret this result. nd q 2. ketch the four possile electric-field configurtions. () Using the sketches from prt () nd the direction of deduce the signs of q nd q (c) Determine the mgnitude of E Answers Chpter Opening Question? Wter molecules hve permnent electric dipole moment: One end of the molecule hs positive chrge nd the other end hs negtive chrge. These ends ttrct negtive nd positive ions, respectively, holding the ions prt in solution. Wter is less effective s solvent for mterils whose molecules do not ionize (clled nonionic sustnces), such s oils. Test Your Understnding Questions 21.1 Answers: () the plstic rod weighs more, () the glss rod weighs less, (c) the fur weighs less, (d) the silk weighs more The plstic rod gets negtive chrge y tking electrons from the fur, so the rod weighs little more nd the fur weighs little less fter the ruing. By contrst, the glss rod gets positive chrge y giving electrons to the silk. Hence, fter they re rued together, the glss rod weighs little less nd the silk weighs little more. The weight chnge is very smll: The numer of electrons trnsferred is smll frction of mole, nd mole of electrons hs mss of only * electrons * kg>electron2 = 5.48 * 10-7 kg = milligrm! 21.2 Answers: () (i), () (ii) Before the two spheres touch, the negtively chrged sphere exerts repulsive force on the electrons in the other sphere, cusing zones of positive nd negtive induced chrge (see Fig. 21.7). The positive zone is closer to the negtively chrged sphere thn the negtive zone, so there is net force of ttrction tht pulls the spheres together, like the com nd insultor in Fig Once the two metl spheres touch, some of the excess electrons on the negtively chrged sphere will flow onto the other sphere (ecuse metls re conductors). Then oth spheres will hve net negtive chrge nd will repel ech other Answer: (iv) The force exerted y q 1 on Q is still s in Exmple The mgnitude of the force exerted y q 2 on Q is still equl to F 1 on Q, ut the direction of the force is now towrd q 2 t n ngle elow the x-xis. Hence the x-components of the two forces cncel while the (negtive) y-components dd together, nd the totl electric force is in the negtive y-direction Answers: () (ii), () (i) The electric field produced y positive point chrge points directly wy from the chrge (see Fig ) nd hs mgnitude tht depends on the distnce r from the chrge to the field point. Hence second, negtive point chrge q 6 0 will feel force F qe tht points directly towrd the positive chrge nd hs mgnitude tht depends on the distnce r etween the two chrges. If the negtive chrge moves directly towrd the positive chrge, the direction of the force remins the sme ut the force mgnitude increses s the distnce r decreses. If the negtive chrge moves in circle round the positive chrge, the force mgnitude stys the sme (ecuse the distnce r is constnt) ut the force direction chnges Answer: (iv) Think of pir of segments of length dy, one t coordinte y 7 0 nd the other t coordinte -y 6 0. The upper segment hs positive chrge nd produces n electric field de t P tht points wy from the segment, so this de hs positive x-component nd negtive y-component, like the vector de in Fig The lower segment hs the sme mount of negtive chrge. It produces de tht hs the sme mgnitude ut points towrd the lower segment, so it hs negtive x-component nd negtive y-component. By symmetry, the two x-components re equl ut opposite, so they cncel. Thus the totl electric field hs only negtive y-component Answer: yes If the field lines re stright, E must point in the sme direction throughout the region. Hence the force F = qe on prticle of chrge q is lwys in the sme direction. A prticle relesed from rest ccelertes in stright line in the direction of F, nd so its trjectory is stright line long field line Answer: (ii) Equtions (21.17) nd (21.18) tell us tht the potentil energy for dipole in n electric field is U = -p # E = -pe cosf, where f is the ngle etween the directions of p nd E. If p nd E point in opposite directions, so tht f = 180, we hve cos f = -1 nd U = pe. This is the mximum vlue tht U cn hve. From our discussion of energy digrms in ection 7.5, it follows tht this is sitution of unstle equilirium. Bridging Prolem Answer: E = 2kQ>p 2 in the y-direction E

39 GAU LAW 22 LEARNING GOAL By studying this chpter, you will lern:? This child cquires n electric chrge y touching the chrged metl sphere. The chrged hirs on the child s hed repel nd stnd out. If the child stnds inside lrge, chrged metl sphere, will her hir stnd on end? Often, there re oth n esy wy nd hrd wy to do jo; the esy wy my involve nothing more thn using the right tools. In physics, n importnt tool for simplifying prolems is the symmetry properties of systems. Mny physicl systems hve symmetry; for exmple, cylindricl ody doesn t look ny different fter you ve rotted it round its xis, nd chrged metl sphere looks just the sme fter you ve turned it out ny xis through its center. Guss s lw is prt of the key to using symmetry considertions to simplify electric-field clcultions. For exmple, the field of stright-line or plne-sheet chrge distriution, which we derived in ection 21.5 using some firly strenuous integrtions, cn e otined in few lines with the help of Guss s lw. But Guss s lw is more thn just wy to mke certin clcultions esier. Indeed, it is fundmentl sttement out the reltionship etween electric chrges nd electric fields. Among other things, Guss s lw cn help us understnd how electric chrge distriutes itself over conducting odies. Here s wht Guss s lw is ll out. Given ny generl distriution of chrge, we surround it with n imginry surfce tht encloses the chrge. Then we look t the electric field t vrious points on this imginry surfce. Guss s lw is reltionship etween the field t ll the points on the surfce nd the totl chrge enclosed within the surfce. This my sound like rther indirect wy of expressing things, ut it turns out to e tremendously useful reltionship. Aove nd eyond its use s clcultionl tool, Guss s lw cn help us gin deeper insights into electric fields. We will mke use of these insights repetedly in the next severl chpters s we pursue our study of electromgnetism Chrge nd Electric Flux In Chpter 21 we sked the question, Given chrge distriution, wht is the electric field produced y tht distriution t point P? We sw tht the nswer could e found y representing the distriution s n ssemly of point chrges, How you cn determine the mount of chrge within closed surfce y exmining the electric field on the surfce. Wht is ment y electric flux, nd how to clculte it. How Guss s lw reltes the electric flux through closed surfce to the chrge enclosed y the surfce. How to use Guss s lw to clculte the electric field due to symmetric chrge distriution. Where the chrge is locted on chrged conductor. The discussion of Guss s lw in this section is sed on nd inspired y the innovtive ides of Ruth W. Chy nd Bruce A. herwood in Electric nd Mgnetic Interctions (John Wiley & ons, 1994). 725

40 726 CHAPTER 22 Guss s Lw ActivPhysics 11.7: Electric Flux 22.1 How cn you mesure the chrge inside ox without opening it? () A ox contining n unknown mount of chrge E q? () Using test chrge outside the ox to proe the mount of chrge inside the ox E E E Test chrge q 0 E E E E E ech of which produces n electric field given y Eq. (21.7). The totl field t P is then the vector sum of the fields due to ll the point chrges. But there is n lterntive reltionship etween chrge distriutions nd electric fields. To discover this reltionship, let s stnd the question of Chpter 21 on its hed nd sk, If the electric field pttern is known in given region, wht cn we determine out the chrge distriution in tht region? Here s n exmple. Consider the ox shown in Fig. 22.1, which my or my not contin electric chrge. We ll imgine tht the ox is mde of mteril tht hs no effect on ny electric fields; it s of the sme reed s the mssless rope nd the frictionless incline. Better still, let the ox represent n imginry surfce tht my or my not enclose some chrge. We ll refer to the ox s closed surfce ecuse it completely encloses volume. How cn you determine how much (if ny) electric chrge lies within the ox? Knowing tht chrge distriution produces n electric field nd tht n electric field exerts force on test chrge, you move test chrge q 0 round the vicinity of the ox. By mesuring the force F experienced y the test chrge t different positions, you mke three-dimensionl mp of the electric field E F >q 0 outside the ox. In the cse shown in Fig. 22.1, the mp turns out to e the sme s tht of the electric field produced y positive point chrge (Fig ). From the detils of the mp, you cn find the exct vlue of the point chrge inside the ox. To determine the contents of the ox, we ctully need to mesure E only on the surfce of the ox. In Fig there is single positive point chrge inside the ox, nd in Fig there re two such chrges. The field ptterns on the surfces of the oxes re different in detil, ut in ech cse the electric field points out of the ox. Figures 22.2c nd 22.2d show cses with one nd two negtive point chrges, respectively, inside the ox. Agin, the detils of E re different for the two cses, ut the electric field points into ech ox. E Electric Flux nd Enclosed Chrge In ection 21.4 we mentioned the nlogy etween electric-field vectors nd the velocity vectors of fluid in motion. This nlogy cn e helpful, even though n electric field does not ctully flow. Using this nlogy, in Figs nd 22.2, in which the electric field vectors point out of the surfce, we sy tht there is n outwrd electric flux. (The word flux comes from Ltin word mening flow. ) In Figs. 22.2c nd 22.2d the E vectors point into the surfce, nd the electric flux is inwrd. Figure 22.2 suggests simple reltionship: Positive chrge inside the ox goes with n outwrd electric flux through the ox s surfce, nd negtive chrge inside goes with n inwrd electric flux. Wht hppens if there is zero chrge 22.2 The electric field on the surfce of oxes contining () single positive point chrge, () two positive point chrges, (c) single negtive point chrge, or (d) two negtive point chrges. () Positive chrge inside ox, outwrd flux () Positive chrges inside ox, outwrd flux (c) Negtive chrge inside ox, inwrd flux (d) Negtive chrges inside ox, inwrd flux E E E E q q q q 2q 2q

41 22.1 Chrge nd Electric Flux 727 inside the ox? In Fig the ox is empty nd E 0 everywhere, so there is no electric flux into or out of the ox. In Fig. 22.3, one positive nd one negtive point chrge of equl mgnitude re enclosed within the ox, so the net chrge inside the ox is zero. There is n electric field, ut it flows into the ox on hlf of its surfce nd flows out of the ox on the other hlf. Hence there is no net electric flux into or out of the ox. The ox is gin empty in Fig. 22.3c. However, there is chrge present outside the ox; the ox hs een plced with one end prllel to uniformly chrged infinite sheet, which produces uniform electric field perpendiculr to the sheet (s we lerned in Exmple of ection 21.5). On one end of the ox, E points into the ox; on the opposite end, E points out of the ox; nd on the sides, E is prllel to the surfce nd so points neither into nor out of the ox. As in Fig. 22.3, the inwrd electric flux on one prt of the ox exctly compenstes for the outwrd electric flux on the other prt. o in ll of the cses shown in Fig. 22.3, there is no net electric flux through the surfce of the ox, nd no net chrge is enclosed in the ox. Figures 22.2 nd 22.3 demonstrte connection etween the sign (positive, negtive, or zero) of the net chrge enclosed y closed surfce nd the sense (outwrd, inwrd, or none) of the net electric flux through the surfce. There is lso connection etween the mgnitude of the net chrge inside the closed surfce nd the strength of the net flow of E over the surfce. In oth Figs nd 22.4 there is single point chrge inside the ox, ut in Fig the mgnitude of the chrge is twice s gret, nd so E is everywhere twice s gret in mgnitude s in Fig If we keep in mind the fluid-flow nlogy, this mens tht the net outwrd electric flux is lso twice s gret in Fig s in Fig This suggests tht the net electric flux through the surfce of the ox is directly proportionl to the mgnitude of the net chrge enclosed y the ox. This conclusion is independent of the size of the ox. In Fig. 22.4c the point chrge q is enclosed y ox with twice the liner dimensions of the ox in Fig The mgnitude of the electric field of point chrge decreses with distnce ccording to 1>r 2, so the verge mgnitude of E on ech fce of the 1 lrge ox in Fig. 22.4c is just 4 of the verge mgnitude on the corresponding fce in Fig But ech fce of the lrge ox hs exctly four times the re of the corresponding fce of the smll ox. Hence the outwrd electric flux is the sme for the two oxes if we define electric flux s follows: For ech fce of the ox, tke the product of the verge perpendiculr component of E nd the re of tht fce; then dd up the results from ll fces of the ox. With this definition the net electric flux due to single point chrge inside the ox is independent of the size of the ox nd depends only on the net chrge inside the ox Three cses in which there is zero net chrge inside ox nd no net electric flux through the surfce of the ox. () An empty ox with E 0. () A ox contining one positive nd one equl-mgnitude negtive point chrge. (c) An empty ox immersed in uniform electric field. () No chrge inside ox, zero flux () Zero net chrge inside ox, inwrd flux cncels outwrd flux. (c) No chrge inside ox, inwrd flux cncels outwrd flux. E 5 0 E 1s Uniformly chrged sheet 1q 2q E

42 728 CHAPTER 22 Guss s Lw 22.4 () A ox enclosing positive point chrge q. () Douling the chrge cuses the mgnitude of E to doule, nd it doules the electric flux through the surfce. (c) If the chrge stys the sme ut the dimensions of the ox re douled, the flux stys the sme. The mgnitude of E on the 1 surfce decreses y fctor of ut the re through which E 4, flows increses y fctor of 4. () A ox contining chrge 1q E To summrize, for the specil cses of closed surfce in the shpe of rectngulr ox nd chrge distriutions mde up of point chrges or infinite chrged sheets, we hve found: 1. Whether there is net outwrd or inwrd electric flux through closed surfce depends on the sign of the enclosed chrge. 2. Chrges outside the surfce do not give net electric flux through the surfce. 3. The net electric flux is directly proportionl to the net mount of chrge enclosed within the surfce ut is otherwise independent of the size of the closed surfce. These oservtions re qulittive sttement of Guss s lw. Do these oservtions hold true for other kinds of chrge distriutions nd for closed surfces of ritrry shpe? The nswer to these questions will prove to e yes. But to explin why this is so, we need precise mthemticl sttement of wht we men y electric flux. We develop this in the next section. () Douling the enclosed chrge doules the flux. E Test Your Understnding of ection 22.1 If ll of the dimensions of the ox in Fig re incresed y fctor of 3, wht effect will this chnge hve on the electric flux through the ox? (i) The flux will e 3 2 = 9 times greter; (ii) the flux will e 3 times greter; (iii) the flux will e unchnged; (iv) the flux 1 will e s gret; (v) the flux will e A 1 3 B 2 = s gret; (vi) not enough informtion is given to decide. 12q 22.2 Clculting Electric Flux (c) Douling the ox dimensions does not chnge the flux. E 1q In the preceding section we introduced the concept of electric flux. We used this to give rough qulittive sttement of Guss s lw: The net electric flux through closed surfce is directly proportionl to the net chrge inside tht surfce. To e le to mke full use of this lw, we need to know how to clculte electric flux. To do this, let s gin mke use of the nlogy etween n electric field E nd the field of velocity vectors v in flowing fluid. (Agin, keep in mind tht this is only n nlogy; n electric field is not flow.) Flux: Fluid-Flow Anlogy Figure 22.5 shows fluid flowing stedily from left to right. Let s exmine the volume flow rte dv>dt (in, sy, cuic meters per second) through the wire rectngle with re A. When the re is perpendiculr to the flow velocity v (Fig. 22.5) nd the flow velocity is the sme t ll points in the fluid, the volume flow rte dv>dt is the re A multiplied y the flow speed v: When the rectngle is tilted t n ngle f (Fig. 22.5) so tht its fce is not perpendiculr to v, the re tht counts is the silhouette re tht we see when we look in the direction of v. This re, which is outlined in red nd leled A in Fig. 22.5, is the projection of the re A onto surfce perpendiculr to v. Two sides of the projected rectngle hve the sme length s the originl one, ut the other two re foreshortened y fctor of cos f, so the projected re A is equl to A cos f. Then the volume flow rte through A is dv dt dv dt = va = vacos f If f = 90, dv>dt = 0; the wire rectngle is edge-on to the flow, nd no fluid psses through the rectngle.

43 22.2 Clculting Electric Flux 729 Also, vcos f is the component of the vector v perpendiculr to the plne of the re A. Clling this component v, we cn rewrite the volume flow rte s We cn express the volume flow rte more compctly y using the concept of vector re A, vector quntity with mgnitude A nd direction perpendiculr to the plne of the re we re descriing. The vector re A descries oth the size of n re nd its orienttion in spce. In terms of A, we cn write the volume flow rte of fluid through the rectngle in Fig s sclr (dot) product: dv dt dv dt = v A = v # A Flux of Uniform Electric Field Using the nlogy etween electric field nd fluid flow, we now define electric flux in the sme wy s we hve just defined the volume flow rte of fluid; we simply replce the fluid velocity v y the electric field E. The symol tht we use for electric flux is E (the cpitl Greek letter phi; the suscript E is reminder tht this is electric flux). Consider first flt re A perpendiculr to uniform electric field E (Fig. 22.6). We define the electric flux through this re to e the product of the field mgnitude E nd the re A: 22.5 The volume flow rte of fluid through the wire rectngle () is va when the re of the rectngle is perpendiculr to v nd () is vacos f when the rectngle is tilted t n ngle f. () A wire rectngle in fluid A () The wire rectngle tilted y n ngle f A ' 5 A cos f A f v A v E = EA Roughly speking, we cn picture E in terms of the field lines pssing through A. Incresing the re mens tht more lines of E pss through the re, incresing the flux; stronger field mens more closely spced lines of E nd therefore more lines per unit re, so gin the flux increses. If the re A is flt ut not perpendiculr to the field E, then fewer field lines pss through it. In this cse the re tht counts is the silhouette re tht we see when looking in the direction of E. This is the re A in Fig nd is equl to Acos f (compre to Fig. 22.5). We generlize our definition of electric flux for uniform electric field to (electric flux for uniform E E = EA cos f, flt surfce) (22.1) ince Ecos f is the component of perpendiculr to the re, we cn rewrite Eq. (22.1) s E = E A (electric flux for uniform E, flt surfce) (22.2) In terms of the vector re perpendiculr to the re, we cn write the electric flux s the sclr product of E nd A : A E E = E # A (electric flux for uniform E, flt surfce) (22.3) Appliction Flux Through Bsking hrk s Mouth Unlike ggressive crnivorous shrks such s gret whites, sking shrk feeds pssively on plnkton in the wter tht psses through the shrk s gills s it swims. To survive on these tiny orgnisms requires huge flux of wter through sking shrk s immense mouth, which cn e up to meter cross. The wter flux the product of the shrk s speed through the wter nd the re of its mouth cn e up to 0.5 m 3 /s (500 liters per second, or lmost 5 * 10 5 gllons per hour). In similr wy, the flux of electric field through surfce depends on the mgnitude of the field nd the re of the surfce (s well s the reltive orienttion of the field nd surfce). Equtions (22.1), (22.2), nd (22.3) express the electric flux for flt surfce nd uniform electric field in different ut equivlent wys. The I unit for electric flux is 1 N # m 2 >C. Note tht if the re is edge-on to the field, E nd A re perpendiculr nd the flux is zero (Fig. 22.6c). We cn represent the direction of vector re A y using unit vector nn perpendiculr to the re; nn stnds for norml. Then A AnN (22.4) A surfce hs two sides, so there re two possile directions for nn nd A. We must lwys specify which direction we choose. In ection 22.1 we relted the chrge inside closed surfce to the electric flux through the surfce. With closed surfce we will lwys choose the direction of nn to e outwrd, nd we

44 730 CHAPTER 22 Guss s Lw 22.6 A flt surfce in uniform electric field. The electric flux through the surfce equls the sclr product of the electric field nd the re vector A E. E () urfce is fce-on to electric field: E nd A re prllel (the ngle etween E nd A is f50). The flux F E 5 E A 5 EA. () urfce is tilted from fce-on orienttion y n ngle f: The ngle etween E nd A is f. The flux F E 5 E A 5 EA cos f. (c) urfce is edge-on to electric field: E nd A re perpendiculr (the ngle etween E nd A is f 5 90 ). The flux F E 5 E A 5 EA cos A E f50 A E f f A E A f590 A A A will spek of the flux out of closed surfce. Thus wht we clled outwrd electric flux in ection 22.1 corresponds to positive vlue of E, nd wht we clled inwrd electric flux corresponds to negtive vlue of E. Flux of Nonuniform Electric Field Wht hppens if the electric field isn t uniform ut vries from point to point over the re A? Or wht if A is prt of curved surfce? Then we divide A into mny smll elements da, ech of which hs unit vector nn perpendiculr to it nd vector re da nn da. We clculte the electric flux through ech element nd integrte the results to otin the totl flux: E E = L Ecos f da = L E da = L E # da (generl definition of electric flux) (22.5) We cll this integrl the surfce integrl of the component E over the re, or the surfce integrl of E # da. In specific prolems, one form of the integrl is sometimes more convenient thn nother. Exmple 22.3 t the end of this section illustrtes the use of Eq. (22.5). In Eq. (22.5) the electric flux 1 E da is equl to the verge vlue of the perpendiculr component of the electric field, multiplied y the re of the surfce. This is the sme definition of electric flux tht we were led to in ection 22.1, now expressed more mthemticlly. In the next section we will see the connection etween the totl electric flux through ny closed surfce, no mtter wht its shpe, nd the mount of chrge enclosed within tht surfce. Exmple 22.1 Electric flux through disk A disk of rdius 0.10 m is oriented with its norml unit vector nn t 30 to uniform electric field E of mgnitude 2.0 * 10 3 N>C (Fig. 22.7). (ince this isn t closed surfce, it hs no inside or outside. Tht s why we hve to specify the direction of nn in the figure.) () Wht is the electric flux through the disk? () Wht is the flux through the disk if it is turned so tht nn is perpendiculr to E? (c) Wht is the flux through the disk if nn is prllel to E? 22.7 The electric flux E through disk depends on the ngle etween its norml nn nd the electric field E. r m n^ 30 E

45 22.2 Clculting Electric Flux 731 OLUTION IDENTIFY nd ET UP: This prolem is out flt surfce in uniform electric field, so we cn pply the ides of this section. We clculte the electric flux using Eq. (22.1). EXECUTE: () The re is A = p10.10 m2 2 = m 2 nd the ngle etween E nd A AnN is f = 30, so from Eq. (22.1), E = EA cos f = 12.0 * 10 3 N>C m 2 21cos 30 2 = 54 N # m 2 >C () The norml to the disk is now perpendiculr to E, so f = 90, cos f = 0, nd E = 0. (c) The norml to the disk is prllel to E, so f = 0 nd cos f = 1: E = EA cos f = 12.0 * 10 3 N>C m = 63 N # m 2 >C EVALUATE: As check on our results, note tht our nswer to prt () is smller thn tht to prt (), which is in turn smller thn tht to prt (c). Is ll this s it should e? Exmple 22.2 Electric flux through cue An imginry cuicl surfce of side L is in region of uniform nd nn 2 is 0, nd the ngle etween nd ech of the other four electric field E. Find the electric flux through ech fce of the cue nd the totl flux through the cue when () it is oriented with two of its fces perpendiculr to E (Fig. 22.8) nd () the cue is turned y n ngle u out verticl xis (Fig. 22.8). OLUTION unit vectors is 90. Ech fce of the cue hs re L 2, so the fluxes through the fces re E1 = E # nn 1 A = EL 2 cos180 = -EL 2 E2 = E # nn 2 A = EL 2 cos0 = EL 2 E3 = E4 = E5 = E6 = EL 2 cos90 = 0 IDENTIFY nd ET UP: ince E is uniform nd ech of the six The flux is negtive on fce 1, where E is directed into the cue, fces of the cue is flt, we find the flux Ei through ech fce nd positive on fce 2, where E is directed out of the cue. The using Eqs. (22.3) nd (22.4). The totl flux through the cue is the totl flux through the cue is sum of the six individul fluxes. E = E1 E2 E3 E4 E5 E6 EXECUTE: () Figure 22.8 shows the unit vectors nn 1 through nn 6 for = -EL ech fce; ech unit vector points outwrd from the cue s closed EL = 0 surfce. The ngle etween E nd nn is 180, the ngle etween E 1 () The field E is directed into fces 1 nd 3, so the fluxes through them re negtive; E is directed out of fces 2 nd 4, so the fluxes through them re positive. We find 22.8 Electric flux of uniform field through cuicl ox of side L in two orienttions. () ^n 3 ^n 1 ^n 5 ^n 6 ^n 4 E r ^n 2 () E ^n 5 ^n 3 ^n 2 u ^n 1 ^n 4 ^n 6 E r 90 2u E1 = E # nn 1 A = EL 2 cos u2 = -EL 2 cos u E2 = E # nn 2 A = EL 2 cos u E3 = E # nn 3 A = EL 2 cos 190 u2 = -EL 2 sin u E4 = E # nn 4 A = EL 2 cos u2 = EL 2 sin u E5 = E6 = EL 2 cos90 = 0 The totl flux E = E1 E2 E3 E4 E5 E6 through the surfce of the cue is gin zero. EVALUATE: We cme to the sme conclusion in our discussion of Fig. 22.3c: There is zero net flux of uniform electric field through closed surfce tht contins no electric chrge. E Exmple 22.3 Electric flux through sphere A point chrge q = 3.0 mc is surrounded y n imginry sphere of rdius r = 0.20 m centered on the chrge (Fig. 22.9). Find the resulting electric flux through the sphere. OLUTION IDENTIFY nd ET UP: The surfce is not flt nd the electric field is not uniform, so to clculte the electric flux (our trget vrile) we must use the generl definition, Eq. (22.5). We use Eq. (22.5) to clculte the electric flux (our trget vrile). Becuse the sphere is centered on the point chrge, t ny point on the sphericl surfce, E is directed out of the sphere perpendiculr to the surfce. The positive direction for oth nn nd E is outwrd, so E nd the flux through surfce element is E = E da # da = EdA. This gretly simplifies the integrl in Eq. (22.5). Continued

46 732 CHAPTER 22 Guss s Lw 22.9 Electric flux through sphere centered on point chrge. da r q E EXECUTE: We must evlute the integrl of Eq. (22.5), E = 1 EdA. At ny point on the sphere of rdius r the electric field hs the sme mgnitude E = q>4pp 0 r 2. Hence E cn e tken outside the integrl, which ecomes E = E 1 da = EA, where A is the re of the sphericl surfce: through the sphere is q E = EA = 4pP 0 r 2 4pr 2 = q P 0 = A = 4pr 2. Hence the totl flux 3.0 * 10-6 C 8.85 * C 2 >N # m 2 = 3.4 * 105 N # m 2 >C EVALUATE: The rdius r of the sphere cncels out of the result for E. We would hve otined the sme flux with sphere of rdius 2.0 m or 200 m. We cme to essentilly the sme conclusion in our discussion of Fig in ection 22.1, where we considered rectngulr closed surfces of two different sizes enclosing point chrge. There we found tht the flux of E ws independent of the size of the surfce; the sme result holds true for sphericl surfce. Indeed, the flux through ny surfce enclosing single point chrge is independent of the shpe or size of the surfce, s we ll soon see. Test Your Understnding of ection 22.2 Rnk the following surfces in order from most positive to most negtive electric flux. (i) flt rectngulr surfce with vector re A 16.0 m 2 2ıN in uniform electric field E 14.0 N>C2 N; (ii) flt circulr surfce with vector re A 13.0 m 2 2 N in uniform electric field E 14.0 N>C2ıN 12.0 N>C2 N; (iii) flt squre surfce with vector re A 13.0 m 2 2ıN 17.0 m 2 2 N in uniform electric field E 14.0 N>C2ıN 12.0 N>C2 N; (iv) flt ovl surfce with vector re A 13.0 m 2 2ıN 17.0 m 2 2 N in uniform electric field E 14.0 N>C2ıN 12.0 N>C2 N Guss s Lw Crl Friedrich Guss helped develop severl rnches of mthemtics, including differentil geometry, rel nlysis, nd numer theory. The ell curve of sttistics is one of his inventions. Guss lso mde stte-of-the-rt investigtions of the erth s mgnetism nd clculted the orit of the first steroid to e discovered. Guss s lw is n lterntive to Coulom s lw. While completely equivlent to Coulom s lw, Guss s lw provides different wy to express the reltionship etween electric chrge nd electric field. It ws formulted y Crl Friedrich Guss ( ), one of the gretest mthemticins of ll time (Fig ). Point Chrge Inside phericl urfce Guss s lw sttes tht the totl electric flux through ny closed surfce ( surfce enclosing definite volume) is proportionl to the totl (net) electric chrge inside the surfce. In ection 22.1 we oserved this reltionship qulittively for certin specil cses; now we ll develop it more rigorously. We ll strt with the field of single positive point chrge q. The field lines rdite out eqully in ll directions. We plce this chrge t the center of n imginry sphericl surfce with rdius R. The mgnitude E of the electric field t every point on the surfce is given y At ech point on the surfce, is perpendiculr to the surfce, nd its mgnitude is the sme t every point, just s in Exmple 22.3 (ection 22.2). The totl electric flux is the product of the field mgnitude E nd the totl re A = 4pR 2 of the sphere: E 1 q E = 4pP 0 1 q E = EA = (22.6) 4pP 0 R 2 14pR2 2 = q P 0 The flux is independent of the rdius R of the sphere. It depends only on the chrge q enclosed y the sphere. R 2

47 22.3 Guss s Lw 733 We cn lso interpret this result in terms of field lines. Figure shows two spheres with rdii R nd 2R centered on the point chrge q. Every field line tht psses through the smller sphere lso psses through the lrger sphere, so the totl flux through ech sphere is the sme. Wht is true of the entire sphere is lso true of ny portion of its surfce. In Fig n re da is outlined on the sphere of rdius R nd then projected onto the sphere of rdius 2R y drwing lines from the center through points on the oundry of da. The re projected on the lrger sphere is clerly 4 da. But since the electric field due to point chrge is inversely proportionl to r 2, the 1 field mgnitude is 4 s gret on the sphere of rdius 2R s on the sphere of rdius R. Hence the electric flux is the sme for oth res nd is independent of the rdius of the sphere Projection of n element of re da of sphere of rdius R onto concentric sphere of rdius 2R. The projection multiplies ech liner dimension y 2, so the re element on the lrger sphere is 4 da. The sme numer of field lines nd the sme flux pss through oth of these re elements. E Point Chrge Inside Nonsphericl urfce This projection technique shows us how to extend this discussion to nonsphericl surfces. Insted of second sphere, let us surround the sphere of rdius R y surfce of irregulr shpe, s in Fig Consider smll element of re da on the irregulr surfce; we note tht this re is lrger thn the corresponding element on sphericl surfce t the sme distnce from q. If norml to da mkes n ngle f with rdil line from q, two sides of the re projected onto the sphericl surfce re foreshortened y fctor cos f (Fig ). The other two sides re unchnged. Thus the electric flux through the sphericl surfce element is equl to the flux EdAcos f through the corresponding irregulr surfce element. We cn divide the entire irregulr surfce into elements da, compute the electric flux EdAcos f for ech, nd sum the results y integrting, s in Eq. (22.5). Ech of the re elements projects onto corresponding sphericl surfce element. Thus the totl electric flux through the irregulr surfce, given y ny of the forms of Eq. (22.5), must e the sme s the totl flux through sphere, which Eq. (22.6) shows is equl to q>p 0. Thus, for the irregulr surfce, 2R R q da 4 da E = C E # da = q P 0 (22.7) Eqution (22.7) holds for surfce of ny shpe or size, provided only tht it is closed surfce enclosing the chrge q. The circle on the integrl sign reminds us tht the integrl is lwys tken over closed surfce. The re elements da nd the corresponding unit vectors nn lwys point out of the volume enclosed y the surfce. The electric flux is then positive in res () The outwrd norml to the surfce mkes n ngle f with the direction of E. E ' f E () E ' f E Clculting the electric flux through nonsphericl surfce. da da r f da cos f R q q The projection of the re element da onto the sphericl surfce is da cos f.

48 734 CHAPTER 22 Guss s Lw where the electric field points out of the surfce nd negtive where it points inwrd. Also, E is positive t points where E points out of the surfce nd negtive t points where E points into the surfce. If the point chrge in Fig is negtive, the E field is directed rdilly inwrd; the ngle f is then greter thn 90, its cosine is negtive, nd the integrl in Eq. (22.7) is negtive. But since q is lso negtive, Eq. (22.7) still holds. For closed surfce enclosing no chrge, E = C E # da = A point chrge outside closed surfce tht encloses no chrge. If n electric field line from the externl chrge enters the surfce t one point, it must leve t nother. E This is the mthemticl sttement tht when region contins no chrge, ny field lines cused y chrges outside the region tht enter on one side must leve gin on the other side. (In ection 22.1 we cme to the sme conclusion y considering the specil cse of rectngulr ox in uniform field.) Figure illustrtes this point. Electric field lines cn egin or end inside region of spce only when there is chrge in tht region. Field line entering surfce me field line leving surfce Generl Form of Guss s Lw Now comes the finl step in otining the generl form of Guss s lw. uppose the surfce encloses not just one point chrge q ut severl chrges q 1, q 2, q The totl (resultnt) electric field E t ny point is the vector sum of the E 3, Á. fields of the individul chrges. Let e the totl chrge enclosed y the surfce: Q Also let E encl = q 1 q 2 q 3 Á Q encl. e the totl field t the position of the surfce re element da, nd let E e its component perpendiculr to the plne of tht element (tht is, prllel to da ). Then we cn write n eqution like Eq. (22.7) for ech chrge nd its corresponding field nd dd the results. When we do, we otin the generl sttement of Guss s lw: E = C E # da = Q encl P 0 (Guss s lw) (22.8) The totl electric flux through closed surfce is equl to the totl (net) electric chrge inside the surfce, divided y `0. CAUTION Gussin surfces re imginry Rememer tht the closed surfce in Guss s lw is imginry; there need not e ny mteril oject t the position of the surfce. We often refer to closed surfce used in Guss s lw s Gussin surfce. Using the definition of Q encl nd the vrious wys to express electric flux given in Eq. (22.5), we cn express Guss s lw in the following equivlent forms: E = C Ecos f da = C E da = C E # da = Q encl P 0 (vrious forms of Guss s lw) (22.9) As in Eq. (22.5), the vrious forms of the integrl ll express the sme thing, the totl electric flux through the Gussin surfce, in different terms. One form is sometimes more convenient thn nother. As n exmple, Fig shows sphericl Gussin surfce of rdius r round positive point chrge q. The electric field points out of the Gussin surfce, so t every point on the surfce E is in the sme direction s da, f = 0, nd E is equl to the field mgnitude E = q>4pp 0 r 2. ince E is the sme t ll points

49 22.3 Guss s Lw 735 () Gussin surfce round positive chrge: positive (outwrd) flux da () Gussin surfce round negtive chrge: negtive (inwrd) flux da phericl Gussin surfces round () positive point chrge nd () negtive point chrge. 1q r 2q r E E on the surfce, we cn tke it outside the integrl in Eq. (22.9). Then the remining integrl is 1 da = A = 4pr 2, the re of the sphere. Hence Eq. (22.9) ecomes q E = E da = C C 4pP 0 r 2 da = q 4pP 0 r 2 C da = q 4pP 0 r 2 4pr 2 = q P 0 The enclosed chrge Q encl is just the chrge q, so this grees with Guss s lw. If the Gussin surfce encloses negtive point chrge s in Fig , then E points into the surfce t ech point in the direction opposite da. Then f = 180 nd E is equl to the negtive of the field mgnitude: E = -E = - ƒ -qƒ>4pp 0 r 2 = -q>4pp 0 r 2. Eqution (22.9) then ecomes -q E = E da = C C 4pP 0 r 2 da = -q 4pP 0 r 2 C da = -q 4pP 0 r 2 4pr 2 = -q This gin grees with Guss s lw ecuse the enclosed chrge in Fig is Q encl = -q. In Eqs. (22.8) nd (22.9), Q encl is lwys the lgeric sum of ll the positive nd negtive chrges enclosed y the Gussin surfce, nd E is the totl field t ech point on the surfce. Also note tht in generl, this field is cused prtly y chrges inside the surfce nd prtly y chrges outside. But s Fig shows, the outside chrges do not contriute to the totl (net) flux through the surfce. o Eqs. (22.8) nd (22.9) re correct even when there re chrges outside the surfce tht contriute to the electric field t the surfce. When Q encl = 0, the totl flux through the Gussin surfce must e zero, even though some res my hve positive flux nd others my hve negtive flux (see Fig. 22.3). Guss s lw is the definitive nswer to the question we posed t the eginning of ection 22.1: If the electric field pttern is known in given region, wht cn we determine out the chrge distriution in tht region? It provides reltionship etween the electric field on closed surfce nd the chrge distriution within tht surfce. But in some cses we cn use Guss s lw to nswer the reverse question: If the chrge distriution is known, wht cn we determine out the electric field tht the chrge distriution produces? Guss s lw my seem like n unppeling wy to ddress this question, since it my look s though evluting the integrl in Eq. (22.8) is hopeless tsk. ometimes it is, ut other times it is surprisingly esy. Here s n exmple in which no integrtion is involved t ll; we ll work out severl more exmples in the next section. P 0 Conceptul Exmple 22.4 Electric flux nd enclosed chrge Figure shows the field produced y two point chrges q OLUTION nd -q (n electric dipole). Find the electric flux through ech of the closed surfces A, B, C, nd D. Guss s lw, Eq. (22.8), sys tht the totl electric flux through closed surfce is equl to the totl enclosed chrge divided y P 0. In Continued

50 736 CHAPTER 22 Guss s Lw Fig , surfce A (shown in red) encloses the positive chrge, so Q encl = q; surfce B (in lue) encloses the negtive chrge, so Q encl = -q; surfce C (in purple) encloses oth chrges, so Q encl = q 1-q2 = 0; nd surfce D (in yellow) encloses no chrges, so Q encl = 0. Hence, without hving to do ny integrtion, we hve EA = q>p 0, EB = -q>p 0, nd EC = ED = 0. These results depend only on the chrges enclosed within ech Gussin surfce, not on the precise shpes of the surfces. We cn drw similr conclusions y exmining the electric field lines. All the field lines tht cross surfce A re directed out of the surfce, so the flux through A must e positive. imilrly, the flux through B must e negtive since ll of the field lines tht cross tht surfce point inwrd. For oth surfce C nd surfce D, there re s mny field lines pointing into the surfce s there re field lines pointing outwrd, so the flux through ech of these surfces is zero The net numer of field lines leving closed surfce is proportionl to the totl chrge enclosed y tht surfce. 2q B 1q E D A C Five Gussin surfces nd six point chrges mc mc 19.0 mc 18.0 mc 11.0 mc mc Test Your Understnding of ection 22.3 Figure shows six point chrges tht ll lie in the sme plne. Five Gussin surfces 1, 2, 3, 4, nd 5 ech enclose prt of this plne, nd Fig shows the intersection of ech surfce with the plne. Rnk these five surfces in order of the electric flux through them, from most positive to most negtive Applictions of Guss s Lw Under electrosttic conditions (chrges not in motion), ny excess chrge on solid conductor resides entirely on the conductor s surfce. Gussin surfce A inside conductor (shown in cross section) Chrge on surfce of conductor Conductor (shown in cross section) Guss s lw is vlid for ny distriution of chrges nd for ny closed surfce. Guss s lw cn e used in two wys. If we know the chrge distriution, nd if it hs enough symmetry to let us evlute the integrl in Guss s lw, we cn find the field. Or if we know the field, we cn use Guss s lw to find the chrge distriution, such s chrges on conducting surfces. In this section we present exmples of oth kinds of pplictions. As you study them, wtch for the role plyed y the symmetry properties of ech system. We will use Guss s lw to clculte the electric fields cused y severl simple chrge distriutions; the results re collected in tle in the chpter summry. In prcticl prolems we often encounter situtions in which we wnt to know the electric field cused y chrge distriution on conductor. These clcultions re ided y the following remrkle fct: When excess chrge is plced on solid conductor nd is t rest, it resides entirely on the surfce, not in the interior of the mteril. (By excess we men chrges other thn the ions nd free electrons tht mke up the neutrl conductor.) Here s the proof. We know from ection 21.4 tht in n electrosttic sitution (with ll chrges t rest) the electric field E t every point in the interior of conducting mteril is zero. If E were not zero, the excess chrges would move. uppose we construct Gussin surfce inside the conductor, such s surfce A in Fig Becuse E 0 everywhere on this surfce, Guss s lw requires tht the net chrge inside the surfce is zero. Now imgine shrinking the surfce like collpsing lloon until it encloses region so smll tht we my consider it s point P; then the chrge t tht point must e zero. We cn do this nywhere inside the conductor, so there cn e no excess chrge t ny point within solid conductor; ny excess chrge must reside on the conductor s surfce. (This result is for solid conductor. In the next section we ll discuss wht cn hppen if the conductor hs cvities in its interior.) We will mke use of this fct frequently in the exmples tht follow.

51 22.4 Applictions of Guss s Lw 737 Prolem-olving trtegy 22.1 Guss s Lw IDENTIFY the relevnt concepts: Guss s lw is most useful when the chrge distriution hs sphericl, cylindricl, or plnr symmetry. In these cses the symmetry determines the direction of E. Then Guss s lw yields the mgnitude of E if we re given the chrge distriution, nd vice vers. In either cse, egin the nlysis y sking the question: Wht is the symmetry? ET UP the prolem using the following steps: 1. List the known nd unknown quntities nd identify the trget vrile. 2. elect the pproprite closed, imginry Gussin surfce. For sphericl symmetry, use concentric sphericl surfce. For cylindricl symmetry, use coxil cylindricl surfce with flt ends perpendiculr to the xis of symmetry (like soup cn). For plnr symmetry, use cylindricl surfce (like tun cn) with its flt ends prllel to the plne. EXECUTE the solution s follows: 1. Determine the pproprite size nd plcement of your Gussin surfce. To evlute the field mgnitude t prticulr point, the surfce must include tht point. It my help to plce one end of cn-shped surfce within conductor, where E nd therefore re zero, or to plce its ends equidistnt from chrged plne. 2. Evlute the integrl A E da in Eq. (22.9). In this eqution E is the perpendiculr component of the totl electric field t ech point on the Gussin surfce. A well-chosen Gussin surfce should mke integrtion trivil or unnecessry. If the surfce comprises severl seprte surfces, such s the sides nd ends of cylinder, the integrl A E da over the entire closed surfce is the sum of the integrls 1 E da over the seprte surfces. Consider points 36 s you work. 3. If E is perpendiculr (norml) t every point to surfce with re A, if it points outwrd from the interior of the surfce, nd if it hs the sme mgnitude t every point on the surfce, then E = E = constnt, nd 1 E over tht surfce is equl to (If E da EA. is inwrd, then E = -E nd 1 E da = -EA.) This should e the cse for prt or ll of your Gussin surfce. If E is tngent to surfce t every point, then E = 0 nd the integrl over tht surfce is zero. This my e the cse for prts of cylindricl Gussin surfce. If E 0 t every point on surfce, the integrl is zero. 4. Even when there is no chrge within Gussin surfce, the field t ny given point on the surfce is not necessrily zero. In tht cse, however, the totl electric flux through the surfce is lwys zero. 5. The flux integrl A E da cn e pproximted s the difference etween the numers of electric lines of force leving nd entering the Gussin surfce. In this sense the flux gives the sign of the enclosed chrge, ut is only proportionl to it; zero flux corresponds to zero enclosed chrge. 6. Once you hve evluted A E da, use Eq. (22.9) to solve for your trget vrile. EVALUATE your nswer: If your result is function tht descries how the mgnitude of the electric field vries with position, ensure tht it mkes sense. Exmple 22.5 Field of chrged conducting sphere We plce totl positive chrge q on solid conducting sphere with rdius R (Fig ). Find E t ny point inside or outside the sphere Clculting the electric field of conducting sphere with positive chrge q. Outside the sphere, the field is the sme s if ll of the chrge were concentrted t the center of the sphere. 1 E1R2 5 4pP 0 q R 2 Inside the sphere, the electric field is zero: E 5 0. E1R2/4 E1R2/9 R O E Outside the sphere, the mgnitude of the electric field decreses with the squre of the rdil distnce from the center of the sphere: 1 q E 5 4pP r 2 0 R 2R 3R Gussin surfces t r 5 2R nd r 5 3R r OLUTION IDENTIFY nd ET UP: As we discussed erlier in this section, ll of the chrge must e on the surfce of the sphere. The chrge is free to move on the conductor, nd there is no preferred position on the surfce; the chrge is therefore distriuted uniformly over the surfce, nd the system is sphericlly symmetric. To exploit this symmetry, we tke s our Gussin surfce sphere of rdius r centered on the conductor. We cn clculte the field inside or outside the conductor y tking r 6 R or r 7 R, respectively. In either cse, the point t which we wnt to clculte E lies on the Gussin surfce. EXECUTE: The sphericl symmetry mens tht the direction of the electric field must e rdil; tht s ecuse there is no preferred direction prllel to the surfce, so E cn hve no component prllel to the surfce. There is lso no preferred orienttion of the sphere, so the field mgnitude E cn depend only on the distnce r from the center nd must hve the sme vlue t ll points on the Gussin surfce. For r 7 R the entire conductor is within the Gussin surfce, so the enclosed chrge is q. The re of the Gussin surfce is 4pr 2, nd E is uniform over the surfce nd perpendiculr to it t ech point. The flux integrl A E da is then just E14pr 2 2, nd Eq. (22.8) gives Continued

52 738 CHAPTER 22 Guss s Lw E14pr 2 2 = q nd P 0 1 q E = 4pP 0 This expression is the sme s tht for point chrge; outside the chrged sphere, its field is the sme s though the entire chrge were concentrted t its center. Just outside the surfce of the sphere, where r = R, 1 q E = (t the surfce of chrged conducting sphere) 4pP 0 R 2 r 2 (outside chrged conducting sphere) CAUTION Flux cn e positive or negtive Rememer tht we hve chosen the chrge q to e positive. If the chrge is negtive, the electric field is rdilly inwrd insted of rdilly outwrd, nd the electric flux through the Gussin surfce is negtive. The electric-field mgnitudes outside nd t the surfce of the sphere re given y the sme expressions s ove, except tht q denotes the mgnitude (solute vlue) of the chrge. For r 6 R we gin hve E14pr 2 2 = Q encl >P 0. But now our Gussin surfce (which lies entirely within the conductor) encloses no chrge, so Q encl = 0. The electric field inside the conductor is therefore zero. EVALUATE: We lredy knew tht E 0 inside solid conductor (whether sphericl or not) when the chrges re t rest. Figure shows E s function of the distnce r from the center of the sphere. Note tht in the limit s R 0, the sphere ecomes point chrge; there is then only n outside, nd the field is everywhere given y E = q>4pp 0 r 2. Thus we hve deduced Coulom s lw from Guss s lw. (In ection 22.3 we deduced Guss s lw from Coulom s lw; the two lws re equivlent.) We cn lso use this method for conducting sphericl shell ( sphericl conductor with concentric sphericl hole inside) if there is no chrge inside the hole. We use sphericl Gussin surfce with rdius r less thn the rdius of the hole. If there were field inside the hole, it would hve to e rdil nd sphericlly symmetric s efore, so E = Q encl >4pP 0 r 2. But now there is no enclosed chrge, so Q encl = 0 nd E = 0 inside the hole. Cn you use this sme technique to find the electric field in the region etween chrged sphere nd concentric hollow conducting sphere tht surrounds it? Exmple 22.6 Field of uniform line chrge Electric chrge is distriuted uniformly long n infinitely long, thin wire. The chrge per unit length is l (ssumed positive). Find the electric field using Guss s lw. OLUTION IDENTIFY nd ET UP: We found in Exmple (ection 21.5) tht the field E of uniformly chrged, infinite wire is rdilly outwrd if l is positive nd rdilly inwrd if l is negtive, nd tht the field mgnitude E depends only on the rdil distnce from the wire. This suggests tht we use cylindricl Gussin surfce, of rdius r nd ritrry length l, coxil with the wire nd with its ends perpendiculr to the wire (Fig ). EXECUTE: The flux through the flt ends of our Gussin surfce is zero ecuse the rdil electric field is prllel to these ends, nd so E # nn = 0. On the cylindricl prt of our surfce we hve E # nn = E = E everywhere. (If l were negtive, we would hve A coxil cylindricl Gussin surfce is used to find the electric field outside n infinitely long, chrged wire. E ' 5 0 r E ' 5 E da l Gussin surfce E # nn = E = -E everywhere.) The re of the cylindricl surfce is 2prl, so the flux through it nd hence the totl flux E through the Gussin surfce is EA = 2prlE. The totl enclosed chrge is Q encl = ll, nd so from Guss s lw, Eq. (22.8), E = 2prlE = ll E = 1 l 2pP 0 r P 0 nd (field of n infinite line of chrge) We found this sme result in Exmple with much more effort. If l is negtive, E is directed rdilly inwrd, nd in the ove expression for E we must interpret l s the solute vlue of the chrge per unit length. EVALUATE: We sw in Exmple tht the entire chrge on the wire contriutes to the field t ny point, nd yet we consider only tht prt of the chrge Q encl = ll within the Gussin surfce when we pply Guss s lw. There s nothing inconsistent here; it tkes the entire chrge to give the field the properties tht llow us to clculte E so esily, nd Guss s lw lwys pplies to the enclosed chrge only. If the wire is short, the symmetry of the infinite wire is lost, nd E is not uniform over coxil, cylindricl Gussin surfce. Guss s lw then cnnot e used to find E ; we must solve the prolem the hrd wy, s in Exmple We cn use the Gussin surfce in Fig to show tht the field outside long, uniformly chrged cylinder is the sme s though ll the chrge were concentrted on line long its xis (see Prolem 22.42). We cn lso clculte the electric field in the spce etween chrged cylinder nd coxil hollow conducting cylinder surrounding it (see Prolem 22.39).

53 22.4 Applictions of Guss s Lw 739 Exmple 22.7 Field of n infinite plne sheet of chrge Use Guss s lw to find the electric field cused y thin, flt, infinite sheet with uniform positive surfce chrge density s. OLUTION IDENTIFY nd ET UP: In Exmple (ection 21.5) we found tht the field E of uniformly chrged infinite sheet is norml to the sheet, nd tht its mgnitude is independent of the distnce from the sheet. To tke dvntge of these symmetry properties, we use cylindricl Gussin surfce with ends of re A nd with its xis perpendiculr to the sheet of chrge (Fig ) A cylindricl Gussin surfce is used to find the field of n infinite plne sheet of chrge. E ' 5 E A E Gussin surfce EXECUTE: The flux through the cylindricl prt of our Gussin surfce is zero ecuse E # nn = 0 everywhere. The flux through ech flt end of the surfce is EA ecuse E # nn = E = E everywhere, so the totl flux through oth ends nd hence the totl flux E through the Gussin surfce is 2EA. The totl enclosed chrge is Q encl = sa, nd so from Guss s lw, 2EA = sa P 0 E = s 2P 0 nd (field of n infinite sheet of chrge) In Exmple we found this sme result using much more complex clcultion. If s is negtive, E is directed towrd the sheet, the flux through the Gussin surfce in Fig is negtive, nd s in the expression E = s>2p 0 denotes the mgnitude (solute vlue) of the chrge density. EVALUATE: Agin we see tht, given fvorle symmetry, we cn deduce electric fields using Guss s lw much more esily thn using Coulom s lw. Exmple 22.8 Field etween oppositely chrged prllel conducting pltes Two lrge plne prllel conducting pltes re given chrges of equl mgnitude nd opposite sign; the surfce chrge densities re s nd -s. Find the electric field in the region etween the pltes. OLUTION IDENTIFY nd ET UP: Figure shows the field. Becuse opposite chrges ttrct, most of the chrge ccumultes t the opposing fces of the pltes. A smll mount of chrge resides on the outer surfces of the pltes, nd there is some spreding or fringing of the field t the edges. But if the pltes re very lrge in comprison to the distnce etween them, the mount of chrge on the outer surfces is negligily smll, nd the fringing cn e neglected except ner the edges. In this cse we cn ssume tht the field is uniform in the interior region etween the pltes, s in Fig , nd tht the chrges re distriuted uniformly over the opposing surfces. To exploit this symmetry, we cn use the shded Gussin surfces 1, 2, 3, nd 4. These surfces re cylinders with flt ends of re A; one end of ech surfce lies within plte Electric field etween oppositely chrged prllel pltes. () Relistic drwing () Idelized model 1 2 Between the two pltes the electric field is nerly uniform, pointing from the positive plte towrd the negtive one. E E 1 In the idelized cse we ignore fringing t the plte edges nd tret the field etween the pltes s uniform. Cylindricl Gussin surfces (seen from the side) E E E 1 E r E 2 c r E 1 Continued

54 740 CHAPTER 22 Guss s Lw EXECUTE: The left-hnd end of surfce 1 is within the positive plte 1. ince the field is zero within the volume of ny solid conductor under electrosttic conditions, there is no electric flux through this end. The electric field etween the pltes is perpendiculr to the right-hnd end, so on tht end, E is equl to E nd the flux is EA; this is positive, since E is directed out of the Gussin surfce. There is no flux through the side wlls of the cylinder, since these wlls re prllel to E. o the totl flux integrl in Guss s lw is EA. The net chrge enclosed y the cylinder is sa, so Eq. (22.8) yields EA = sa>p 0 ; we then hve E = s (field etween oppositely chrged conducting pltes) P 0 The field is uniform nd perpendiculr to the pltes, nd its mgnitude is independent of the distnce from either plte. The Gussin surfce 4 yields the sme result. urfces 2 nd 3 yield E = 0 to the left of plte 1 nd to the right of plte 2, respectively. We leve these clcultions to you (see Exercise 22.29). EVALUATE: We otined the sme results in Exmple y using the principle of superposition of electric fields. The fields due to the two sheets of chrge (one on ech plte) re E nd E 1 2; from Exmple 22.7, oth of these hve mgnitude s>2p 0. The totl electric field t ny point is the vector sum E E At points nd in Fig , E nd E 1 E 2. c 1 point in opposite directions, nd their sum is zero. At point, E 2 nd E 1 2 re in the sme direction; their sum hs mgnitude E = s>p 0, just s we found ove using Guss s lw. Exmple 22.9 Field of uniformly chrged sphere Positive electric chrge Q is distriuted uniformly throughout the volume of n insulting sphere with rdius R. Find the mgnitude of the electric field t point P distnce r from the center of the sphere. OLUTION IDENTIFY nd ET UP: As in Exmple 22.5, the system is sphericlly symmetric. Hence we cn use the conclusions of tht exmple out the direction nd mgnitude of E. To mke use of the sphericl symmetry, we choose s our Gussin surfce sphere with rdius r, concentric with the chrge distriution. EXECUTE: From symmetry, the direction of E is rdil t every point on the Gussin surfce, so E = E nd the field mgnitude E is the sme t every point on the surfce. Hence the totl electric flux through the Gussin surfce is the product of E nd the totl re of the surfce A = 4pr 2 tht is, E = 4pr 2 E. The mount of chrge enclosed within the Gussin surfce depends on r. To find E inside the sphere, we choose r 6 R. The volume chrge density r is the chrge Q divided y the volume of the entire chrged sphere of rdius R: Q r = 4pR 3 >3 4 The volume V enclosed y the Gussin surfce is 3 pr 3 encl, so the totl chrge Q encl enclosed y tht surfce is Q Q encl = rv encl = 4pR 3 >3 A 4 3 pr 3 B = Q r 3 R 3 Then Guss s lw, Eq. (22.8), ecomes 4pr 2 E = Q r 3 P 0 R 3 or 1 Qr E = 4pP 0 R 3 The field mgnitude is proportionl to the distnce r of the field point from the center of the sphere (see the grph of E versus r in Fig ). To find E outside the sphere, we tke r 7 R. This surfce encloses the entire chrged sphere, so Q encl = Q, nd Guss s lw gives 4pr 2 E = Q or P 0 1 Q E = 4pP 0 r 2 (field inside uniformly chrged sphere) (field outside uniformly chrged sphere) The mgnitude of the electric field of uniformly chrged insulting sphere. Compre this with the field for conducting sphere (see Fig ). r r R 1 Q E(R) 5 4pP R 2 0 O E 1 Qr E 5 4pP R 3 0 R phericl insultor Gussin surfce Q E 5 1 4pP r 2 0 The field outside ny sphericlly symmetric chrged ody vries s 1>r 2, s though the entire chrge were concentrted t the center. This is grphed in Fig If the chrge is negtive, E is rdilly inwrd nd in the expressions for E we interpret Q s the solute vlue of the chrge. EVALUATE: Notice tht if we set r = R in either expression for E, we get the sme result E = Q>4pP 0 R 2 for the mgnitude of the field t the surfce of the sphere. This is ecuse the mgnitude E is continuous function of r. By contrst, for the chrged conducting sphere of Exmple 22.5 the electric-field mgnitude is discontinuous t r = R (it jumps from E = 0 just inside the sphere to E = Q>4pP 0 R 2 just outside the sphere). In generl, the electric field E is discontinuous in mgnitude, direction, or oth wherever there is sheet of chrge, such s t the surfce of chrged conducting sphere (Exmple 22.5), t the surfce of n infinite chrged sheet (Exmple 22.7), or t the surfce of chrged conducting plte (Exmple 22.8). The pproch used here cn e pplied to ny sphericlly symmetric distriution of chrge, even if it is not rdilly uniform, s it ws here. uch chrge distriutions occur within mny toms nd tomic nuclei, so Guss s lw is useful in tomic nd nucler physics. r

55 22.5 Chrges on Conductors 741 Exmple Chrge on hollow sphere A thin-wlled, hollow sphere of rdius m hs n unknown chrge distriuted uniformly over its surfce. At distnce of m from the center of the sphere, the electric field points rdilly inwrd nd hs mgnitude 1.80 * 10 2 N>C. How much chrge is on the sphere? OLUTION IDENTIFY nd ET UP: The chrge distriution is sphericlly symmetric. As in Exmples 22.5 nd 22.9, it follows tht the electric field is rdil everywhere nd its mgnitude is function only of the rdil distnce r from the center of the sphere. We use sphericl Gussin surfce tht is concentric with the chrge distriution nd hs rdius r = m. Our trget vrile is Q encl = q. EXECUTE: The chrge distriution is the sme s if the chrge were on the surfce of m-rdius conducting sphere. Hence we cn orrow the results of Exmple We note tht the electric field here is directed towrd the sphere, so tht q must e negtive. Furthermore, the electric field is directed into the Gussin surfce, so tht E nd -E14pr 2 = -E E = A E da = 2. By Guss s lw, the flux is equl to the chrge q on the sphere (ll of which is enclosed y the Gussin surfce) divided y P 0. olving for q, we find q = -E14pP 0 r 2 2 = * 10 2 N>C214p2 * * C 2 >N # m m2 2 = * 10-9 C = nc EVALUATE: To determine the chrge, we hd to know the electric field t ll points on the Gussin surfce so tht we could clculte the flux integrl. This ws possile here ecuse the chrge distriution is highly symmetric. If the chrge distriution is irregulr or lcks symmetry, Guss s lw is not very useful for clculting the chrge distriution from the field, or vice vers. Test Your Understnding of ection 22.4 You plce known mount of chrge Q on the irregulrly shped conductor shown in Fig If you know the size nd shpe of the conductor, cn you use Guss s lw to clculte the electric field t n ritrry position outside the conductor? 22.5 Chrges on Conductors We hve lerned tht in n electrosttic sitution (in which there is no net motion of chrge) the electric field t every point within conductor is zero nd tht ny excess chrge on solid conductor is locted entirely on its surfce (Fig ). But wht if there is cvity inside the conductor (Fig )? If there is no chrge within the cvity, we cn use Gussin surfce such s A (which lies completely within the mteril of the conductor) to show tht the net chrge on the surfce of the cvity must e zero, ecuse E 0 everywhere on the Gussin surfce. In fct, we cn prove in this sitution tht there cn t e ny chrge nywhere on the cvity surfce. We will postpone detiled proof of this sttement until Chpter 23. uppose we plce smll ody with chrge q inside cvity within conductor (Fig c). The conductor is unchrged nd is insulted from the chrge q. Agin E 0 everywhere on surfce A, so ccording to Guss s lw the totl chrge inside this surfce must e zero. Therefore there must e chrge -q distriuted on the surfce of the cvity, drwn there y the chrge q inside the cvity. The totl chrge on the conductor must remin zero, so chrge q must pper Appliction Chrge Distriution Inside Nerve Cell The interior of humn nerve cell contins oth positive potssium ions ( K ) nd negtively chrged protein molecules ( Pr - ). Potssium ions cn flow out of the cell through the cell memrne, ut the much lrger protein molecules cnnot. The result is tht the interior of the cell hs net negtive chrge. (The fluid outside the cell hs positive chrge tht lnces this.) The fluid within the cell is good conductor, so the Pr - molecules distriute themselves on the outer surfce of the fluid tht is, on the inner surfce of the cell memrne, which is n insultor. This is true no mtter wht the shpe of the cell Finding the electric field within chrged conductor. () olid conductor with chrge q C q C E 5 0 within conductor The chrge q C resides entirely on the surfce of the conductor. The sitution is electrosttic, so E 5 0 within the conductor. () The sme conductor with n internl cvity q C Cvity Aritrry Gussin surfce A Becuse E 5 0 t ll points within the conductor, the electric field t ll points on the Gussin surfce must e zero. (c) An isolted chrge q plced in the cvity q C 1 q q For E to e zero t ll points on the Gussin surfce, the surfce of the cvity must hve totl chrge 2q.

56 742 CHAPTER 22 Guss s Lw either on its outer surfce or inside the mteril. But we showed tht in n electrosttic sitution there cn t e ny excess chrge within the mteril of conductor. o we conclude tht the chrge q must pper on the outer surfce. By the sme resoning, if the conductor originlly hd chrge q C, then the totl chrge on the outer surfce must e q C q fter the chrge q is inserted into the cvity. Conceptul Exmple A conductor with cvity A solid conductor with cvity crries totl chrge of 7 nc. Within the cvity, insulted from the conductor, is point chrge of -5 nc. How much chrge is on ech surfce (inner nd outer) of the conductor? Our sketch for this prolem. There is zero electric field inside the ulk conductor nd hence zero flux through the Gussin surfce shown, so the chrge on the cvity wll must e the opposite of the point chrge. OLUTION Figure shows the sitution. If the chrge in the cvity is q = -5 nc, the chrge on the inner cvity surfce must e -q = -1-5 nc2 = 5 nc. The conductor crries totl chrge of 7 nc, none of which is in the interior of the mteril. If 5 ncis on the inner surfce of the cvity, then there must e 17 nc2-15 nc2 = 2nCon the outer surfce of the conductor. Testing Guss s Lw Experimentlly We cn now consider historic experiment, shown in Fig We mount conducting continer on n insulting stnd. The continer is initilly unchrged. Then we hng chrged metl ll from n insulting thred (Fig ), lower it into the continer, nd put the lid on (Fig ). Chrges re induced on the wlls of the continer, s shown. But now we let the ll touch the inner wll (Fig c). The surfce of the ll ecomes prt of the cvity surfce. The sitution is now the sme s Fig ; if Guss s lw is correct, the net chrge on the cvity surfce must e zero. Thus the ll must lose ll its chrge. Finlly, we pull the ll out; we find tht it hs indeed lost ll its chrge. This experiment ws performed in the 19th century y the English scientist Michel Frdy, using metl icepil with lid, nd it is clled Frdy s icepil experiment. The result confirms the vlidity of Guss s lw nd therefore of () A chrged conducting ll suspended y n insulting thred outside conducting continer on n insulting stnd. () The ll is lowered into the continer, nd the lid is put on. (c) The ll is touched to the inner surfce of the continer. () Insulting thred Metl continer Chrged conducting ll Insulting stnd () Metl lid (c) Metl lid Chrged ll induces chrges on the interior nd exterior of the continer. Once the ll touches the continer, it is prt of the interior surfce; ll the chrge moves to the continer s exterior.

57 22.5 Chrges on Conductors Cutwy view of the essentil prts of Vn de Grff electrosttic genertor. The electron sink t the ottom drws electrons from the elt, giving it positive chrge; t the top the elt ttrcts electrons wy from the conducting shell, giving the shell positive chrge. Conducting shell Insulting elt Insulting support Electron sink Motor for elt Coulom s lw. Frdy s result ws significnt ecuse Coulom s experimentl method, using torsion lnce nd dividing of chrges, ws not very precise; it is very difficult to confirm the 1>r 2 dependence of the electrosttic force y direct force mesurements. By contrst, experiments like Frdy s test the vlidity of Guss s lw, nd therefore of Coulom s lw, with much greter precision. Modern versions of this experiment hve shown tht the exponent 2 in the 1>r 2 of Coulom s lw does not differ from precisely 2 y more thn o there is no reson to elieve it is nything other thn exctly 2. The sme principle ehind Frdy s icepil experiment is used in Vn de Grff electrosttic genertor (Fig ). A chrged elt continuously crries chrge to the inside of conducting shell. By Guss s lw, there cn never e ny chrge on the inner surfce of this shell, so the chrge is immeditely crried wy to the outside surfce of the shell. As result, the chrge on the shell nd the electric field round it cn ecome very lrge very rpidly. The Vn de Grff genertor is used s n ccelertor of chrged prticles nd for physics demonstrtions. This principle lso forms the sis for electrosttic shielding. uppose we hve very sensitive electronic instrument tht we wnt to protect from? stry electric fields tht might cuse erroneous mesurements. We surround the instrument with conducting ox, or we line the wlls, floor, nd ceiling of the room with conducting mteril such s sheet copper. The externl electric field redistriutes the free electrons in the conductor, leving net positive chrge on the outer surfce in some regions nd net negtive chrge in others (Fig ). This chrge distriution cuses n dditionl electric field such tht the totl field t every point inside the ox is zero, s Guss s lw sys it must e. The chrge distriution on the ox lso lters the shpes of the field lines ner the ox, s the figure shows. uch setup is often clled Frdy cge. The sme physics tells

58 744 CHAPTER 22 Guss s Lw () A conducting ox ( Frdy cge) immersed in uniform electric field. The field of the induced chrges on the ox comines with the uniform field to give zero totl field inside the ox. () This person is inside Frdy cge, nd so is protected from the powerful electric dischrge. () () Field pushes electrons towrd left side. Net positive chrge remins on right side. E E 5 0 E Field perpendiculr to conductor surfce you tht one of the sfest plces to e in lightning storm is inside n utomoile; if the cr is struck y lightning, the chrge tends to remin on the metl skin of the vehicle, nd little or no electric field is produced inside the pssenger comprtment The field just outside chrged conductor is perpendiculr to the surfce, nd its perpendiculr component E is equl to s/p 0. Outer surfce of chrged conductor Gussin surfce A A E E E 0 E 0 Field t the urfce of Conductor Finlly, we note tht there is direct reltionship etween the E field t point just outside ny conductor nd the surfce chrge density s t tht point. In generl, s vries from point to point on the surfce. We will show in Chpter 23 tht t ny such point, the direction of E is lwys perpendiculr to the surfce. (You cn see this effect in Fig ) To find reltionship etween s t ny point on the surfce nd the perpendiculr component of the electric field t tht point, we construct Gussin surfce in the form of smll cylinder (Fig ). One end fce, with re A, lies within the conductor nd the other lies just outside. The electric field is zero t ll points within the conductor. Outside the conductor the component of E perpendiculr to the side wlls of the cylinder is zero, nd over the end fce the perpendiculr component is equl to E. (If s is positive, the electric field points out of the conductor nd E is positive; if s is negtive, the field points inwrd nd E is negtive.) Hence the totl flux through the surfce is E A. The chrge enclosed within the Gussin surfce is sa, so from Guss s lw, E A = sa P 0 nd E = s P 0 (field t the surfce of conductor) (22.10) We cn check this with the results we hve otined for sphericl, cylindricl, nd plne surfces. We showed in Exmple 22.8 tht the field mgnitude etween two infinite flt oppositely chrged conducting pltes lso equls s>p 0. In this cse the field mgnitude E is the sme t ll distnces from the pltes, ut in ll other cses E decreses with incresing distnce from the surfce.

59 22.5 Chrges on Conductors 745 Conceptul Exmple Field t the surfce of conducting sphere Verify Eq. (22.10) for conducting sphere with rdius R nd totl chrge q. The surfce chrge density is uniform nd equl to q divided y the surfce re of the sphere: OLUTION s = q 4pR 2 In Exmple 22.5 (ection 22.4) we showed tht the electric field Compring these two expressions, we see tht E = s>p 0, which just outside the surfce is verifies Eq. (22.10). E = 1 q 4pP 0 R 2 Exmple Electric field of the erth The erth ( conductor) hs net electric chrge. The resulting electric field ner the surfce hs n verge vlue of out 150 N>C, directed towrd the center of the erth. () Wht is the corresponding surfce chrge density? () Wht is the totl surfce chrge of the erth? OLUTION IDENTIFY nd ET UP: We re given the electric-field mgnitude t the surfce of the conducting erth. We cn clculte the surfce chrge density s using Eq. (22.10). The totl chrge Q on the erth s surfce is then the product of s nd the erth s surfce re. EXECUTE: () The direction of the field mens tht s is negtive (corresponding to E eing directed into the surfce, so E is negtive). From Eq. (22.10), s =P 0 E = * C 2 >N # m N>C2 = * 10-9 C>m 2 = nc>m 2 () The erth s surfce re is 4pR 2 where R E = 6.38 * 10 6 E, m is the rdius of the erth (see Appendix F). The totl chrge Q is the product 4pRE 2 s, or Q = 4p16.38 * 10 6 m * 10-9 C>m 2 2 = -6.8 * 10 5 C = -680 kc EVALUATE: You cn check our result in prt () using the result of Exmple olving for Q, we find Q = 4pP 0 R 2 E 1 = 9.0 * 10 9 N # m 2 >C * 106 m N>C2 = -6.8 * 10 5 C One electron hs chrge of * C. Hence this much excess negtive electric chrge corresponds to there eing * 10 5 C2> * C2 = 4.2 * excess electrons on the erth, or out 7 moles of excess electrons. This is compensted y n equl deficiency of electrons in the erth s upper tmosphere, so the comintion of the erth nd its tmosphere is electriclly neutrl. Test Your Understnding of ection 22.5 A hollow conducting sphere hs no net chrge. There is positive point chrge q t the center of the sphericl cvity within the sphere. You connect conducting wire from the outside of the sphere to ground. Will you mesure n electric field outside the sphere?

60 CHAPTER 22 UMMARY Electric flux: Electric flux is mesure of the flow of electric field through surfce. It is equl to the product of n re element nd the perpendiculr component of E, integrted over surfce. (ee Exmples ) E = L Ecos f da = L E da = L E # da (22.5) E f f A A A Guss s lw: Guss s lw sttes tht the totl electric flux through closed surfce, which cn e written s the surfce integrl of the component of E norml to the surfce, equls constnt times the totl chrge Q encl enclosed y the surfce. Guss s lw is logiclly equivlent to Coulom s lw, ut its use gretly simplifies prolems with high degree of symmetry. (ee Exmples ) When excess chrge is plced on conductor nd is t rest, it resides entirely on the surfce, nd E 0 everywhere in the mteril of the conductor. (ee Exmples ) E = C Ecos f da = C E da = C E # da = Q encl P 0 (22.8), (22.9) Outwrd norml to surfce E f E r r R q da Electric field of vrious symmetric chrge distriutions: The following tle lists electric fields cused y severl symmetric chrge distriutions. In the tle, q, Q, l, nd s refer to the mgnitudes of the quntities. Point in Electric Field Chrge Distriution Electric Field Mgnitude ingle point chrge q Distnce r from q 1 q E = 4pP 0 r 2 Chrge q on surfce of conducting sphere with rdius R Outside sphere, r 7 R Infinite wire, chrge per unit length l Inside sphere, r 6 R Distnce r from wire Infinite conducting cylinder with rdius R, chrge per Outside cylinder, r 7 R unit length l Inside cylinder, r 6 R olid insulting sphere with rdius R, chrge Q distriuted Outside sphere, r 7 R uniformly throughout volume Infinite sheet of chrge with uniform chrge per unit re s Inside sphere, r 6 R Any point 1 q E = 4pP 0 r 2 E = 0 E = E = E = 0 E = 1 l 2pP 0 r 1 l 2pP 0 r 1 Q 4pP 0 r 2 1 Qr E = 4pP 0 R 3 E = s 2P 0 Two oppositely chrged conducting pltes with surfce chrge densities s nd -s Any point etween pltes E = s P 0 Chrged conductor Just outside the conductor E = s P 0 746

61 Discussion Questions 747 BRIDGING PROBLEM Electric Field Inside Hydrogen Atom A hydrogen tom is mde up of proton of chrge Q = 1.60 * C nd n electron of chrge -Q = * C. The proton my e regrded s point chrge t r = 0, the center of the tom. The motion of the electron cuses its chrge to e smered out into sphericl distriution round the proton, so tht the electron is equivlent to chrge per unit volume of r1r2 = -1Q>p 3, where 0 = 5.29 * e -2r> 0 m is clled the Bohr rdius. () Find the totl mount of the hydrogen tom s chrge tht is enclosed within sphere with rdius r centered on the proton. () Find the electric field (mgnitude nd direction) cused y the chrge of the hydrogen tom s function of r. (c) Mke grph s function of r of the rtio of the electric-field mgnitude E to the mgnitude of the field due to the proton lone. OLUTION GUIDE ee MsteringPhysics study re for Video Tutor solution. IDENTIFY nd ET UP 1. The chrge distriution in this prolem is sphericlly symmetric, just s in Exmple 22.9, so you cn solve it using Guss s lw. 2. The chrge within sphere of rdius r includes the proton chrge Q plus the portion of the electron chrge distriution tht lies within the sphere. The difference from Exmple 22.9 is tht the electron chrge distriution is not uniform, so the chrge enclosed within sphere of rdius r is not simply the chrge density multiplied y the volume 4pr 3 >3 of the sphere. Insted, you ll hve to do n integrl. 3. Consider thin sphericl shell centered on the proton, with rdius r nd infinitesiml thickness dr. ince the shell is so thin, every point within the shell is t essentilly the sme rdius from the proton. Hence the mount of electron chrge within this shell is equl to the electron chrge density r1r 2 t this rdius multiplied y the volume dv of the shell. Wht is dv in terms of r? 4. The totl electron chrge within rdius r equls the integrl of r1r 2dV from r =0 to r =r. et up this integrl (ut don t solve it yet), nd use it to write n expression for the totl chrge (including the proton) within sphere of rdius r. EXECUTE 5. Integrte your expression from step 4 to find the chrge within rdius r. Hint: Integrte y sustitution: Chnge the integrtion vrile from r to x = 2r > 0. You cn clculte the integrl 1 x 2 e -x dx using integrtion y prts, or you cn look it up in tle of integrls or on the World Wide We. 6. Use Guss s lw nd your results from step 5 to find the electric field t distnce r from the proton. 7. Find the rtio referred to in prt (c) nd grph it versus r. (You ll ctully find it simplest to grph this function versus the quntity r> 0.) EVALUATE 8. How do your results for the enclosed chrge nd the electricfield mgnitude ehve in the limit r 0? In the limit r q? Explin your results. Prolems For instructor-ssigned homework, go to : Prolems of incresing difficulty. CP: Cumultive prolems incorporting mteril from erlier chpters. CALC: Prolems requiring clculus. BIO: Biosciences prolems. DICUION QUETION Q22.1 A ruer lloon hs single point chrge in its interior. Does the electric flux through the lloon depend on whether or not it is fully inflted? Explin your resoning. Q22.2 uppose tht in Fig oth chrges were positive. Wht would e the fluxes through ech of the four surfces in the exmple? Q22.3 In Fig , suppose third point chrge were plced outside the purple Gussin surfce C. Would this ffect the electric flux through ny of the surfces A, B, C, or D in the figure? Why or why not? Q22.4 A certin region of spce ounded y n imginry closed surfce contins no chrge. Is the electric field lwys zero everywhere on the surfce? If not, under wht circumstnces is it zero on the surfce? Q22.5 A sphericl Gussin surfce encloses point chrge q. If the point chrge is moved from the center of the sphere to point wy from the center, does the electric field t point on the surfce chnge? Does the totl flux through the Gussin surfce chnge? Explin. Q22.6 You find seled ox on your doorstep. You suspect tht the ox contins severl chrged metl spheres pcked in insulting mteril. How cn you determine the totl net chrge inside the ox without opening the ox? Or isn t this possile? Q22.7 A solid copper sphere hs net positive chrge. The chrge is distriuted uniformly over the surfce of the sphere, nd the electric field inside the sphere is zero. Then negtive point chrge outside the sphere is rought close to the surfce of the sphere. Is ll the net chrge on the sphere still on its surfce? If so, is this chrge still distriuted uniformly over the surfce? If it is not uniform, how is it distriuted? Is the electric field inside the sphere still zero? In ech cse justify your nswers. Q22.8 If the electric field of point chrge were proportionl to 1>r 3 insted of 1>r 2, would Guss s lw still e vlid? Explin your resoning. (Hint: Consider sphericl Gussin surfce centered on single point chrge.) Q22.9 In conductor, one or more electrons from ech tom re free to rom throughout the volume of the conductor. Does this contrdict the sttement tht ny excess chrge on solid conductor must reside on its surfce? Why or why not?

62 748 CHAPTER 22 Guss s Lw Q22.10 You chrge up the vn de Grff genertor shown in Fig , nd then ring n identicl ut unchrged hollow conducting sphere ner it, without letting the two spheres touch. ketch the distriution of chrges on the second sphere. Wht is the net flux through the second sphere? Wht is the electric field inside the second sphere? Q22.11 A lightning rod is rounded copper rod mounted on top of uilding nd welded to hevy copper cle running down into the ground. Lightning rods re used to protect houses nd rns from lightning; the lightning current runs through the copper rther thn through the uilding. Why? Why should the end of the rod e rounded? Q22.12 A solid conductor hs cvity in its interior. Would the presence of point chrge inside the cvity ffect the electric field outside the conductor? Why or why not? Would the presence of point chrge outside the conductor ffect the electric field inside the cvity? Agin, why or why not? Q22.13 Explin this sttement: In sttic sitution, the electric field t the surfce of conductor cn hve no component prllel to the surfce ecuse this would violte the condition tht the chrges on the surfce re t rest. Would this sme sttement e vlid for the electric field t the surfce of n insultor? Explin your nswer nd the reson for ny differences etween the cses of conductor nd n insultor. Q22.14 In certin region of spce, the electric field E is uniform. () Use Guss s lw to prove tht this region of spce must e electriclly neutrl; tht is, the volume chrge density r must e zero. () Is the converse true? Tht is, in region of spce where there is no chrge, must E e uniform? Explin. Q22.15 () In certin region of spce, the volume chrge density r hs uniform positive vlue. Cn E e uniform in this region? Explin. () uppose tht in this region of uniform positive r there is ule within which r = 0. Cn E e uniform within this ule? Explin. EXERCIE ection 22.2 Clculting Electric Flux A flt sheet of pper of re m 2 is oriented so tht the norml to the sheet is t n ngle of 60 to uniform electric field of mgnitude 14 N>C. () Find the mgnitude of the electric flux through the sheet. () Does the nswer to prt () depend on the shpe of the sheet? Why or why not? (c) For wht ngle f etween the norml to the sheet nd the electric field is the mgnitude of the flux through the sheet (i) lrgest nd (ii) smllest? Explin your nswers A flt sheet is in the shpe of rectngle with sides of lengths m nd m. The sheet is immersed in uniform electric field of mgnitude 75.0 N>C tht is directed t 20 from the plne of the sheet (Fig. E22.2). Find the mgnitude of the electric flux through the sheet. Figure E E m m You mesure n electric field of 1.25 * 10 6 N>C t distnce of m from point chrge. There is no other source of electric field in the region other thn this point chrge. () Wht is the electric flux through the surfce of sphere tht hs this chrge t its center nd tht hs rdius m? () Wht is the mgnitude of this chrge? It ws shown in Exmple (ection 21.5) tht the electric field due to n infinite line of chrge is perpendiculr to the line nd hs mgnitude E = l>2pp 0 r. Consider n imginry cylinder with rdius r = m nd length l = m tht hs n infinite line of positive chrge running long its xis. The chrge per unit length on the line is l = 3.00 mc>m. () Wht is the electric flux through the cylinder due to this infinite line of chrge? () Wht is the flux through the cylinder if its rdius is incresed to r = m? (c) Wht is the flux through the cylinder if its length is incresed to l = m? A hemisphericl surfce with rdius r in region of uniform electric field E hs its xis ligned prllel to the direction of the field. Clculte the flux through the surfce The cue in Fig. E22.6 hs sides of length L =10.0 cm. Figure E22.6 The electric field is uniform, z hs mgnitude E = 4.00 * 10 3 N>C, nd is prllel to the xy-plne t n ngle of 53.1 mesured from the x-xis towrd the y-xis. () Wht is the electric flux through ech of the six cue fces 1, 2, 3, 4, 5, nd 6? () Wht is the totl electric flux through ll fces of the cue? 5 (front) 3 (right side) ection 22.3 Guss s Lw BIO As discussed in ection 22.5, humn nerve cells hve net negtive chrge nd the mteril in the interior of the cell is good conductor. If cell hs net chrge of 8.65 pc, wht re the mgnitude nd direction (inwrd or outwrd) of the net flux through the cell oundry? The three smll spheres shown in Fig. E22.8 crry chrges q 1 = 4.00 nc, q 2 = nc, nd q 3 = 2.40 nc. Find the net electric flux through ech of the following closed surfces shown in cross section in the figure: () 1 ; () 2 ; (c) 3 ; (d) 4 ; (e) 5. (f) Do your nswers to prts ()(e) depend on how the chrge is distriuted over ech smll sphere? Why or why not? Figure E q 3 3 q 1 q (left side) A chrged pint is spred in very thin uniform lyer over the surfce of plstic sphere of dimeter 12.0 cm, giving it chrge of mc. Find the electric field () just inside the pint lyer; () just outside the pint lyer; (c) 5.00 cm outside the surfce of the pint lyer A point chrge q 1 = 4.00 nc is locted on the x-xis t x = 2.00 m, nd second point chrge q 2 = nc is on the y-xis t y = 1.00 m. Wht is the totl electric flux due to these two point chrges through sphericl surfce centered t the origin nd with rdius () m, () 1.50 m, (c) 2.50 m? 5 x L urfce L 2 (top) L Wht it encloses q 1 q 2 q 1 nd q 2 q 1 nd q 3 q 1 nd q 2 nd q 3 6 (ck) 4 (ottom) y

63 Exercises A 6.20-mC point chrge is t the center of cue with sides of length m. () Wht is the electric flux through one of the six fces of the cue? () How would your nswer to prt () chnge if the sides were m long? Explin Electric Fields in n Atom. The nuclei of lrge toms, such s urnium, with 92 protons, cn e modeled s sphericlly symmetric spheres of chrge. The rdius of the urnium nucleus is pproximtely 7.4 * m. () Wht is the electric field this nucleus produces just outside its surfce? () Wht mgnitude of electric field does it produce t the distnce of the electrons, which is out 1.0 * m? (c) The electrons cn e modeled s forming uniform shell of negtive chrge. Wht net electric field do they produce t the loction of the nucleus? A point chrge of 5.00 mc is locted on the x-xis t x = 4.00 m, next to sphericl surfce of rdius 3.00 m centered t the origin. () Clculte the mgnitude of the electric field t x = 3.00 m. () Clculte the mgnitude of the electric field t x = m. (c) According to Guss s lw, the net flux through the sphere is zero ecuse it contins no chrge. Yet the field due to the externl chrge is much stronger on the ner side of the sphere (i.e., t x = 3.00 m) thn on the fr side (t x = m). How, then, cn the flux into the sphere (on the ner side) equl the flux out of it (on the fr side)? Explin. A sketch will help. ection 22.4 Applictions of Guss s Lw nd ection 22.5 Chrges on Conductors A solid metl sphere with rdius m crries net chrge of nc. Find the mgnitude of the electric field () t point m outside the surfce of the sphere nd () t point inside the sphere, m elow the surfce Two very long uniform lines of chrge re prllel nd re seprted y m. Ech line of chrge hs chrge per unit length 5.20 mc>m. Wht mgnitude of force does one line of chrge exert on m section of the other line of chrge? ome plnetry scientists hve suggested tht the plnet Mrs hs n electric field somewht similr to tht of the erth, producing net electric flux of 3.63 * N # m 2 >C t the plnet s surfce, directed towrd the center of the plnet. Clculte: () the totl electric chrge on the plnet; () the electric field t the plnet s surfce (refer to the stronomicl dt inside the ck cover); (c) the chrge density on Mrs, ssuming ll the chrge is uniformly distriuted over the plnet s surfce How mny excess electrons must e dded to n isolted sphericl conductor 32.0 cm in dimeter to produce n electric field of 1150 N>C just outside the surfce? The electric field m from very long uniform line of chrge is 840 N>C. How much chrge is contined in 2.00-cm section of the line? A very long uniform line of chrge hs chrge per unit length 4.80 mc>m nd lies long the x-xis. A second long uniform line of chrge hs chrge per unit length mc>m nd is prllel to the x-xis t y = m. Wht is the net electric field (mgnitude nd direction) t the following points on the y-xis: () y = m nd () y = m? () At distnce of cm from the center of chrged conducting sphere with rdius cm, the electric field is 480 N>C. Wht is the electric field cm from the center of the sphere? () At distnce of cm from the xis of very long chrged conducting cylinder with rdius cm, the electric field is 480 N>C. Wht is the electric field cm from the xis of the cylinder? (c) At distnce of cm from lrge uniform sheet of chrge, the electric field is 480 N>C. Wht is the electric field 1.20 cm from the sheet? A hollow, conducting sphere with n outer rdius of m nd n inner rdius of m hs uniform surfce chrge density of 6.37 * 10-6 C>m 2. A chrge of mc is now introduced into the cvity inside the sphere. () Wht is the new chrge density on the outside of the sphere? () Clculte the strength of the electric field just outside the sphere. (c) Wht is the electric flux through sphericl surfce just inside the inner surfce of the sphere? A point chrge of mc is locted in the center of sphericl cvity of rdius 6.50 cm inside n insulting chrged solid. The chrge density in the solid is r = 7.35 * 10-4 C>m 3. Clculte the electric field inside the solid t distnce of 9.50 cm from the center of the cvity The electric field t distnce of m from the surfce of solid insulting sphere with rdius m is 1750 N>C. () Assuming the sphere s chrge is uniformly distriuted, wht is the chrge density inside it? () Clculte the electric field inside the sphere t distnce of m from the center CP A very smll oject with mss 8.20 * 10-9 kg nd positive chrge 6.50 * 10-9 C is projected directly towrd very lrge insulting sheet of positive chrge tht hs uniform surfce chrge density 5.90 * 10-8 C>m 2. The oject is initilly m from the sheet. Wht initil speed must the oject hve in order for its closest distnce of pproch to the sheet to e m? CP At time t = 0 proton is distnce of m from very lrge insulting sheet of chrge nd is moving prllel to the sheet with speed 9.70 * 10 2 m>s. The sheet hs uniform surfce chrge density 2.34 * 10-9 C>m 2. Wht is the speed of the proton t t = 5.00 * 10-8 s? CP An electron is relesed from rest t distnce of m from lrge insulting sheet of chrge tht hs uniform surfce chrge density 2.90 * C>m 2. () How much work is done on the electron y the electric field of the sheet s the electron moves from its initil position to point m from the sheet? () Wht is the speed of the electron when it is m from the sheet? CP CALC An insulting sphere of rdius R = m hs uniform chrge density r = 7.20 * 10-9 C>m 3. A smll oject tht cn e treted s point chrge is relesed from rest just outside the surfce of the sphere. The smll oject hs positive chrge q = 3.40 * 10-6 C. How much work does the electric field of the sphere do on the oject s the oject moves to point very fr from the sphere? A conductor with n inner cvity, like tht shown in Fig c, crries totl chrge of 5.00 nc. The chrge within the cvity, insulted from the conductor, is nc. How much chrge is on () the inner surfce of the conductor nd () the outer surfce of the conductor? Apply Guss s lw to the Gussin surfces 2, 3, nd 4 in Fig to clculte the electric field etween nd outside the pltes A squre insulting sheet 80.0 cm on side is held horizontlly. The sheet hs 7.50 nc of chrge spred uniformly over its re. () Clculte the electric field t point mm ove the center of the sheet. () Estimte the electric field t point 100 m ove the center of the sheet. (c) Would the nswers to prts () nd () e different if the sheet were mde of conducting mteril? Why or why not? An infinitely long cylindricl conductor hs rdius R nd uniform surfce chrge density s. () In terms of s nd R, wht is the chrge per unit length l for the cylinder? () In terms of s, wht is the mgnitude of the electric field produced y the chrged cylinder t distnce r 7 R from its xis? (c) Express the result of prt () in terms of l nd show tht the electric field outside the cylinder is the

64 750 CHAPTER 22 Guss s Lw sme s if ll the chrge were on the xis. Compre your result to the result for line of chrge in Exmple 22.6 (ection 22.4) Two very lrge, nonconducting plstic sheets, ech 10.0 cm thick, Figure E22.32 crry uniform chrge densities s 1, s 1 s 2 s 3 s 4 s 2, s 3, nd s 4 on their surfces, s shown in Fig. E These surfce chrge densities hve the vlues s 1 = mc>m 2 s 2 = 5.00 mc>m 2 A B C -6.00,, s 3 = 2.00 mc>m 2, nd s 4 = 4.00 mc>m 2. Use Guss s lw to find the mgnitude nd direction of the electric 10 cm 12 cm 10 cm field t the following points, fr from the edges of these sheets: () point A, 5.00 cm from the left fce of the left-hnd sheet; () point B, 1.25 cm from the inner surfce of the right-hnd sheet; (c) point C, in the middle of the right-hnd sheet A negtive chrge -Q is plced inside the cvity of hollow metl solid. The outside of the solid is grounded y connecting conducting wire etween it nd the erth. () Is there ny excess chrge induced on the inner surfce of the piece of metl? If so, find its sign nd mgnitude. () Is there ny excess chrge on the outside of the piece of metl? Why or why not? (c) Is there n electric field in the cvity? Explin. (d) Is there n electric field within the metl? Why or why not? Is there n electric field outside the piece of metl? Explin why or why not. (e) Would someone outside the solid mesure n electric field due to the chrge -Q? Is it resonle to sy tht the grounded conductor hs shielded the region from the effects of the chrge -Q? In principle, could the sme thing e done for grvity? Why or why not? PROBLEM A cue hs sides of length L = m. It is plced with one corner t the origin s shown in Fig. E22.6. The electric field is not uniform ut is given y E N>C # m2xın N>C # m2zk N. () Find the electric flux through ech of the six cue fces 1, 2, 3, 4, 5, nd 6. () Find the totl electric chrge inside the cue The electric field E in Fig. P22.35 is everywhere prllel Figure P22.35 z to the x-xis, so the components E y nd E z re zero. The x-component of the field E x depends on x ut not 3.0 m on y nd z. At points in the yz-plne y (where x = 0), E x = 125 N>C. () Wht is the electric flux through surfce I in Fig. P22.35? () Wht is the electric flux through surfce 2.0 m E II? (c) The volume shown in the figure is smll section of 1.0 m O x very lrge insulting sl 1.0 m thick. If there is totl chrge of nc within the volume shown, wht re the mgnitude nd direction of E t the fce opposite surfce I? (d) Is the electric field produced only y chrges within the sl, or is the field lso due to chrges outside the sl? How cn you tell? CALC In region of spce there is n electric field E tht is in the z-direction nd tht hs mgnitude E = 1964 N>1C # m22x. Find the flux for this field through squre in the xy-plne t z = 0 nd with side length m. One side of the squre is long the x-xis nd nother side is long the y-xis. II I The electric field E 1 t Figure P22.37 one fce of prllelepiped is uniform over the entire fce nd is 2 E directed out of the fce. At the opposite fce, the electric field E 2 is lso 6.00 cm uniform over the entire fce nd is 5.00 E directed into tht fce (Fig. P22.37) cm The two fces in question re inclined t 30.0 from the horizontl, while E nd E re oth horizontl; E 1 2 hs mgnitude of nd E * 10 4 N>C, hs mgnitude of 7.00 * N>C. () Assuming tht no other electric field lines cross the surfces of the prllelepiped, determine the net chrge contined within. () Is the electric field produced only y the chrges within the prllelepiped, or is the field lso due to chrges outside the prllelepiped? How cn you tell? A long line crrying uniform liner chrge density 50.0 mc>m runs prllel to nd 10.0 cm from the surfce of lrge, flt plstic sheet tht hs uniform surfce chrge density of -100 mc>m 2 on one side. Find the loction of ll points where n prticle would feel no force due to this rrngement of chrged ojects The Coxil Cle. A long coxil cle consists of n inner cylindricl conductor with rdius nd n outer coxil cylinder with inner rdius nd outer rdius c. The outer cylinder is mounted on insulting supports nd hs no net chrge. The inner cylinder hs uniform positive chrge per unit length l. Clculte the electric field () t ny point etween the cylinders distnce r from the xis nd () t ny point outside the outer cylinder. (c) Grph the mgnitude of the electric field s function of the distnce r from the xis of the cle, from r = 0 to r = 2c. (d) Find the chrge per unit length on the inner surfce nd on the outer surfce of the outer cylinder A very long conducting tue (hollow cylinder) hs inner rdius nd outer rdius. It crries chrge per unit length, where is positive constnt with units of C>m. A line of chrge lies long the xis of the tue. The line of chrge hs chrge per unit length. () Clculte the electric field in terms of nd the distnce r from the xis of the tue for (i) r 6 ; (ii) 6 r 6 ; (iii) r 7. how your results in grph of E s function of r. () Wht is the chrge per unit length on (i) the inner surfce of the tue nd (ii) the outer surfce of the tue? Repet Prolem 22.40, ut now let the conducting tue hve chrge per unit length -. As in Prolem 22.40, the line of chrge hs chrge per unit length A very long, solid cylinder with rdius R hs positive chrge uniformly distriuted throughout it, with chrge per unit volume r. () Derive the expression for the electric field inside the volume t distnce r from the xis of the cylinder in terms of the chrge density r. () Wht is the electric field t point outside the volume in terms of the chrge per unit length l in the cylinder? Figure P22.43 (c) Compre the nswers to prts () nd () for r = R. (d) Grph the electric-field mgnitude s function of r from r = 0 to u r = 3R CP A smll sphere with mss of 4.00 * 10-6 kg nd crrying chrge of 5.00 * 10-8 C hngs from thred ner very lrge, chrged insulting

65 Prolems 751 sheet, s shown in Fig. P The chrge density on the surfce of the sheet is uniform nd equl to 2.50 * 10-9 C>m 2. Find the ngle of the thred A phere in phere. A solid conducting sphere crrying chrge q hs rdius. It is inside concentric hollow conducting sphere with inner rdius nd outer rdius c. The hollow sphere hs no net chrge. () Derive expressions for the electricfield mgnitude in terms of the distnce r from the center for the regions r 6, 6 r 6, 6 r 6 c, nd r 7 c. () Grph the mgnitude of the electric field s function of r from r = 0 to r = 2c. (c) Wht is the chrge on the inner surfce of the hollow sphere? (d) On the outer surfce? (e) Represent the chrge of the smll sphere y four plus signs. ketch the field lines of the system within sphericl volume of rdius 2c A solid conducting sphere with rdius R tht crries positive chrge Q is concentric with very thin insulting shell of rdius 2R tht lso crries chrge Q. The chrge Q is distriuted uniformly over the insulting shell. () Find the electric field (mgnitude nd direction) in ech of the regions 0 6 r 6 R, R 6 r 6 2R, nd r 7 2R. () Grph the electric-field mgnitude s function of r A conducting sphericl shell with inner rdius nd outer rdius hs positive point Figure chrge Q locted t its center. The totl chrge on P22.46 the shell is -3Q, nd it is insulted from its surroundings (Fig. P22.46). () Derive expressions for the electric-field mgnitude in terms of the Q distnce r from the center for the regions r 6, 23Q 6 r 6, nd r 7. () Wht is the surfce chrge density on the inner surfce of the conducting shell? (c) Wht is the surfce chrge density on the outer surfce of the conducting shell? (d) ketch the electric field lines nd the loction of ll chrges. (e) Grph the electric-field mgnitude s function of r Concentric phericl hells. A smll conducting sphericl shell with inner Figure P22.47 rdius nd outer rdius is concentric with lrger conducting sphericl shell with inner rdius c nd outer rdius d (Fig. P22.47). The inner shell hs totl chrge 2q, nd the outer shell hs chrge 4q. () Clculte the electric field (mgnitude nd direc- c tion) in terms of q nd the distnce r from d the common center of the two shells for (i) r 6 ; (ii) 6 r 6 ; (iii) 6 r 6 c; (iv) c 6 r 6 d; (v) r 7 d. how your results in grph of the rdil component of E s function of r. () Wht is the totl chrge on the (i) inner surfce of the smll shell; (ii) outer surfce of the smll shell; (iii) inner surfce of the lrge shell; (iv) outer surfce of the lrge shell? Repet Prolem 22.47, ut now let the outer shell hve chrge -2q. As in Prolem 22.47, the inner shell hs chrge 2q Repet Prolem 22.47, ut now let the outer shell hve chrge -4q. As in Prolem 22.47, the inner shell hs chrge 2q A solid conducting sphere with rdius R crries positive totl chrge Q. The sphere is surrounded y n insulting shell with inner rdius R nd outer rdius 2R. The insulting shell hs uniform chrge density r. () Find the vlue of r so tht the net chrge of the entire system is zero. () If r hs the vlue found in prt (), find the electric field (mgnitude nd direction) in ech of the regions 0 6 r 6 R, R 6 r 6 2R, nd r 7 2R. how your results in grph of the rdil component of E s function of r. (c) As generl rule, the electric field is discontinuous only t loctions where there is thin sheet of chrge. Explin how your results in prt () gree with this rule Negtive chrge -Q is distriuted uniformly over the surfce of thin sphericl insulting shell with rdius R. Clculte the force (mgnitude nd direction) tht the shell exerts on positive point chrge q locted () distnce r 7 R from the center of the shell (outside the shell) nd () distnce r 6 R from the center of the shell (inside the shell) () How mny excess electrons must e distriuted uniformly within the volume of n isolted plstic sphere 30.0 cm in dimeter to produce n electric field of 1390 N>C just outside the surfce of the sphere? () Wht is the electric field t point 10.0 cm outside the surfce of the sphere? CALC An insulting hollow sphere hs inner rdius nd outer rdius. Within the insulting mteril the volume chrge density is given y r1r2 = r, where is positive constnt. () In terms of nd, wht is the mgnitude of the electric field t distnce r from the center of the shell, where 6 r 6? () A point chrge q is plced t the center of the hollow spce, t r = 0. In terms of nd, wht vlue must q hve (sign nd mgnitude) in order for the electric field to e constnt in the region 6 r 6, nd wht then is the vlue of the constnt field in this region? CP Thomson s Model of the Atom. In the erly yers of the 20th century, leding model of the structure of the tom ws tht of the English physicist J. J. Thomson (the discoverer of the electron). In Thomson s model, n tom consisted of sphere of positively chrged mteril in which were emedded negtively chrged electrons, like chocolte chips in ll of cookie dough. Consider such n tom consisting of one electron with mss m nd chrge -e, which my e regrded s point chrge, nd uniformly chrged sphere of chrge e nd rdius R. () Explin why the equilirium position of the electron is t the center of the nucleus. () In Thomson s model, it ws ssumed tht the positive mteril provided little or no resistnce to the motion of the electron. If the electron is displced from equilirium y distnce less thn R, show tht the resulting motion of the electron will e simple hrmonic, nd clculte the frequency of oscilltion. (Hint: Review the definition of simple hrmonic motion in ection If it cn e shown tht the net force on the electron is of this form, then it follows tht the motion is simple hrmonic. Conversely, if the net force on the electron does not follow this form, the motion is not simple hrmonic.) (c) By Thomson s time, it ws known tht excited toms emit light wves of only certin frequencies. In his model, the frequency of emitted light is the sme s the oscilltion frequency of the electron or electrons in the tom. Wht would the rdius of Thomson-model tom hve to e for it to produce red light of frequency 4.57 * Hz? Compre your nswer to the rdii of rel toms, which re of the order of m (see Appendix F for dt out the electron). (d) If the electron were displced from equilirium y distnce greter thn R, would the electron oscillte? Would its motion e simple hrmonic? Explin your resoning. (Historicl note: In 1910, the tomic nucleus ws discovered, proving the Thomson model to e incorrect. An tom s positive chrge is not spred over its volume s Figure P22.55 Thomson supposed, ut is concentrted in the tiny nucleus of rdius to 12e m. ) Thomson s Model of the Atom, d d Continued. Using Thomson s (outdted) 2e 2e model of the tom descried in Prolem 22.54, consider n tom consisting of two R electrons, ech of chrge -e, emedded in sphere of chrge 2e nd rdius R. In

66 752 CHAPTER 22 Guss s Lw equilirium, ech electron is distnce d from the center of the tom (Fig. P22.55). Find the distnce d in terms of the other properties of the tom A Uniformly Chrged l. A sl of insulting mteril hs thickness 2d nd is oriented so tht its fces re prllel to the yz-plne nd given y the plnes x = d nd x = -d. The y- nd z-dimensions of the sl re very lrge compred to d nd my e treted s essentilly infinite. The sl hs uniform positive chrge density r. () Explin why the electric field due to the sl is zero t the center of the sl 1x = 02. () Using Guss s lw, find the electric field due to the sl (mgnitude nd direction) t ll points in spce CALC A Nonuniformly Chrged l. Repet Prolem 22.56, ut now let the chrge density of the sl e given y r1x2 = r 0 1x>d2 2, where r 0 is positive constnt CALC A nonuniform, ut sphericlly symmetric, distriution of chrge hs chrge density r1r2 given s follows: r1r2 = r r>3R2 for r R r1r2 = 0 for r Ú R where r 0 is positive constnt. () Find the totl chrge contined in the chrge distriution. () Otin n expression for the electric field in the region r Ú R. (c) Otin n expression for the electric field in the region r R. (d) Grph the electric-field mgnitude E s function of r. (e) Find the vlue of r t which the electric field is mximum, nd find the vlue of tht mximum field CP CALC Guss s Lw for Grvittion. The grvittionl force etween two point msses seprted y distnce r is proportionl to 1>r 2, just like the electric force etween two point chrges. Becuse of this similrity etween grvittionl nd electric interctions, there is lso Guss s lw for grvittion. () Let g e the ccelertion due to grvity cused y point mss m t the origin, so tht g -1Gm>r 2 2rn. Consider sphericl Gussin surfce with rdius r centered on this point mss, nd show tht the flux of g through this surfce is given y # C g da = -4pGm () By following the sme logicl steps used in ection 22.3 to otin Guss s lw for the electric field, show tht the flux of g through ny closed surfce is given y g # da = -4pGMencl C where M encl is the totl mss enclosed within the closed surfce CP Applying Guss s Lw for Grvittion. Using Guss s lw for grvittion (derived in prt () of Prolem 22.59), show tht the following sttements re true: () For ny sphericlly symmetric mss distriution with totl mss M, the ccelertion due to grvity outside the distriution is the sme s though ll the mss were concentrted t the center. (Hint: ee Exmple 22.5 in ection 22.4.) () At ny point inside sphericlly symmetric shell of mss, the ccelertion due to grvity is zero. (Hint: ee Exmple 22.5.) (c) If we could drill hole through sphericlly symmetric plnet to its center, nd if the density were uniform, we would find tht the mgnitude of g is directly proportionl to the distnce r from the center. (Hint: ee Exmple 22.9 in ection 22.4.) We proved these results in ection 13.6 using some firly strenuous nlysis; the proofs using Guss s lw for grvittion re much esier () An insulting sphere with rdius hs uniform chrge density r. The sphere is not centered t the origin ut t r. how tht the electric field inside the sphere is given y E r1r - 2>3P 0. () An insulting sphere Figure P22.61 of rdius R hs sphericl hole of rdius locted within its volume nd centered distnce from the center of the sphere, where R 6 6 R ( cross section of the sphere is shown in Fig. P22.61). The solid prt of the Chrge density r sphere hs uniform volume chrge density r. Find the mgnitude nd direction of the electric field E inside the hole, nd show tht E is uniform over the entire hole. [Hint: Use the principle of superposition nd the result of prt ().] A very long, solid insulting Figure P22.62 cylinder with rdius R hs cylindricl hole with rdius ored long its entire length. The xis of the hole is distnce from the xis of the cylinder, where 6 6 R (Fig. P22.62). The solid mteril R of the cylinder hs uniform volume chrge density r. Find the mgnitude nd Chrge direction of the electric field E density r inside the hole, nd show tht E is uniform over the entire hole. (Hint: ee Prolem ) Positive chrge Q is Figure P22.63 distriuted uniformly over ech of two sphericl volumes with y rdius R. One sphere of chrge is centered t the origin nd the other t x = 2R (Fig. P22.63). R R R x O Find the mgnitude nd direction of the net electric field due Q Q to these two distriutions of chrge t the following points on the x-xis: () x = 0; () x = R>2; (c) x = R; (d) x = 3R Repet Prolem 22.63, ut now let the left-hnd sphere hve positive chrge Q nd let the right-hnd sphere hve negtive chrge -Q CALC A nonuniform, ut sphericlly symmetric, distriution of chrge hs chrge density r1r2 given s follows: r1r2 = r r>r2 for r R r1r2 = 0 where r 0 = 3Q>pR 3 is positive constnt. () how tht the totl chrge contined in the chrge distriution is Q. () how tht the electric field in the region r Ú R is identicl to tht produced y point chrge Q t r = 0. (c) Otin n expression for the electric field in the region r R. (d) Grph the electric-field mgnitude E s function of r. (e) Find the vlue of r t which the electric field is mximum, nd find the vlue of tht mximum field. CHALLENGE PROBLEM CP CALC A region in spce contins totl positive chrge Q tht is distriuted sphericlly such tht the volume chrge density r1r2 is given y r1r2 = r1r2 = r>r2 r1r2 = 0 for r Ú R for r R>2 for R>2 r R for r Ú R Here is positive constnt hving units of C>m 3. () Determine in terms of Q nd R. () Using Guss s lw, derive n expression for the mgnitude of E s function of r. Do this seprtely for ll

67 Answers 753 three regions. Express your nswers in terms of the totl chrge Q. Be sure to check tht your results gree on the oundries of the regions. (c) Wht frction of the totl chrge is contined within the region r R>2? (d) If n electron with chrge q =-e is oscillting ck nd forth out r = 0 (the center of the distriution) with n mplitude less thn R>2, show tht the motion is simple hrmonic. (Hint: Review the discussion of simple hrmonic motion in ection If, nd only if, the net force on the electron is proportionl to its displcement from equilirium, then the motion is simple hrmonic.) (e) Wht is the period of the motion in prt (d)? (f) If the mplitude of the motion descried in prt (e) is greter thn R>2, is the motion still simple hrmonic? Why or why not? CP CALC A region in spce contins totl positive chrge Q tht is distriuted sphericlly such tht the volume chrge density r1r2 is given y r1r2 = 3r>12R2 r1r2 = 31-1r>R2 2 4 r1r2 = 0 for r R>2 for R>2 r R for r Ú R Here is positive constnt hving units of C>m 3. () Determine in terms of Q nd R. () Using Guss s lw, derive n expression for the mgnitude of the electric field s function of r. Do this seprtely for ll three regions. Express your nswers in terms of the totl chrge Q. (c) Wht frction of the totl chrge is contined within the region R>2 r R? (d) Wht is the mgnitude of E t r = R>2? (e) If n electron with chrge q =-e is relesed from rest t ny point in ny of the three regions, the resulting motion will e oscilltory ut not simple hrmonic. Why? (ee Chllenge Prolem ) Answers Chpter Opening Question? No. The electric field inside cvity within conductor is zero, so there is no electric effect on the child. (ee ection 22.5.) Test Your Understnding Questions 22.1 Answer: (iii) Ech prt of the surfce of the ox will e three times frther from the chrge q, so the electric field will e = 1 s strong. But the re of the ox will increse y fctor of = 9. Hence the electric flux will e multiplied y fctor of = 1. In other words, the flux will e unchnged Answer: (iv), (ii), (i), (iii) In ech cse the electric field is uniform, so the flux is E = E # A. We use the reltionships for the sclr products of unit vectors: ın # ın = N # N = 1, ın# N = 0. In cse (i) we hve E = 14.0 N>C216.0 m 2 2ıN # N = 0 (the electric field nd vector re re perpendiculr, so there is zero flux). In cse (ii) we hve 13.0 m 12.0 N>C2# 6.0 N # 2 E N>C2ıN 12.0 N>C2 N4# 2 N = 13.0 m 2 2 = m 2 >C. imilrly, in cse (iii) we hve m 2 2ıN 17.0 m -2 N # 2 E = N>C2ıN 12.0 N>C2 N4 # 2 N4 = 14.0 N>C213.0 m N>C217.0 m 2 2 = m nd in cse (iv) we hve m 12.0 N>C2 # 2 E = N>C2ıN 12.0 N>C2 N4 # 2 >C, 2ıN 17.0 m 26 N # 2 2 N4 = 14.0 N>C213.0 m m 2 2 = m 2 >C Answer: 2, 5, 4, 1 nd 3 (tie) Guss s lw tells us tht the flux through closed surfce is proportionl to the mount of chrge enclosed within tht surfce. o n ordering of these surfces y their fluxes is the sme s n ordering y the mount of enclosed chrge. urfce 1 encloses no chrge, surfce 2 encloses 9.0 mc 5.0 mc mc2 = 7.0 mc, surfce 3 encloses 9.0 mc 1.0 mc mc2 = 0, surfce 4 encloses 8.0 mc mc2 = 1.0 mc, nd surfce 5 encloses 8.0 mc mc mc mc mc mc2 = 6.0 mc Answer: no You might e tempted to drw Gussin surfce tht is n enlrged version of the conductor, with the sme shpe nd plced so tht it completely encloses the conductor. While you know the flux through this Gussin surfce (y Guss s lw, it s E = Q>P 0 ), the direction of the electric field need not e perpendiculr to the surfce nd the mgnitude of the field need not e the sme t ll points on the surfce. It s not possile to do the flux integrl A E da, nd we cn t clculte the electric field. Guss s lw is useful for clculting the electric field only when the chrge distriution is highly symmetric Answer: no Before you connect the wire to the sphere, the presence of the point chrge will induce chrge -q on the inner surfce of the hollow sphere nd chrge q on the outer surfce (the net chrge on the sphere is zero). There will e n electric field outside the sphere due to the chrge on the outer surfce. Once you touch the conducting wire to the sphere, however, electrons will flow from ground to the outer surfce of the sphere to neutrlize the chrge there (see Fig. 21.7c). As result the sphere will hve no chrge on its outer surfce nd no electric field outside. Bridging Prolem Answers: () Q1r2 = Qe -2r> 0 321r> r> () E = kqe-2r> 0 r 2 321r> r> (c) E/E proton r/

68 23 ELECTRIC POTENTIAL LEARNING GOAL By studying this chpter, you will lern: How to clculte the electric potentil energy of collection of chrges. The mening nd significnce of electric potentil. How to clculte the electric potentil tht collection of chrges produces t point in spce. How to use equipotentil surfces to visulize how the electric potentil vries in spce. How to use electric potentil to clculte the electric field.? In one type of welding, electric chrge flows etween the welding tool nd the metl pieces tht re to e joined together. This produces glowing rc whose high temperture fuses the pieces together. Why must the tool e held close to the pieces eing welded? This chpter is out energy ssocited with electricl interctions. Every time you turn on light, listen to n MP3 plyer, or tlk on moile phone, you re using electricl energy, n indispensle ingredient of our technologicl society. In Chpters 6 nd 7 we introduced the concepts of work nd energy in the context of mechnics; now we ll comine these concepts with wht we ve lerned out electric chrge, electric forces, nd electric fields. Just s we found for mny prolems in mechnics, using energy ides mkes it esier to solve vriety of prolems in electricity. When chrged prticle moves in n electric field, the field exerts force tht cn do work on the prticle. This work cn lwys e expressed in terms of electric potentil energy. Just s grvittionl potentil energy depends on the height of mss ove the erth s surfce, electric potentil energy depends on the position of the chrged prticle in the electric field. We ll descrie electric potentil energy using new concept clled electric potentil, or simply potentil. In circuits, difference in potentil from one point to nother is often clled voltge. The concepts of potentil nd voltge re crucil to understnding how electric circuits work nd hve eqully importnt pplictions to electron ems used in cncer rdiotherpy, high-energy prticle ccelertors, nd mny other devices Electric Potentil Energy The concepts of work, potentil energy, nd conservtion of energy proved to e extremely useful in our study of mechnics. In this section we ll show tht these concepts re just s useful for understnding nd nlyzing electricl interctions. 754

69 23.1 Electric Potentil Energy 755 Let s egin y reviewing three essentil points from Chpters 6 nd 7. First, when force F cts on prticle tht moves from point to point, the work done y the force is given y line integrl: W W = F # dl L = F cos f dl L (work done y force) (23.1) where d l is n infinitesiml displcement long the prticle s pth nd f is the ngle etween F nd d l t ech point long the pth. econd, if the force F is conservtive, s we defined the term in ection 7.3, the work done y F cn lwys e expressed in terms of potentil energy U. When the prticle moves from point where the potentil energy is U to point where it is U, the chnge in potentil energy is U = U - U nd the work done y the force is W W = U - U = -1U - U 2 = - U (work done y conservtive force) (23.2) When W is positive, U is greter thn U, U is negtive, nd the potentil energy decreses. Tht s wht hppens when sell flls from high point () to lower point () under the influence of the erth s grvity; the force of grvity does positive work, nd the grvittionl potentil energy decreses (Fig. 23.1). When tossed ll is moving upwrd, the grvittionl force does negtive work during the scent, nd the potentil energy increses. Third, the workenergy theorem sys tht the chnge in kinetic energy K = K - K during displcement equls the totl work done on the prticle. If only conservtive forces do work, then Eq. (23.2) gives the totl work, nd K - K = -1U - U 2. We usully write this s K U = K U (23.3) Tht is, the totl mechnicl energy (kinetic plus potentil) is conserved under these circumstnces. Electric Potentil Energy in Uniform Field Let s look t n electricl exmple of these sic concepts. In Fig pir of chrged prllel metl pltes sets up uniform, downwrd electric field with mgnitude E. The field exerts downwrd force with mgnitude F = q 0 E on positive test chrge q 0. As the chrge moves downwrd distnce d from point to point, the force on the test chrge is constnt nd independent of its loction. o the work done y the electric field is the product of the force mgnitude nd the component of displcement in the (downwrd) direction of the force: W = Fd = q 0 Ed (23.4) This work is positive, since the force is in the sme direction s the net displcement of the test chrge. The y-component of the electric force, F y = -q 0 E, is constnt, nd there is no x- or z-component. This is exctly nlogous to the grvittionl force on mss m ner the erth s surfce; for this force, there is constnt y-component F y = -mg nd the x- nd z-components re zero. Becuse of this nlogy, we cn conclude tht the force exerted on q 0 y the uniform electric field in Fig is conservtive, just s is the grvittionl force. This mens tht the work W done y the field is independent of the pth the prticle tkes from to. We cn represent this work with potentil-energy function U, just s we did for grvittionl potentil energy 23.1 The work done on sell moving in uniform grvittionl field. Oject moving in uniform grvittionl field w 5 mg Point chrge moving in uniform electric field h The work done y the grvittionl force is the sme for ny pth from to : W 52DU 5 mgh 23.2 The work done on point chrge moving in uniform electric field. Compre with Fig E q 0 The work done y the electric force is the sme for ny pth from to : W 52DU 5 q 0 Ed y O F 5 q 0 E y d

70 756 CHAPTER 23 Electric Potentil 23.3 A positive chrge moving () in the direction of the electric field nd () in the direction opposite E. E () Positive chrge moves in the direction of E: Field does positive work on chrge. U decreses. y () Positive chrge moves opposite E: Field does negtive work on chrge. U increses. y E F 5 q 0 E E y y O y y F 5 q 0 E O in ection 7.1. The potentil energy for the grvittionl force F y = -mg ws U = mgy; hence the potentil energy for the electric force F y = -q 0 E is U = q 0 Ey (23.5) When the test chrge moves from height y to height y, the work done on the chrge y the field is given y W = - U = -1U - U 2 = -1q 0 Ey - q 0 Ey 2 = q 0 E1y - y 2 (23.6) When y is greter thn y (Fig. 23.3), the positive test chrge q 0 moves downwrd, in the sme direction s E ; the displcement is in the sme direction s the force F q 0 E, so the field does positive work nd U decreses. [In prticulr, if y - y = d s in Fig. 23.2, Eq. (23.6) gives W = q 0 Ed, in greement with Eq. (23.4).] When y is less thn y (Fig. 23.3), the positive test chrge q 0 moves upwrd, in the opposite direction to E ; the displcement is opposite the force, the field does negtive work, nd U increses. If the test chrge q 0 is negtive, the potentil energy increses when it moves with the field nd decreses when it moves ginst the field (Fig. 23.4). Whether the test chrge is positive or negtive, the following generl rules pply: U increses if the test chrge q 0 moves in the direction opposite the electric force F q 0 E (Figs nd 23.4); U decreses if moves in the sme 23.4 A negtive chrge moving () in the direction of the electric field nd () in the direction opposite E. Compre with Fig () Negtive chrge moves in the direction of E: Field does negtive work on chrge. U increses. y q 0 E () Negtive chrge moves opposite E: Field does positive work on chrge. U decreses. y E F 5 q 0 E E y y F 5 q 0 E y O y O

71 23.1 Electric Potentil Energy 757 direction s F q 0 E (Figs nd 23.4). This is the sme ehvior s for grvittionl potentil energy, which increses if mss m moves upwrd (opposite the direction of the grvittionl force) nd decreses if m moves downwrd (in the sme direction s the grvittionl force). CAUTION Electric potentil energy The reltionship etween electric potentil energy chnge nd motion in n electric field is n importnt one tht we ll use often, ut tht tkes some effort to truly understnd. Tke the time to crefully study the preceding prgrph s well s Figs nd Doing so now will help you tremendously lter! Electric Potentil Energy of Two Point Chrges The ide of electric potentil energy isn t restricted to the specil cse of uniform electric field. Indeed, we cn pply this concept to point chrge in ny electric field cused y sttic chrge distriution. Recll from Chpter 21 tht we cn represent ny chrge distriution s collection of point chrges. Therefore it s useful to clculte the work done on test chrge q 0 moving in the electric field cused y single, sttionry point chrge q. We ll consider first displcement long the rdil line in Fig The force on is given y Coulom s lw, nd its rdil component is q 0 (23.7) If q nd q 0 hve the sme sign 1 or -2 the force is repulsive nd F r is positive; if the two chrges hve opposite signs, the force is ttrctive nd F r is negtive. The force is not constnt during the displcement, nd we hve to integrte to clculte the work W done on q 0 y this force s q 0 moves from to : r W = F r dr = L L r (23.8) The work done y the electric force for this prticulr pth depends only on the endpoints. Now let s consider more generl displcement (Fig. 23.6) in which nd do not lie on the sme rdil line. From Eq. (23.1) the work done on q 0 during this displcement is given y r W = Fcos f dl = L L But Fig shows tht cos f dl = dr. Tht is, the work done during smll displcement d l depends only on the chnge dr in the distnce r etween the chrges, which is the rdil component of the displcement. Thus Eq. (23.8) is vlid even for this more generl displcement; the work done on q 0 y the electric field E produced y q depends only on r nd r, not on the detils of the pth. Also, if q 0 returns to its strting point y different pth, the totl work done in the round-trip displcement is zero (the integrl in Eq. (23.8) is from r ck to r 2. These re the needed chrcteristics for conservtive force, s we defined it in ection 7.3. Thus the force on q 0 is conservtive force. We see tht Eqs. (23.2) nd (23.8) re consistent if we define the potentil energy to e U = qq 0 >4pP 0 r when q 0 is distnce r from q, nd to e U = qq 0 >4pP 0 r when q 0 is distnce r from q. Thus the potentil energy U when the test chrge is t ny distnce r from chrge q is q 0 r F r = r 1 r 1 4pP 0 qq 0 r 2 qq 0 4pP 0 r 2 dr = qq pP 0 r r r 1 r qq 0 cos f dl 4pP 2 0 r 23.5 Test chrge q 0 moves long stright line extending rdilly from chrge q. As it moves from to, the distnce vries from r to r. Test chrge q 0 moves from to long rdil line from q. q E r q 0 r 23.6 The work done on chrge q 0 y the electric field of chrge q does not depend on the pth tken, ut only on the distnces r nd r. Test chrge q 0 moves from to long n ritrry pth. r dr q 0 r q f r F r dl dr E E U = 1 qq 0 4pP 0 r (electric potentil energy of two point chrges ) q nd q 0 (23.9)

72 758 CHAPTER 23 Electric Potentil 23.7 Grphs of the potentil energy U of two point chrges q nd q 0 versus their seprtion r. () q nd q 0 hve the sme sign. U q q 0 q q 0 or r r Eqution (23.9) is vlid no mtter wht the signs of the chrges q nd q 0. The potentil energy is positive if the chrges q nd q 0 hve the sme sign (Fig. 23.7) nd negtive if they hve opposite signs (Fig. 23.7). CAUTION Electric potentil energy vs. electric force Don t confuse Eq. (23.9) for the potentil energy of two point chrges with the similr expression in Eq. (23.7) for the rdil component of the electric force tht one chrge exerts on the other. Potentil energy U is proportionl to 1>r, while the force component is proportionl to 1>r 2. F r O O U U. 0 As r 0, U 1`. As r `, U 0. () q nd q 0 hve opposite signs. q r q 0 or q U, 0 As r 0, U 2`. As r `, U 0. r r r q 0 Potentil energy is lwys defined reltive to some reference point where U = 0. In Eq. (23.9), U is zero when q nd q 0 re infinitely fr prt nd r = q. Therefore U represents the work tht would e done on the test chrge q 0 y the field of q if q 0 moved from n initil distnce r to infinity. If q nd q 0 hve the sme sign, the interction is repulsive, this work is positive, nd U is positive t ny finite seprtion (Fig. 23.7). If the chrges hve opposite signs, the interction is ttrctive, the work done is negtive, nd U is negtive (Fig. 23.7). We emphsize tht the potentil energy U given y Eq. (23.9) is shred property of the two chrges. If the distnce etween q nd q 0 is chnged from r to r, the chnge in potentil energy is the sme whether q is held fixed nd q 0 is moved or q 0 is held fixed nd q is moved. For this reson, we never use the phrse the electric potentil energy of point chrge. (Likewise, if mss m is t height h ove the erth s surfce, the grvittionl potentil energy is shred property of the mss m nd the erth. We emphsized this in ections 7.1 nd 13.3.) Eqution (23.9) lso holds if the chrge q 0 is outside sphericlly symmetric chrge distriution with totl chrge q; the distnce r is from q 0 to the center of the distriution. Tht s ecuse Guss s lw tells us tht the electric field outside such distriution is the sme s if ll of its chrge q were concentrted t its center (see Exmple 22.9 in ection 22.4). Exmple 23.1 Conservtion of energy with electric forces A positron (the electron s ntiprticle) hs mss 9.11 * kg EXECUTE: () Both prticles hve positive chrge, so the positron nd chrge q 0 = e = 1.60 * C. uppose positron speeds up s it moves wy from the prticle. From the energyconservtion eqution, Eq. (23.3), the finl kinetic energy is moves in the vicinity of n (lph) prticle, which hs chrge q = 2e = 3.20 * C nd mss 6.64 * kg. The K = 1 2 prticle s mss is more thn 7000 times tht of the positron, so we mv 2 = K U - U ssume tht the prticle remins t rest. When the positron is In this expression, 1.00 * m from the prticle, it is moving directly wy from the prticle t 3.00 * 10 6 m>s. () Wht is the positron s K = 1 2 mv 2 = * kg * 10 6 m>s2 2 speed when the prticles re 2.00 * m prt? () Wht is = 4.10 * J the positron s speed when it is very fr from the prticle? (c) uppose the initil conditions re the sme ut the moving prticle is n electron (with the sme mss s the positron ut chrge 4pP 0 r 1 qq 0 U = = 19.0 * 10 9 N # m 2 >C 2 2 q 0 = -e). Descrie the susequent motion. * * C * C2 OLUTION 1.00 * m IDENTIFY nd ET UP: The electric force etween positron (or = 4.61 * J n electron) nd n prticle is conservtive, so mechnicl 1 qq 0 U energy (kinetic plus potentil) is conserved. Eqution (23.9) gives the = = 2.30 * J 4pP 0 r potentil energy U t ny seprtion r: The potentil-energy function Hence the positron kinetic energy nd speed t r = r = 2.00 * for prts () nd () looks like tht of Fig. 23.7, nd the function for m re prt (c) looks like tht of Fig We re given the positron speed v = 3.00 * 10 6 m>s when the seprtion etween the prticles is K = 1 r = 1.00 * mv 2 = 4.10 * J 4.61 * J * J m. In prts () nd () we use Eqs. (23.3) nd (23.9) to find the speed for r = r = 2.00 * = 6.41 * J m nd r = r c q, respectively. In prt (c) we replce the positron with n 2K v = electron nd reconsider the prolem. A m = * J2 A 9.11 * kg = 3.8 * 106 m>s

73 23.1 Electric Potentil Energy 759 () When the positron nd prticle re very fr prt so tht r = r c q, the finl potentil energy U c pproches zero. Agin from energy conservtion, the finl kinetic energy nd speed of the positron in this cse re K c = K U - U c = 4.10 * J 4.61 * J - 0 = 8.71 * J v c = A 2K c m = * J2 A 9.11 * kg = 4.4 * 106 m>s (c) The electron nd prticle hve opposite chrges, so the force is ttrctive nd the electron slows down s it moves wy. Chnging the moving prticle s sign from e to -e mens tht the initil potentil energy is now U = * J, which mkes the totl mechnicl energy negtive: K U = * J * J2 = * J The totl mechnicl energy would hve to e positive for the electron to move infinitely fr wy from the prticle. Like rock thrown upwrd t low speed from the erth s surfce, it will rech mximum seprtion r = r d from the prticle efore reversing direction. At this point its speed nd its kinetic energy K d re zero, so t seprtion we hve r d U d = K U - K d = * J2-0 1 qq 0 U d = = * J 4pP 0 r d r d = 1 U d qq 0 4pP 0 = 19.0 * 109 N # m 2 >C * * C * C2 J = 9.0 * m For r we hve U = * = 2.00 * m J, so the electron kinetic energy nd speed t this point re K = 1 2 mv 2 v = A 2K m = 4.10 * J * J * J2 = 1.79 * J = * J2 A 9.11 * kg = 2.0 * 106 m>s EVALUATE: Both prticles ehve s expected s they move wy from the prticle: The positron speeds up, nd the electron slows down nd eventully turns round. How fst would n electron hve to e moving t r = 1.00 * m to trvel infinitely fr from the prticle? (Hint: ee Exmple 13.4 in ection 13.3.) Electric Potentil Energy with everl Point Chrges uppose the electric field in which chrge q 0 moves is cused y severl point chrges q 1, q 2, q 3, Á t distnces r 1, r 2, r 3, Á from q 0, s in Fig For exmple, q 0 could e positive ion moving in the presence of other ions (Fig. 23.9). The totl electric field t ech point is the vector sum of the fields due to the individul chrges, nd the totl work done on q 0 during ny displcement is the sum of the contriutions from the individul chrges. From Eq. (23.9) we conclude tht the potentil energy ssocited with the test chrge q 0 t point in Fig is the lgeric sum (not vector sum): E 23.8 The potentil energy ssocited with chrge q 0 t point depends on the other chrges q 1, q 2, nd q 3 nd on their distnces r 1, r 2, nd from point. r 3 q 1 q 2 r 1 U = q 0 4pP 0 q 1 r 1 q 2 r 2 q 3 r 3 Á = q 0 4pP 0 i q i r i (point chrge q 0 nd collection (23.10) of chrges q i ) q0 r 3 r 2 q 3 When q 0 is t different point, the potentil energy is given y the sme expression, ut r 1, r 2, Á re the distnces from q 1, q 2, Á to point. The work done on chrge q 0 when it moves from to long ny pth is equl to the difference U - U etween the potentil energies when q 0 is t nd t. We cn represent ny chrge distriution s collection of point chrges, so Eq. (23.10) shows tht we cn lwys find potentil-energy function for ny sttic electric field. It follows tht for every electric field due to sttic chrge distriution, the force exerted y tht field is conservtive. Equtions (23.9) nd (23.10) define U to e zero when ll the distnces r 1, r 2, Á re infinite tht is, when the test chrge q 0 is very fr wy from ll the chrges tht produce the field. As with ny potentil-energy function, the point where U = 0 is ritrry; we cn lwys dd constnt to mke U equl zero t ny point we choose. In electrosttics prolems it s usully simplest to choose this point to e t infinity. When we nlyze electric circuits in Chpters 25 nd 26, other choices will e more convenient.

74 760 CHAPTER 23 Electric Potentil 23.9 This ion engine for spcecrft uses electric forces to eject strem of positive xenon ions 1Xe 2 t speeds in excess of 30 km/s. The thrust produced is very low (out 0.09 newton) ut cn e mintined continuously for dys, in contrst to chemicl rockets, which produce lrge thrust for short time (see Fig. 8.33). uch ion engines hve een used for mneuvering interplnetry spcecrft. Eqution (23.10) gives the potentil energy ssocited with the presence of the test chrge q in the E 0 field produced y q 1, q 2, q 3,.... But there is lso potentil energy involved in ssemling these chrges. If we strt with chrges q 1, q 2, q 3, Á ll seprted from ech other y infinite distnces nd then ring them together so tht the distnce etween q i nd q j is r ij, the totl potentil energy U is the sum of the potentil energies of interction for ech pir of chrges. We cn write this s 1 q i q j U = (23.11) 4pP 0 i6j r ij This sum extends over ll pirs of chrges; we don t let i = j (ecuse tht would e n interction of chrge with itself ), nd we include only terms with i 6 j to mke sure tht we count ech pir only once. Thus, to ccount for the interction etween q 3 nd q 4, we include term with i = 3 nd j = 4 ut not term with i = 4 nd j = 3. Interpreting Electric Potentil Energy As finl comment, here re two viewpoints on electric potentil energy. We hve defined it in terms of the work done y the electric field on chrged prticle moving in the field, just s in Chpter 7 we defined potentil energy in terms of the work done y grvity or y spring. When prticle moves from point to point, the work done on it y the electric field is W = U - U. Thus the potentil-energy difference U - U equls the work tht is done y the electric force when the prticle moves from to. When U is greter thn U, the field does positive work on the prticle s it flls from point of higher potentil energy () to point of lower potentil energy (). An lterntive ut equivlent viewpoint is to consider how much work we would hve to do to rise prticle from point where the potentil energy is U to point where it hs greter vlue U (pushing two positive chrges closer together, for exmple). To move the prticle slowly (so s not to give it ny kinetic energy), we need to exert n dditionl externl force F ext tht is equl nd opposite to the electric-field force nd does positive work. The potentilenergy difference U - U is then defined s the work tht must e done y n externl force to move the prticle slowly from to ginst the electric force. Becuse F ext is the negtive of the electric-field force nd the displcement is in the opposite direction, this definition of the potentil difference U - U is equivlent to tht given ove. This lterntive viewpoint lso works if U is less thn U, corresponding to lowering the prticle; n exmple is moving two positive chrges wy from ech other. In this cse, U - U is gin equl to the work done y the externl force, ut now this work is negtive. We will use oth of these viewpoints in the next section to interpret wht is ment y electric potentil, or potentil energy per unit chrge. Exmple 23.2 A system of point chrges Two point chrges re locted on the x-xis, q 1 = -e t x = 0 nd infinity to x = 2. We do this y using Eq. (23.10) to find the q 2 = e t x =. () Find the work tht must e done y n potentil energy ssocited with q 3 in the presence of q 1 nd q 2. In externl force to ring third point chrge q 3 = e from infinity prt () we use Eq. (23.11), the expression for the potentil energy to x = 2. () Find the totl potentil energy of the system of three of collection of point chrges, to find the totl potentil energy of chrges. the system. OLUTION IDENTIFY nd ET UP: Figure shows the finl rrngement of the three chrges. In prt () we need to find the work W tht must e done on y n externl force to ring in from q 3 F ext q Our sketch of the sitution fter the third chrge hs een rought in from infinity.

75 23.2 Electric Potentil 761 EXECUTE: () The work W equls the difference etween (i) the potentil energy U ssocited with q 3 when it is t x = 2 nd (ii) the potentil energy when it is infinitely fr wy. The second of these is zero, so the work required is equl to U. The distnces etween the chrges re r 13 = 2 nd r 23 =, so from Eq. (23.10), W = U = q 3 4pP 0 q 1 r 13 q 2 r 23 = e 4pP 0 -e 2 e = e2 8pP 0 This is positive, just s we should expect. If we ring q 3 in from infinity long the x-xis, it is ttrcted y q 1 ut is repelled more strongly y q 2. Hence we must do positive work to push q 3 to the position t x = 2. () From Eq. (23.11), the totl potentil energy of the threechrge system is U = = 1 4pP 0 i6j q i q j r ij = 1 c 1-e21e2 1-e21e2 4pP pP 0 q 1q 2 r 12 q 1q 3 r 13 q 2q 3 r 23 1e21e2 d = -e2 8pP 0 EVALUATE: Our negtive result in prt () mens tht the system hs lower potentil energy thn it would if the three chrges were infinitely fr prt. An externl force would hve to do negtive work to ring the three chrges from infinity to ssemle this entire rrngement nd would hve to do positive work to move the three chrges ck to infinity. Test Your Understnding of ection 23.1 Consider the system of three point chrges in Exmple 21.4 (ection 21.3) nd shown in Fig () Wht is the sign of the totl potentil energy of this system? (i) positive; (ii) negtive; (iii) zero. () Wht is the sign of the totl mount of work you would hve to do to move these chrges infinitely fr from ech other? (i) positive; (ii) negtive; (iii) zero Electric Potentil In ection 23.1 we looked t the potentil energy U ssocited with test chrge q 0 in n electric field. Now we wnt to descrie this potentil energy on per unit chrge sis, just s electric field descries the force per unit chrge on chrged prticle in the field. This leds us to the concept of electric potentil, often clled simply potentil. This concept is very useful in clcultions involving energies of chrged prticles. It lso fcilittes mny electric-field clcultions ecuse electric potentil is closely relted to the electric field E. When we need to determine n electric field, it is often esier to determine the potentil first nd then find the field from it. Potentil is potentil energy per unit chrge. We define the potentil V t ny point in n electric field s the potentil energy U per unit chrge ssocited with test chrge t tht point: q 0 PhET: Chrges nd Fields ActivPhysics 11.13: Electricl Potentil Energy nd Potentil V = U q 0 or U = q 0 V (23.12) Potentil energy nd chrge re oth sclrs, so potentil is sclr. From Eq. (23.12) its units re the units of energy divided y those of chrge. The I unit of potentil, clled one volt (1 V) in honor of the Itlin electricl experimenter Alessndro Volt ( ), equls 1 joule per coulom: Let s put Eq. (23.2), which equtes the work done y the electric force during displcement from to to the quntity - U = -1U - U 2, on work per unit chrge sis. We divide this eqution y q 0, otining W q 0 = - U q 0 1 V = 1 volt = 1 J>C = 1 joule>coulom = - U q 0 - U q 0 = -1V - V 2 = V - V (23.13) where V = U >q 0 is the potentil energy per unit chrge t point nd similrly for V. We cll V nd V the potentil t point nd potentil t point, respectively. Thus the work done per unit chrge y the electric force when chrged ody moves from to is equl to the potentil t minus the potentil t.

76 762 CHAPTER 23 Electric Potentil The voltge of this ttery equls the difference in potentil V = V - V etween its positive terminl (point ) nd its negtive terminl (point ). Point (positive terminl) Point (negtive terminl) V volts The difference V - V is clled the potentil of with respect to ; we sometimes revite this difference s V = V - V (note the order of the suscripts). This is often clled the potentil difference etween nd, ut tht s miguous unless we specify which is the reference point. In electric circuits, which we will nlyze in lter chpters, the potentil difference etween two points is often clled voltge (Fig ). Eqution (23.13) then sttes: V, the potentil of with respect to, equls the work done y the electric force when UNIT chrge moves from to. Another wy to interpret the potentil difference V in Eq. (23.13) is to use the lterntive viewpoint mentioned t the end of ection In tht viewpoint, U - U is the mount of work tht must e done y n externl force to move prticle of chrge q 0 slowly from to ginst the electric force. The work tht must e done per unit chrge y the externl force is then 1U - U 2>q 0 = V - V = V. In other words: V, the potentil of with respect to, equls the work tht must e done to move UNIT chrge slowly from to ginst the electric force. An instrument tht mesures the difference of potentil etween two points is clled voltmeter. (In Chpter 26 we ll discuss how these devices work.) Voltmeters tht cn mesure potentil difference of 1 mv re common, nd sensitivities down to V cn e ttined. Clculting Electric Potentil To find the potentil V due to single point chrge q, we divide Eq. (23.9) y q 0 : V = U q 0 = 1 q 4pP 0 r (potentil due to point chrge) (23.14) where r is the distnce from the point chrge q to the point t which the potentil is evluted. If q is positive, the potentil tht it produces is positive t ll points; if q is negtive, it produces potentil tht is negtive everywhere. In either cse, V is equl to zero t r = q, n infinite distnce from the point chrge. Note tht potentil, like electric field, is independent of the test chrge q 0 tht we use to define it. imilrly, we divide Eq. (23.10) y q 0 to find the potentil due to collection of point chrges: Appliction Electrocrdiogrphy The electrodes used in n electrocrdiogrm EKG or ECG for short mesure the potentil differences (typiclly no greter thn 1 mv = 10-3 V) etween different prts of the ptient s skin. These re indictive of the potentil differences etween regions of the hert, nd so provide sensitive wy to detect ny normlities in the electricl ctivity tht drives crdic function. V = U q 0 = 1 4pP 0 i q i r i (potentil due to collection of point chrge) (23.15) In this expression, r i is the distnce from the ith chrge, q i, to the point t which V is evluted. Just s the electric field due to collection of point chrges is the vector sum of the fields produced y ech chrge, the electric potentil due to collection of point chrges is the sclr sum of the potentils due to ech chrge. When we hve continuous distriution of chrge long line, over surfce, or through volume, we divide the chrge into elements dq, nd the sum in Eq. (23.15) ecomes n integrl: V = 1 4pP 0 L dq r (potentil due to continuous distriution of chrge) (23.16) where r is the distnce from the chrge element dq to the field point where we re finding V. We ll work out severl exmples of such cses. The potentil defined y Eqs. (23.15) nd (23.16) is zero t points tht re infinitely fr wy from ll the chrges. Lter we ll encounter cses in which the chrge distriution itself

77 23.2 Electric Potentil 763 extends to infinity. We ll find tht in such cses we cnnot set V = 0 t infinity, nd we ll need to exercise cre in using nd interpreting Eqs. (23.15) nd (23.16). CAUTION Wht is electric potentil? Before getting too involved in the detils of how to clculte electric potentil, you should stop nd remind yourself wht potentil is. The electric potentil t certin point is the potentil energy tht would e ssocited with unit chrge plced t tht point. Tht s why potentil is mesured in joules per coulom, or volts. Keep in mind, too, tht there doesn t hve to e chrge t given point for potentil V to exist t tht point. (In the sme wy, n electric field cn exist t given point even if there s no chrge there to respond to it.) Finding Electric Potentil from Electric Field When we re given collection of point chrges, Eq. (23.15) is usully the esiest wy to clculte the potentil V. But in some prolems in which the electric field is known or cn e found esily, it is esier to determine V from E. The force F on test chrge q cn e written s F q 0 E 0, so from Eq. (23.1) the work done y the electric force s the test chrge moves from to is given y If we divide this y q 0 W = F # dl = q 0 E # dl L L nd compre the result with Eq. (23.13), we find V - V = E # dl L = Ecos f dl L (potentil difference s n integrl of E ) (23.17) The vlue of V - V is independent of the pth tken from to, just s the vlue of W is independent of the pth. To interpret Eq. (23.17), rememer tht E is the electric force per unit chrge on test chrge. If the line integrl 1 # E d l is positive, the electric field does positive work on positive test chrge s it moves from to. In this cse the electric potentil energy decreses s the test chrge moves, so the potentil energy per unit chrge decreses s well; hence V is less thn V nd V - V is positive. As n illustrtion, consider positive point chrge (Fig ). The electric field is directed wy from the chrge, nd V = q/4pp 0 r is positive t ny finite distnce from the chrge. If you move wy from the chrge, in the direction of E, you move towrd lower vlues of V; if you move towrd the chrge, in the direction opposite E, you move towrd greter vlues of V. For the negtive point chrge in Fig , E is directed towrd the chrge nd V = q/4pp 0 r is negtive t ny finite distnce from the chrge. In this cse, if you move towrd the chrge, you re moving in the direction of E nd in the direction of decresing (more negtive) V. Moving wy from the chrge, in the direction opposite E, moves you towrd incresing (less negtive) vlues of V. The generl rule, vlid for ny electric field, is: Moving with the direction of E mens moving in the direction of decresing V, nd moving ginst the direction of E mens moving in the direction of incresing V. Also, positive test chrge q 0 experiences n electric force in the direction of E, towrd lower vlues of V; negtive test chrge experiences force opposite E, towrd higher vlues of V. Thus positive chrge tends to fll from high-potentil region to lower-potentil region. The opposite is true for negtive chrge. Notice tht Eq. (23.17) cn e rewritten s V - V = - E# d l (23.18) L This hs negtive sign compred to the integrl in Eq. (23.17), nd the limits re reversed; hence Eqs. (23.17) nd (23.18) re equivlent. But Eq. (23.18) hs slightly different interprettion. To move unit chrge slowly ginst the electric If you move in the direction of, electric potentil V decreses; if you move in the direction opposite E, V increses. () A positive point chrge V increses s you move inwrd. E () A negtive point chrge V decreses s you move inwrd. E E V decreses s you move outwrd. V increses s you move outwrd.

78 764 CHAPTER 23 Electric Potentil force, we must pply n externl force per unit chrge equl to E, equl nd opposite to the electric force per unit chrge E. Eqution (23.18) sys tht V - V = V, the potentil of with respect to, equls the work done per unit chrge y this externl force to move unit chrge from to. This is the sme lterntive interprettion we discussed under Eq. (23.13). Equtions (23.17) nd (23.18) show tht the unit of potentil difference 11 V2 is equl to the unit of electric field 11 N>C2 multiplied y the unit of distnce 11 m2. Hence the unit of electric field cn e expressed s 1 volt per meter 11 V>m2, s well s 1 N>C: In prctice, the volt per meter is the usul unit of electric-field mgnitude. Electron Volts 1 V>m = 1 volt>meter = 1 N>C = 1 newton>coulom The mgnitude e of the electron chrge cn e used to define unit of energy tht is useful in mny clcultions with tomic nd nucler systems. When prticle with chrge q moves from point where the potentil is V to point where it is V, the chnge in the potentil energy U is U - U = q1v - V 2 = qv If the chrge q equls the mgnitude e of the electron chrge, * C, nd the potentil difference is V = 1 V, the chnge in energy is U - U = * C211 V2 = * J This quntity of energy is defined to e 1 electron volt 11 ev2: 1 ev = * J The multiples mev, kev, MeV, GeV, nd TeV re often used. CAUTION Electron volts vs. volts Rememer tht the electron volt is unit of energy, not unit of potentil or potentil difference! When prticle with chrge e moves through potentil difference of 1 volt, the chnge in potentil energy is 1 ev. If the chrge is some multiple of e sy Ne the chnge in potentil energy in electron volts is N times the potentil difference in volts. For exmple, when n lph prticle, which hs chrge 2e, moves etween two points with potentil difference of 1000 V, the chnge in potentil energy is ev2 = 2000 ev. To confirm this, we write U - U = qv = 12e V2 = * C V2 = * J = 2000 ev Although we hve defined the electron volt in terms of potentil energy, we cn use it for ny form of energy, such s the kinetic energy of moving prticle. When we spek of one-million-electron-volt proton, we men proton with kinetic energy of one million electron volts 11 MeV2, equl to * J2 = * J. The Lrge Hdron Collider ner Genev, witzerlnd, is designed to ccelerte protons to kinetic energy of 7 TeV 17 * ev2. Exmple 23.3 Electric force nd electric potentil A proton (chrge e = * C) moves distnce nitude E = 1.5 * 10 7 V>m = 1.5 * 10 7 N>C in the direction d = 0.50 m in stright line etween points nd in liner ccelertor. The electric field is uniform long this line, with mg- from to. Determine () the force on the proton; () the work done on it y the field; (c) the potentil difference V - V.

79 23.2 Electric Potentil 765 OLUTION IDENTIFY nd ET UP: This prolem uses the reltionship etween electric field nd electric force. It lso uses the reltionship mong force, work, nd potentil-energy difference. We re given the electric field, so it is strightforwrd to find the electric force on the proton. Clculting the work is lso strightforwrd ecuse E is uniform, so the force on the proton is constnt. Once the work is known, we find V - V using Eq. (23.13). EXECUTE: () The force on the proton is in the sme direction s the electric field, nd its mgnitude is F = qe = * C211.5 * 10 7 N>C2 = 2.4 * N () The force is constnt nd in the sme direction s the displcement, so the work done on the proton is W = Fd = 12.4 * N m2 = 1.2 * J = 11.2 * ev J * J = 7.5 * 10 6 ev = 7.5 MeV (c) From Eq. (23.13) the potentil difference is the work per unit chrge, which is V - V = W q We cn get this sme result even more esily y rememering tht 1 electron volt equls 1 volt multiplied y the chrge e. The work done is 7.5 * 10 6 ev nd the chrge is e, so the potentil difference is 17.5 * 10 6 ev2>e = 7.5 * 10 6 V. EVALUATE: We cn check our result in prt (c) y using Eq. (23.17) or Eq. (23.18). The ngle f etween the constnt field E nd the displcement is zero, so Eq. (23.17) ecomes V - V = Ecos f dl = Edl = E dl L L L = 1.2 * J * C = 7.5 * 10 6 J>C = 7.5 * 10 6 V = 7.5 MV The integrl of dl from to is just the distnce d, so we gin find V - V = Ed = 11.5 * 10 7 V>m m2 = 7.5 * 10 6 V Exmple 23.4 Potentil due to two point chrges An electric dipole consists of point chrges q 1 = 12 nc nd q 2 = -12 nc plced 10.0 cm prt (Fig ). Compute the electric potentils t points,, nd c Wht re the potentils t points,, nd c due to this electric dipole? c OLUTION IDENTIFY nd ET UP: This is the sme rrngement s in Exmple 21.8, in which we clculted the electric field t ech point y doing vector sum. Here our trget vrile is the electric potentil V t three points, which we find y doing the lgeric sum in Eq. (23.15). EXECUTE: At point we hve r 1 = m nd r 2 = m, so Eq. (23.15) ecomes 1 V = 4pP 0 i q i r i = 1 4pP 0 q 1 r 1 1 4pP 0 q 2 r 2 = 19.0 * 10 9 N # m 2 >C * 10-9 C m 19.0 * 10 9 N # m 2 >C * 10-9 C m = 1800 N # m>c N # m>c2 = 1800 V V2 = -900 V In similr wy you cn show tht the potentil t point (where r 1 = m nd r 2 = m) is V = 1930 V nd tht the potentil t point c (where r 1 = r 2 = m) is V c = 0. EVALUATE: Let s confirm tht these results mke sense. Point is closer to the 12-nC chrge thn to the 12-nC chrge, so the potentil t is negtive. The potentil is positive t point, which 4.0 cm 13.0 cm 13.0 cm q 1 q cm 4.0 cm is closer to the 12-nC chrge thn the 12-nC chrge. Finlly, point c is equidistnt from the 12-nC chrge nd the 12-nC chrge, so the potentil there is zero. (The potentil is lso equl to zero t point infinitely fr from oth chrges.) Compring this exmple with Exmple 21.8 shows tht it s much esier to clculte electric potentil ( sclr) thn electric field ( vector). We ll tke dvntge of this simplifiction whenever possile.

80 766 CHAPTER 23 Electric Potentil Exmple 23.5 Potentil nd potentil energy Compute the potentil energy ssocited with 4.0-nC point EVALUATE: Note tht zero net work is done on the 4.0-nC chrge if chrge if it is plced t points,, nd c in Fig it moves from point c to infinity y ny pth. In prticulr, let the pth e long the perpendiculr isector of the line joining the other OLUTION two chrges q 1 nd q 2 in Fig As shown in Exmple 21.8 (ection 21.5), t points on the isector, the direction of E is perpendiculr to the isector. Hence the force on the 4.0-nC chrge is IDENTIFY nd ET UP: The potentil energy U ssocited with point chrge q t loction where the electric potentil is V is perpendiculr to the pth, nd no work is done in ny displcement U = qv. We use the vlues of V from Exmple long it. EXECUTE: At the three points we find U = qv = 14.0 * 10-9 C J>C2 = -3.6 * 10-6 J U = qv = 14.0 * 10-9 C J>C2 = 7.7 * 10-6 J U c = qv c = 0 All of these vlues correspond to U nd V eing zero t infinity. Exmple 23.6 Finding potentil y integrtion By integrting the electric field s in Eq. (23.17), find the potentil t distnce r from point chrge q Clculting the potentil y integrting E for single point chrge. OLUTION IDENTIFY nd ET UP: We let point in Eq. (23.17) e t distnce r nd let point e t infinity (Fig ). As usul, we choose the potentil to e zero t n infinite distnce from the chrge q. EXECUTE: To crry out the integrl, we cn choose ny pth we like etween points nd. The most convenient pth is rdil line s shown in Fig , so tht d l is in the rdil direction nd hs mgnitude dr. Writing d l rndr, we hve from Eq. (23.17) q V - 0 = V = E # d l L r q q 1 q = L r 4pP 0 r 2 rn q # rndr = L r 4pP 0 r 2 dr q q = - 4pP 0 r ` q = r 4pP 0 r q V = 4pP 0 r EVALUATE: Our result grees with Eq. (23.14) nd is correct for positive or negtive q. Exmple 23.7 Moving through potentil difference In Fig dust prticle with mss m = 5.0 * 10-9 kg = 5.0 mg nd chrge q 0 = 2.0 nc strts from rest nd moves in stright line from point to point. Wht is its speed v t point? OLUTION IDENTIFY nd ET UP: Only the conservtive electric force cts on the prticle, so mechnicl energy is conserved: K U = K U. We get the potentil energies U from the The prticle moves from point to point ; its ccelertion is not constnt. 3.0 nc 1.0 cm Prticle 1.0 cm 1.0 cm 23.0 nc corresponding potentils V using Eq. (23.12): U = q 0 V nd U = q 0 V.

81 23.3 Clculting Electric Potentil 767 EXECUTE: We hve K nd K = 1 = 0 2 mv2. We sustitute these nd our expressions for U nd U into the energy-conservtion eqution, then solve for v. We find 0 q 0 V = 1 2 mv2 q 0 V 2q v = 0 1V - V 2 A m We clculte the potentils using Eq. (23.15), V = q>4pp 0 r: V = 19.0 * 10 9 N # m 2 >C * 10-9 C m * 10-9 C m = 1350 V Finlly, V = 19.0 * 10 9 N # m 2 >C * 10-9 C m * 10-9 C m = V V - V = V V2 = 2700 V v = A * 10-9 C V2 5.0 * 10-9 kg = 46 m>s EVALUATE: Our result mkes sense: The positive test chrge speeds up s it moves wy from the positive chrge nd towrd the negtive chrge. To check unit consistency in the finl line of the clcultion, note tht 1 V = 1 J>C, so the numertor under the rdicl hs units of J or kg # m 2 >s 2. Test Your Understnding of ection 23.2 If the electric potentil t certin point is zero, does the electric field t tht point hve to e zero? (Hint: Consider point c in Exmple 23.4 nd Exmple 21.8.) 23.3 Clculting Electric Potentil When clculting the potentil due to chrge distriution, we usully follow one of two routes. If we know the chrge distriution, we cn use Eq. (23.15) or (23.16). Or if we know how the electric field depends on position, we cn use Eq. (23.17), defining the potentil to e zero t some convenient plce. ome prolems require comintion of these pproches. As you red through these exmples, compre them with the relted exmples of clculting electric field in ection You ll see how much esier it is to clculte sclr electric potentils thn vector electric fields. The morl is cler: Whenever possile, solve prolems using n energy pproch (using electric potentil nd electric potentil energy) rther thn dynmics pproch (using electric fields nd electric forces). Prolem-olving trtegy 23.1 Clculting Electric Potentil IDENTIFY the relevnt concepts: Rememer tht electric potentil is potentil energy per unit chrge. ET UP the prolem using the following steps: 1. Mke drwing showing the loctions nd vlues of the chrges (which my e point chrges or continuous distriution of chrge) nd your choice of coordinte xes. 2. Indicte on your drwing the position of the point t which you wnt to clculte the electric potentil V. ometimes this position will e n ritrry one (sy, point distnce r from the center of chrged sphere). EXECUTE the solution s follows: 1. To find the potentil due to collection of point chrges, use Eq. (23.15). If you re given continuous chrge distriution, devise wy to divide it into infinitesiml elements nd use Eq. (23.16). Crry out the integrtion, using pproprite limits to include the entire chrge distriution. 2. If you re given the electric field, or if you cn find it using ny of the methods presented in Chpter 21 or 22, it my e esier to find the potentil difference etween points nd using Eq. (23.17) or (23.18). When pproprite, mke use of your freedom to define V to e zero t some convenient plce, nd choose this plce to e point. (For point chrges, this will usully e t infinity. For other distriutions of chrge especilly those tht themselves extend to infinity it my e necessry to define V to e zero t some finite distnce from the chrge distriution.) Then the potentil t ny other point, sy, cn y found from Eq. (23.17) or (23.18) with V = Although potentil V is sclr quntity, you my hve to use components of the vectors E nd d l when you use Eq. (23.17) or (23.18) to clculte V. EVALUATE your nswer: Check whether your nswer grees with your intuition. If your result gives V s function of position, grph the function to see whether it mkes sense. If you know the electric field, you cn mke rough check of your result for V y verifying tht V decreses if you move in the direction of E.

82 768 CHAPTER 23 Electric Potentil Exmple 23.8 A chrged conducting sphere A solid conducting sphere of rdius R hs totl chrge q. Find the electric potentil everywhere, oth outside nd inside the sphere. OLUTION IDENTIFY nd ET UP: In Exmple 22.5 (ection 22.4) we used Guss s lw to find the electric field t ll points for this chrge distriution. We cn use tht result to determine the corresponding potentil. EXECUTE: From Exmple 22.5, the field outside the sphere is the sme s if the sphere were removed nd replced y point chrge q. We tke V = 0 t infinity, s we did for point chrge. Then the potentil t point outside the sphere t distnce r from its center is the sme s tht due to point chrge q t the center: 1 q V = 4pP 0 r The potentil t the surfce of the sphere is V surfce = q>4pp 0 R. Inside the sphere, E is zero everywhere. Hence no work is done on test chrge tht moves from ny point to ny other point inside the sphere. This mens tht the potentil is the sme t every point inside the sphere nd is equl to its vlue q>4pp 0 R t the surfce. EVALUATE: Figure shows the field nd potentil for positive chrge q. In this cse the electric field points rdilly wy from the sphere. As you move wy from the sphere, in the direction of E, V decreses (s it should) Electric-field mgnitude E nd potentil V t points inside nd outside positively chrged sphericl conductor. R E 5 0 O O E V 1 q E 5 4pP R 2 0 q V 5 1 4pP 0 R q E 5 1 4pP r 2 0 V 5 1 4pP 0 q r r r Ioniztion nd Coron Dischrge The results of Exmple 23.8 hve numerous prcticl consequences. One consequence reltes to the mximum potentil to which conductor in ir cn e rised. This potentil is limited ecuse ir molecules ecome ionized, nd ir ecomes conductor, t n electric-field mgnitude of out 3 * 10 6 V>m. Assume for the moment tht q is positive. When we compre the expressions in Exmple 23.8 for the potentil V surfce nd field mgnitude E surfce t the surfce of chrged conducting sphere, we note tht V surfce = E surfce R. Thus, if E m represents the electric-field mgnitude t which ir ecomes conductive (known s the dielectric strength of ir), then the mximum potentil V m to which sphericl conductor cn e rised is V m = RE m For conducting sphere 1 cm in rdius in ir, V 10 6 m = m213 * V>m2 = 30,000 V. No mount of chrging could rise the potentil of conducting sphere of this size in ir higher thn out 30,000 V; ttempting to rise the potentil further y dding extr chrge would cuse the surrounding ir to ecome ionized nd conductive, nd the extr dded chrge would lek into the ir. To ttin even higher potentils, high-voltge mchines such s Vn de Grff genertors use sphericl terminls with very lrge rdii (see Fig nd the photogrph tht opens Chpter 22). For exmple, terminl of rdius R = 2 m hs mximum potentil V m = 12 m213 * 10 6 V>m2 = 6 * 10 6 V = 6 MV. Our result in Exmple 23.8 lso explins wht hppens with chrged conductor with very smll rdius of curvture, such s shrp point or thin wire. Becuse the mximum potentil is proportionl to the rdius, even reltively

83 23.3 Clculting Electric Potentil 769 smll potentils pplied to shrp points in ir produce sufficiently high fields just outside the point to ionize the surrounding ir, mking it ecome conductor. The resulting current nd its ssocited glow (visile in drk room) re clled coron. Lser printers nd photocopying mchines use coron from fine wires to spry chrge on the imging drum (see Fig. 21.2). A lrge-rdius conductor is used in situtions where it s importnt to prevent coron. An exmple is the metl ll t the end of cr rdio ntenn, which prevents the sttic tht would e cused y coron. Another exmple is the lunt end of metl lightning rod (Fig ). If there is n excess chrge in the tmosphere, s hppens during thunderstorms, sustntil chrge of the opposite sign cn uild up on this lunt end. As result, when the tmospheric chrge is dischrged through lightning olt, it tends to e ttrcted to the chrged lightning rod rther thn to other nery structures tht could e dmged. (A conducting wire connecting the lightning rod to the ground then llows the cquired chrge to dissipte hrmlessly.) A lightning rod with shrp end would llow less chrge uildup nd hence would e less effective The metl mst t the top of the Empire tte Building cts s lightning rod. It is struck y lightning s mny s 500 times ech yer. Exmple 23.9 Oppositely chrged prllel pltes Find the potentil t ny height y etween the two oppositely chrged prllel pltes discussed in ection 23.1 (Fig ). OLUTION IDENTIFY nd ET UP: We discussed this sitution in ection From Eq. (23.5), we know the electric potentil energy U for test chrge q 0 is U = q 0 Ey. (We set y 0 nd U 0 t the ottom plte.) We use Eq. (23.12), U = q 0 V, to find the electric potentil V s function of y. EXECUTE: The potentil V1y2 t coordinte y is the potentil energy per unit chrge: V1y2 = U1y2 q 0 The potentil decreses s we move in the direction of from the upper to the lower plte. At point, where y = d nd V1y2 = V, V - V = Ed nd E = V - V d where V is the potentil of the positive plte with respect to the negtive plte. Tht is, the electric field equls the potentil difference etween the pltes divided y the distnce etween them. For given potentil difference V, the smller the distnce d etween the two pltes, the greter the mgnitude E of the electric field. (This reltionship etween E nd holds only for the plnr geometry V = q 0Ey q 0 = Ey = V d E The chrged prllel pltes from Fig E q 0 O y we hve descried. It does not work for situtions such s concentric cylinders or spheres in which the electric field is not uniform.) EVALUATE: Our result shows tht V = 0 t the ottom plte (t y = 0). This is consistent with our choice tht U = q 0 V = 0 for test chrge plced t the ottom plte. CAUTION Zero potentil is ritrry You might think tht if conducting ody hs zero potentil, it must necessrily lso hve zero net chrge. But tht just isn t so! As n exmple, the plte t y = 0 in Fig hs zero potentil 1V = 02 ut hs nonzero chrge per unit re -s. There s nothing prticulrly specil out the plce where potentil is zero; we cn define this plce to e wherever we wnt it to e. y d x? Exmple An infinite line chrge or chrged conducting cylinder Find the potentil t distnce r from very long line of chrge rdil distnce r from long stright-line chrge (Fig ) with liner chrge density (chrge per unit length) l. hs only rdil component given y E r = l>2pp 0 r. We use this expression to find the potentil y integrting E s in OLUTION Eq. (23.17). IDENTIFY nd ET UP: In oth Exmple (ection 21.5) nd EXECUTE: ince the field hs only rdil component, we hve Exmple 22.6 (ection 22.4) we found tht the electric field t E # d l = E r dr. Hence from Eq. (23.17) the potentil of ny point Continued

84 770 CHAPTER 23 Electric Potentil with respect to ny other point, t rdil distnces r nd r from the line of chrge, is r V - V = E # l dr d l = E r dr = L L 2pP 0 Lr r = l ln r 2pP 0 r If we tke point t infinity nd set V = 0, we find tht V is infinite for ny finite distnce r from the line chrge: V = 1l>2pP 0 2ln 1q>r 2 = q. This is not useful wy to define V for this prolem! The difficulty is tht the chrge distriution itself extends to infinity. Insted, s recommended in Prolem-olving trtegy 23.1, we set V = 0 t point t n ritrry ut finite rdil distnce r 0. Then the potentil V = V t point t rdil distnce r is given y V - 0 = 1l>2pP 0 2ln 1r 0 >r2, or V = l ln r 0 2pP 0 r EVALUATE: According to our result, if l is positive, then V decreses s r increses. This is s it should e: V decreses s we move in the direction of E. From Exmple 22.6, the expression for E r with which we strted lso pplies outside long, chrged conducting cylinder with chrge per unit length l (Fig ). Hence our result lso gives the potentil for such cylinder, ut only for vlues Electric field outside () long, positively chrged wire nd () long, positively chrged cylinder. () E r r () R E r of r (the distnce from the cylinder xis) equl to or greter thn the rdius R of the cylinder. If we choose r 0 to e the cylinder rdius R, so tht V = 0 when r = R, then t ny point for which r 7 R, V = l 2pP 0 ln R r Inside the cylinder, E 0, nd V hs the sme vlue (zero) s on the cylinder s surfce. r Exmple A ring of chrge Electric chrge Q is distriuted uniformly round thin ring of rdius (Fig ). Find the potentil t point P on the ring xis t distnce x from the center of the ring. OLUTION IDENTIFY nd ET UP: We divide the ring into infinitesiml segments nd use Eq. (23.16) to find V. All prts of the ring (nd therefore ll elements of the chrge distriution) re t the sme distnce from P. EXECUTE: Figure shows tht the distnce from ech chrge element dq to P is r = 1x 2 2. Hence we cn tke the fctor 1>r outside the integrl in Eq. (23.16), nd 1 dq 1 1 V = = 4pP 0 L r 4pP 0 2x 2 2 L dq = 1 Q 4pP 0 2x 2 2 EVALUATE: When x is much lrger thn, our expression for V ecomes pproximtely V = Q>4pP 0 x, which is the potentil t distnce x from point chrge Q. Very fr wy from chrged All the chrge in ring of chrge Q is the sme distnce r from point P on the ring xis. Q O r 5 x ring, its electric potentil looks like tht of point chrge. We drew similr conclusion out the electric field of ring in Exmple 21.9 (ection 21.5). We know the electric field t ll points long the x-xis from Exmple 21.9 (ection 21.5), so we cn lso find V long this xis y integrting E # d l s in Eq. (23.17). x P Exmple Potentil of line of chrge Positive electric chrge Q is distriuted uniformly long line of length 2 lying long the y-xis etween y = - nd y = (Fig ). Find the electric potentil t point P on the x-xis t distnce x from the origin. OLUTION IDENTIFY nd ET UP: This is the sme sitution s in Exmple (ection 21.5), where we found n expression for the electric

85 23.4 Equipotentil urfces 771 field t n ritrry point on the x-xis. We cn find V t point P y integrting over the chrge distriution using Eq. (23.16). Unlike the sitution in Exmple 23.11, ech chrge element dq is different distnce from point P, so the integrtion will tke little more effort. EXECUTE: As in Exmple 21.10, the element of chrge dq corresponding to n element of length dy on the rod is dq = 1Q>22dy. The distnce from dq to P is 1x 2 y 2, so the contriution dv tht the chrge element mkes to the potentil t P is E To find the potentil t P due to the entire rod, we integrte dv over the length of the rod from y = - to y = : You cn look up the integrl in tle. The finl result is V = dv = V = 1 Q 4pP Q 4pP 0 2 L - dy 2x 2 y 2 dy 2x 2 y 2 1 Q 4pP 0 2 ln 22 x x Our sketch for this prolem. EVALUATE: We cn check our result y letting x pproch infinity. In this limit the point P is infinitely fr from ll of the chrge, so we expect V to pproch zero; you cn verify tht it does. We know the electric field t ll points long the x-xis from Exmple We invite you to use this informtion to find V long this xis y integrting E s in Eq. (23.17). Test Your Understnding of ection 23.3 If the electric field t certin point is zero, does the electric potentil t tht point hve to e zero? (Hint: Consider the center of the ring in Exmple nd Exmple 21.9.) 23.4 Equipotentil urfces Field lines (see ection 21.6) help us visulize electric fields. In similr wy, the potentil t vrious points in n electric field cn e represented grphiclly y equipotentil surfces. These use the sme fundmentl ide s topogrphic mps like those used y hikers nd mountin climers (Fig ). On topogrphic mp, contour lines re drwn through points tht re ll t the sme elevtion. Any numer of these could e drwn, ut typiclly only few contour lines re shown t equl spcings of elevtion. If mss m is moved over the terrin long such contour line, the grvittionl potentil energy mgy does not chnge ecuse the elevtion y is constnt. Thus contour lines on topogrphic mp re relly curves of constnt grvittionl potentil energy. Contour lines re close together where the terrin is steep nd there re lrge chnges in elevtion over smll horizontl distnce; the contour lines re frther prt where the terrin is gently sloping. A ll llowed to roll downhill will experience the gretest downhill grvittionl force where contour lines re closest together. By nlogy to contour lines on topogrphic mp, n equipotentil surfce is three-dimensionl surfce on which the electric potentil V is the sme t every point. If test chrge q 0 is moved from point to point on such surfce, the electric potentil energy q 0 V remins constnt. In region where n electric field is present, we cn construct n equipotentil surfce through ny point. In digrms we usully show only few representtive equipotentils, often with equl potentil differences etween djcent surfces. No point cn e t two different potentils, so equipotentil surfces for different potentils cn never touch or intersect Contour lines on topogrphic mp re curves of constnt elevtion nd hence of constnt grvittionl potentil energy. Equipotentil urfces nd Field Lines Becuse potentil energy does not chnge s test chrge moves over n equipotentil surfce, the electric field cn do no work on such chrge. It follows tht must e perpendiculr to the surfce t every point so tht the electric force q 0 E E is lwys perpendiculr to the displcement of chrge moving on the surfce.

86 772 CHAPTER 23 Electric Potentil Cross sections of equipotentil surfces (lue lines) nd electric field lines (red lines) for ssemlies of point chrges. There re equl potentil differences etween djcent surfces. Compre these digrms to those in Fig , which showed only the electric field lines. () A single positive chrge () An electric dipole (c) Two equl positive chrges V 5130 V V 5150 V V 5170 V V 5230 V V 5130 V V 5250 V V 5 0 V V 5150 V V 5270 V V 5170 V V 5130 V V 5150 V V 5170 V Electric field lines Cross sections of equipotentil surfces Field lines nd equipotentil surfces re lwys mutully perpendiculr. In generl, field lines re curves, nd equipotentils re curved surfces. For the specil cse of uniform field, in which the field lines re stright, prllel, nd eqully spced, the equipotentils re prllel plnes perpendiculr to the field lines. Figure shows three rrngements of chrges. The field lines in the plne of the chrges re represented y red lines, nd the intersections of the equipotentil surfces with this plne (tht is, cross sections of these surfces) re shown s lue lines. The ctul equipotentil surfces re three-dimensionl. At ech crossing of n equipotentil nd field line, the two re perpendiculr. In Fig we hve drwn equipotentils so tht there re equl potentil differences etween djcent surfces. In regions where the mgnitude of E is lrge, the equipotentil surfces re close together ecuse the field does reltively lrge mount of work on test chrge in reltively smll displcement. This is the cse ner the point chrge in Fig or etween the two point chrges in Fig ; note tht in these regions the field lines re lso closer together. This is directly nlogous to the downhill force of grvity eing gretest in regions on topogrphic mp where contour lines re close together. Conversely, in regions where the field is weker, the equipotentil surfces re frther prt; this hppens t lrger rdii in Fig , to the left of the negtive chrge or the right of the positive chrge in Fig , nd t greter distnces from oth chrges in Fig c. (It my pper tht two equipotentil surfces intersect t the center of Fig c, in violtion of the rule tht this cn never hppen. In fct this is single figure-8shped equipotentil surfce.) CAUTION E need not e constnt over n equipotentil surfce On given equipotentil surfce, the potentil V hs the sme vlue t every point. In generl, however, the electricfield mgnitude E is not the sme t ll points on n equipotentil surfce. For instnce, on the equipotentil surfce leled V = -30 V in Fig , the mgnitude E is less to the left of the negtive chrge thn it is etween the two chrges. On the figure-8shped equipotentil surfce in Fig c, E = 0 t the middle point hlfwy etween the two chrges; t ny other point on this surfce, E is nonzero. Equipotentils nd Conductors Here s n importnt sttement out equipotentil surfces: When ll chrges re t rest, the surfce of conductor is lwys n equipotentil surfce.

87 23.4 Equipotentil urfces 773 ince the electric field is lwys perpendiculr to n equipotentil surfce, we cn prove this sttement y proving tht when ll chrges re t rest, the electric field just outside conductor must e perpendiculr to the surfce t every point (Fig ). We know tht E 0 everywhere inside the conductor; otherwise, chrges would move. In prticulr, t ny point just inside the surfce the component of tngent to the surfce is zero. It follows tht the tngentil component of E E is lso zero just outside the surfce. If it were not, chrge could move round rectngulr pth prtly inside nd prtly outside (Fig ) nd return to its strting point with net mount of work hving een done on it. This would violte the conservtive nture of electrosttic fields, so the tngentil component of E just outside the surfce must e zero t every point on the surfce. Thus E is perpendiculr to the surfce t ech point, proving our sttement. It lso follows tht when ll chrges re t rest, the entire solid volume of conductor is t the sme potentil. Eqution (23.17) sttes tht the potentil difference etween two points nd within the conductor s solid volume,, is equl to the line integrl 1 V - V # of the electric field from to. ince E E d l 0 everywhere inside the conductor, the integrl is gurnteed to e zero for ny two such points nd. Hence the potentil is the sme for ny two points within the solid volume of the conductor. We descrie this y sying tht the solid volume of the conductor is n equipotentil volume. Finlly, we cn now prove theorem tht we quoted without proof in ection The theorem is s follows: In n electrosttic sitution, if conductor contins cvity nd if no chrge is present inside the cvity, then there cn e no net chrge nywhere on the surfce of the cvity. This mens tht if you re inside chrged conducting ox, you cn sfely touch ny point on the inside wlls of the ox without eing shocked. To prove this theorem, we first prove tht every point in the cvity is t the sme potentil. In Fig the conducting surfce A of the cvity is n equipotentil surfce, s we hve just proved. uppose point P in the cvity is t different potentil; then we cn construct different equipotentil surfce B including point P. Now consider Gussin surfce, shown in Fig , etween the two equipotentil surfces. Becuse of the reltionship etween E nd the equipotentils, we know tht the field t every point etween the equipotentils is from A towrd B, or else t every point it is from B towrd A, depending on which equipotentil surfce is t higher potentil. In either cse the flux through this Gussin surfce is certinly not zero. But then Guss s lw sys tht the chrge enclosed y the Gussin surfce cnnot e zero. This contrdicts our initil ssumption tht there is no chrge in the cvity. o the potentil t P cnnot e different from tht t the cvity wll. The entire region of the cvity must therefore e t the sme potentil. But for this to e true, the electric field inside the cvity must e zero everywhere. Finlly, Guss s lw shows tht the electric field t ny point on the surfce of conductor is proportionl to the surfce chrge density s t tht point. We conclude tht the surfce chrge density on the wll of the cvity is zero t every point. This chin of resoning my seem tortuous, ut it is worth creful study. E CAUTION Equipotentil surfces vs. Gussin surfces Don t confuse equipotentil surfces with the Gussin surfces we encountered in Chpter 22. Gussin surfces hve relevnce only when we re using Guss s lw, nd we cn choose ny Gussin surfce tht s convenient. We re not free to choose the shpe of equipotentil surfces; the shpe is determined y the chrge distriution. Test Your Understnding of ection 23.4 Would the shpes of the equipotentil surfces in Fig chnge if the sign of ech chrge were reversed? When chrges re t rest, conducting surfce is lwys n equipotentil surfce. Field lines re perpendiculr to conducting surfce. E Cross sections of equipotentil surfces Electric field lines At ll points on the surfce of conductor, the electric field must e perpendiculr to the surfce. If E hd tngentil component, net mount of work would e done on test chrge y moving it round loop s shown here which is impossile ecuse the electric force is conservtive. An impossile electric field If the electric field just outside conductor hd tngentil component E i, chrge could move in loop with net work done. E ' E i E E 5 0 Vcuum Conductor A cvity in conductor. If the cvity contins no chrge, every point in the cvity is t the sme potentil, the electric field is zero everywhere in the cvity, nd there is no chrge nywhere on the surfce of the cvity. A Cross section of equipotentil surfce through P B P Conductor Gussin surfce (in cross section) urfce of cvity

88 774 CHAPTER 23 Electric Potentil 23.5 Potentil Grdient ActivPhysics : Electricl Potentil, Field, nd Force Appliction Potentil Grdient Across Cell Memrne The interior of humn cell is t lower electric potentil V thn the exterior. (The potentil difference when the cell is inctive is out 70 mv in neurons nd out 95 mv in skeletl muscle cells.) Hence there is potentil grdient V tht points from the interior to the exterior of the cell memrne, nd n electric field E - V tht points from the exterior to the interior. This field ffects how ions flow into or out of the cell through specil chnnels in the memrne. Electric field nd potentil re closely relted. Eqution (23.17), restted here, expresses one spect of tht reltionship: V - V = E # d l L If we know t vrious points, we cn use this eqution to clculte potentil differences. In this section we show how to turn this round; if we know the potentil V t vrious points, we cn use it to determine E. Regrding V s function of the coordintes 1x, y, z2 of point in spce, we will show tht the components of E re relted to the prtil derivtives of V with respect to x, y, nd z. In Eq. (23.17), V - V is the potentil of with respect to tht is, the chnge of potentil encountered on trip from to. We cn write this s V - V = dv = - dv L L E where dv is the infinitesiml chnge of potentil ccompnying n infinitesiml element d l of the pth from to. Compring to Eq. (23.17), we hve - dv = E# d l L L These two integrls must e equl for ny pir of limits nd, nd for this to e true the integrnds must e equl. Thus, for ny infinitesiml displcement d l, -dv = E# d l To interpret this expression, we write nd in terms of their components: E ın E nd d l ın dx N dy k N x N E y k N E d l E z dz. Then we hve -dv = E x dx E y dy E z dz uppose the displcement is prllel to the x-xis, so dy = dz = 0. Then -dv = E x dx or E x = -1dV>dx2 y, z constnt, where the suscript reminds us tht only x vries in the derivtive; recll tht V is in generl function of x, y, nd z. But this is just wht is ment y the prtil derivtive 0V>0x. The y- nd z-components of E re relted to the corresponding derivtives of V in the sme wy, so we hve E x = - 0V 0x E y = - 0V 0y E z = - 0V 0z (23.19) This is consistent with the units of electric field eing V>m. In terms of unit vectors we cn write E s E ın 0V 0x N 0V 0V kn 0y 0z 1E in terms of V2 (components of in terms of V) E (23.20) In vector nottion the following opertion is clled the grdient of the function ƒ: (23.21) The opertor denoted y the symol is clled grd or del. Thus in vector nottion, E V (23.22) This is red is the negtive of the grdient of V or equls negtive grd V. The quntity V is clled the potentil grdient. E f ın 0 0x 0 N 0y kn 0 0z ƒ E

89 23.5 Potentil Grdient 775 At ech point, the potentil grdient points in the direction in which V increses most rpidly with chnge in position. Hence t ech point the direction of E is the direction in which V decreses most rpidly nd is lwys perpendiculr to the equipotentil surfce through the point. This grees with our oservtion in ection 23.2 tht moving in the direction of the electric field mens moving in the direction of decresing potentil. Eqution (23.22) doesn t depend on the prticulr choice of the zero point for V. If we were to chnge the zero point, the effect would e to chnge V t every point y the sme mount; the derivtives of V would e the sme. If E is rdil with respect to point or n xis nd r is the distnce from the point or the xis, the reltionship corresponding to Eqs. (23.19) is E r = - 0V 0r 1rdil electric field2 (23.23) Often we cn compute the electric field cused y chrge distriution in either of two wys: directly, y dding the E fields of point chrges, or y first clculting the potentil nd then tking its grdient to find the field. The second method is often esier ecuse potentil is sclr quntity, requiring t worst the integrtion of sclr function. Electric field is vector quntity, requiring computtion of components for ech element of chrge nd seprte integrtion for ech component. Thus, quite prt from its fundmentl significnce, potentil offers very useful computtionl technique in field clcultions. Below, we present two exmples in which knowledge of V is used to find the electric field. We stress once more tht if we know E s function of position, we cn clculte V using Eq. (23.17) or (23.18), nd if we know V s function of position, we cn clculte E using Eq. (23.19), (23.20), or (23.23). Deriving V from E requires integrtion, nd deriving E from V requires differentition. Exmple Potentil nd field of point chrge From Eq. (23.14) the potentil t rdil distnce r from point chrge q is V = q>4pp 0 r. Find the vector electric field from this expression for V. OLUTION IDENTIFY nd ET UP: This prolem uses the generl reltionship etween the electric potentil s function of position nd the electric-field vector. By symmetry, the electric field here hs only rdil component E r. We use Eq. (23.23) to find this component. EXECUTE: From Eq. (23.23), E r = - 0V 0r = - 0 0r 1 q 4pP 0 r = 1 q 4pP 0 r 2 so the vector electric field is E rne r 1 4pP 0 q r 2 rn EVALUATE: Our result grees with Eq. (21.7), s it must. An lterntive pproch is to ignore the rdil symmetry, write the rdil distnce s r = 2x 2 y 2 z 2, nd tke the derivtives of V with respect to x, y, nd z s in Eq. (23.20). We find 0V 0x = 0 0x 1 4pP 0 qx = - 4pP 0 r 3 nd similrly 0V 0y = - qy 4pP 0 r 3 Then from Eq. (23.20), q 2x 2 y 2 z 2 =- 1 4pP 0 qx qy E cın - N - 3 4pP 0 r 4pP 0 r 3 kn qz - 4pP 0 r 3 d 1 q xın y N zkn 4pP 2 0 r r 0V 0z = - qz 4pP 0 r 3 1 4pP 0 q r 2 rn This pproch gives us the sme nswer, ut with more effort. Clerly it s est to exploit the symmetry of the chrge distriution whenever possile. qx 1x 2 y 2 z 2 2 3>2

90 776 CHAPTER 23 Electric Potentil Exmple Potentil nd field of ring of chrge In Exmple (ection 23.3) we found tht for ring of chrge with rdius nd totl chrge Q, the potentil t point P on the ring s symmetry xis distnce x from the center is Find the electric field t P. OLUTION V = 1 Q 4pP 0 2x 2 2 IDENTIFY nd ET UP: Figure shows the sitution. We re given V s function of x long the x-xis, nd we wish to find the electric field t point on this xis. From the symmetry of the chrge distriution, the electric field long the symmetry (x-) xis of the ring cn hve only n x-component. We find it using the first of Eqs. (23.19). EXECUTE: The x-component of the electric field is E x = - 0V 0x = 1 4pP 0 Qx 1x >2 EVALUATE: This grees with our result in Exmple CAUTION Don t use expressions where they don t pply In this exmple, V is not function of y or z on the ring xis, so tht 0V>0y =0V>0z = 0 nd E y = E z = 0. But tht does not men tht it s true everywhere; our expressions for V nd E x re vlid only on the ring xis. If we hd n expression for V vlid t ll points in spce, we could use it to find the components of E t ny point using Eqs. (23.19). Test Your Understnding of ection 23.5 In certin region of spce the potentil is given y V = A Bx Cy 3 Dxy, where A, B, nd D re positive constnts. Which of these sttements out the electric field C,E in this region of spce is correct? (There my e more thn one correct nswer.) (i) Incresing the vlue of A will increse the vlue of E t ll points; (ii) incresing the vlue of A will decrese the vlue of E t ll points; (iii) E hs no z-component; (iv) the electric field is zero t the origin 1x = 0, y = 0, z = 02.

91 CHAPTER 23 UMMARY Electric potentil energy: The electric force cused y ny collection of chrges t rest is conservtive force. The work W done y the electric force on chrged prticle moving in n electric field cn e represented y the chnge in potentil-energy function U. The electric potentil energy for two point chrges q nd q 0 depends on their seprtion r. The electric potentil energy for chrge q 0 in the presence of collection of chrges q 1, q 2, q 3 depends on the distnce from q 0 to ech of these other chrges. (ee Exmples 23.1 nd 23.2.) W = U - U 1 qq 0 U = 4pP 0 r (two point chrges) (23.2) (23.9) U = q 0 q 1 q 2 q 3 4pP 0 r 1 r 2 r Á 3 q 0 q i = (23.10) 4pP 0 i r i ( q 0 in presence of other point chrges) q 1 U 5 q q 1 q 2 q 3 4pP 0 r 1 r 2 r 3 q 2 r 1 r 2 q 3 r 3 q 0 Electric potentil: Potentil, denoted y V, is potentil energy per unit chrge. The potentil difference etween two points equls the mount of work tht would e required to move unit positive test chrge etween those points. The potentil V due to quntity of chrge cn e clculted y summing (if the chrge is collection of point chrges) or y integrting (if the chrge is distriution). (ee Exmples 23.3, 23.4, 23.5, 23.7, 23.11, nd ) The potentil difference etween two points nd, lso clled the potentil of with respect to, is given y the line integrl of E. The potentil t given point cn e found y first finding E nd then crrying out this integrl. (ee Exmples 23.6, 23.8, 23.9, nd ) V = U 1 q = (23.14) q 0 4pP 0 r (due to point chrge) V = U 1 q i = (23.15) q 0 4pP 0 i r i (due to collection of point chrges) 1 dq V = (23.16) 4pP 0 L r (due to chrge distriution) V - V = E# d l = Ecos f dl L L (23.17) q 1 V q 1 q 2 q 3 4pP 0 r 1 r 2 r 3 q 2 r 1 r 2 q 3 r 3 P Equipotentil surfces: An equipotentil surfce is surfce on which the potentil hs the sme vlue t every point. At point where field line crosses n equipotentil surfce, the two re perpendiculr. When ll chrges re t rest, the surfce of conductor is lwys n equipotentil surfce nd ll points in the interior of conductor re t the sme potentil. When cvity within conductor contins no chrge, the entire cvity is n equipotentil region nd there is no surfce chrge nywhere on the surfce of the cvity. Electric field line Cross section of equipotentil surfce Finding electric field from electric potentil: If the potentil V is known s function of the coordintes x, y, nd z, the components of electric field E t ny point re given y prtil derivtives of V. (ee Exmples nd ) E x = - 0V 0x E y = - 0V 0y E z = - 0V 0z (23.19) E ın 0V 0x (vector form) 0V 0V N kn 0y 0z (23.20) 777

92 778 CHAPTER 23 Electric Potentil BRIDGING PROBLEM A Point Chrge nd Line of Chrge Positive electric chrge Q is distriuted uniformly long thin rod of length 2. The rod lies long the x-xis etween x nd x. Clculte how much work you must do to ring positive point chrge q from infinity to the point x L on the x-xis, where L. OLUTION GUIDE ee MsteringPhysics study re for Video Tutor solution. IDENTIFY nd ET UP 1. In this prolem you must first clculte the potentil V t x L due to the chrged rod. You cn then find the chnge in potentil energy involved in ringing the point chrge q from infinity (where V = 0) to x L. 2. To find V, divide the rod into infinitesiml segments of length dx. How much chrge is on such segment? Consider one such segment locted t x = x, where - x. Wht is the potentil dv t x L due to this segment? 3. The totl potentil t x L is the integrl of dv, including contriutions from ll of the segments for x from to. et up this integrl. EXECUTE 4. Integrte your expression from step 3 to find the potentil V t x L. A simple, stndrd sustitution will do the trick; use tle of integrls only s lst resort. 5. Use your result from step 4 to find the potentil energy for point chrge q plced t x L. 6. Use your result from step 5 to find the work you must do to ring the point chrge from infinity to x L. EVALUATE 7. Wht does your result from step 5 ecome in the limit 0? Does this mke sense? 8. uppose the point chrge q were negtive rther thn positive. How would this ffect your result in step 4? In step 5? Prolems For instructor-ssigned homework, go to : Prolems of incresing difficulty. CP: Cumultive prolems incorporting mteril from erlier chpters. CALC: Prolems requiring clculus. BIO: Biosciences prolems. DICUION QUETION Q23.1 A student sked, ince electricl potentil is lwys proportionl to potentil energy, why other with the concept of potentil t ll? How would you respond? Q23.2 The potentil (reltive to point t infinity) midwy etween two chrges of equl mgnitude nd opposite sign is zero. Is it possile to ring test chrge from infinity to this midpoint in such wy tht no work is done in ny prt of the displcement? If so, descrie how it cn e done. If it is not possile, explin why. Q23.3 Is it possile to hve n rrngement of two point chrges seprted y finite distnce such tht the electric potentil energy of the rrngement is the sme s if the two chrges were infinitely fr prt? Why or why not? Wht if there re three chrges? Explin your resoning. Q23.4 ince potentil cn hve ny vlue you wnt depending on the choice of the reference level of zero potentil, how does voltmeter know wht to red when you connect it etween two points? Q23.5 If E is zero everywhere long certin pth tht leds from point A to point B, wht is the potentil difference etween those two points? Does this men tht E is zero everywhere long ny pth from A to B? Explin. Q23.6 If E is zero throughout certin region of spce, is the potentil Figure Q23.7 necessrily lso zero in this region? dl Why or why not? If not, wht cn e sid out the potentil? Q23.7 If you crry out the integrl of E the electric field # 1 E d l for closed pth like tht shown in Fig. Q23.7, the integrl will lwys e equl to zero, independent of the shpe of the pth nd independent of where chrges my e locted reltive to the pth. Explin why. Q23.8 The potentil difference etween the two terminls of n AA ttery (used in flshlights nd portle stereos) is 1.5 V. If two AA tteries re plced end to end with the positive terminl of one ttery touching the negtive terminl of the other, wht is the potentil difference etween the terminls t the exposed ends of the comintion? Wht if the two positive terminls re touching ech other? Explin your resoning. Q23.9 It is esy to produce potentil difference of severl thousnd volts etween your ody nd the floor y scuffing your shoes cross nylon crpet. When you touch metl doorkno, you get mild shock. Yet contct with power line of comprle voltge would proly e ftl. Why is there difference? Q23.10 If the electric potentil t single point is known, cn t tht point e determined? If so, how? If not, why not? Q23.11 Becuse electric field lines nd equipotentil surfces re lwys perpendiculr, two equipotentil surfces cn never cross; if they did, the direction of would e miguous t the crossing points. Yet two equipotentil surfces pper to cross t the center of Fig c. Explin why there is no miguity out the direction of E in this prticulr cse. Q23.12 A uniform electric field is directed due est. Point B is 2.00 m west of point A, point C is 2.00 m est of point A, nd point D is 2.00 m south of A. For ech point, B, C, nd D, is the potentil t tht point lrger, smller, or the sme s t point A? Give the resoning ehind your nswers. Q23.13 We often sy tht if point A is t higher potentil thn point B, A is t positive potentil nd B is t negtive potentil. Does it necessrily follow tht point t positive potentil is positively chrged, or tht point t negtive potentil is negtively chrged? Illustrte your nswers with cler, simple exmples. E E

93 Exercises 779 Q23.14 A conducting sphere is to e chrged y ringing in positive chrge little t time until the totl chrge is Q. The totl work required for this process is lleged to e proportionl to Q 2. Is this correct? Why or why not? Q23.15 Three pirs of prllel Figure Q23.15 metl pltes (A, B, nd C) re connected s shown in Fig. A Q23.15, nd ttery mintins potentil of 1.5 V cross. B Wht cn you sy out the potentil difference cross ech pir of pltes? Why? C Q23.16 A conducting sphere is plced etween two chrged prllel pltes such s those shown in Fig Does the electric field inside the sphere depend on precisely where etween the pltes the sphere is plced? Wht out the electric potentil inside the sphere? Do the nswers to these questions depend on whether or not there is net chrge on the sphere? Explin your resoning. Q23.17 A conductor tht crries net chrge Q hs hollow, empty cvity in its interior. Does the potentil vry from point to point within the mteril of the conductor? Wht out within the cvity? How does the potentil inside the cvity compre to the potentil within the mteril of the conductor? Q23.18 A high-voltge dc power line flls on cr, so the entire metl ody of the cr is t potentil of 10,000 V with respect to the ground. Wht hppens to the occupnts () when they re sitting in the cr nd () when they step out of the cr? Explin your resoning. Q23.19 When thunderstorm is pproching, silors t se sometimes oserve phenomenon clled t. Elmo s fire, luish flickering light t the tips of msts. Wht cuses this? Why does it occur t the tips of msts? Why is the effect most pronounced when the msts re wet? (Hint: ewter is good conductor of electricity.) Q23.20 A positive point chrge is plced ner very lrge conducting plne. A professor of physics sserted tht the field cused y this configurtion is the sme s would e otined y removing the plne nd plcing negtive point chrge of equl mgnitude in the mirror-imge position ehind the initil position of the plne. Is this correct? Why or why not? (Hint: Inspect Fig ) Q23.21 In electronics it is customry to define the potentil of ground (thinking of the erth s lrge conductor) s zero. Is this consistent with the fct tht the erth hs net electric chrge tht is not zero? (Refer to Exercise ) EXERCIE ection 23.1 Electric Potentil Energy A point chrge q 1 = 2.40 mc is held sttionry t the origin. A second point chrge q 2 = mc moves from the point x = m, y = 0 to the point x = m, y = m. How much work is done y the electric force on q 2? A point chrge q 1 is held sttionry t the origin. A second chrge q 2 is plced t point, nd the electric potentil energy of the pir of chrges is 5.4 * 10-8 J. When the second chrge is moved to point, the electric force on the chrge does -1.9 * 10-8 J of work. Wht is the electric potentil energy of the pir of chrges when the second chrge is t point? Energy of the Nucleus. How much work is needed to ssemle n tomic nucleus contining three protons (such s Be) if we model it s n equilterl tringle of side 2.00 * m with proton t ech vertex? Assume the protons strted from very fr wy () How much work would it tke to push two protons very slowly from seprtion of 2.00 * m ( typicl tomic distnce) to 3.00 * m ( typicl nucler distnce)? () If the protons re oth relesed from rest t the closer distnce in prt (), how fst re they moving when they rech their originl seprtion? A smll metl sphere, crrying net chrge of q 1 = mc, is held in sttionry position y insulting supports. A second smll metl sphere, with net chrge of Figure E23.5 v m/s q 2 q 1 q 2 = mc nd mss m 1.50 g, is projected towrd q 1. When the two spheres re m prt, q 2 is moving towrd q 1 with speed 22.0 m>s (Fig. E23.5). Assume tht the two spheres cn e treted s point chrges. You cn ignore the force of grvity. () Wht is the speed of q 2 when the spheres re m prt? () How close does q 2 get to q 1? BIO Energy of DNA Bse Piring, I. (ee Exercise ) () Clculte the electric potentil energy of the deninethymine ond, using the sme comintions of molecules ( O - H - N nd N - H - N) s in Exercise () Compre this energy with the potentil energy of the protonelectron pir in the hydrogen tom BIO Energy of DNA Bse Piring, II. (ee Exercise ) Clculte the electric potentil energy of the gunine cytosine ond, using the sme comintions of molecules ( O - H - O, N - H - N, nd O - H - N) s in Exercise Three equl 1.20-mC point chrges re plced t the corners of n equilterl tringle whose sides re m long. Wht is the potentil energy of the system? (Tke s zero the potentil energy of the three chrges when they re infinitely fr prt.) Two protons re relesed from rest when they re nm prt. () Wht is the mximum speed they will rech? When does this speed occur? () Wht is the mximum ccelertion they will chieve? When does this ccelertion occur? Four electrons re locted t the corners of squre 10.0 nm on side, with n lph prticle t its midpoint. How much work is needed to move the lph prticle to the midpoint of one of the sides of the squre? Three point chrges, which initilly re infinitely fr prt, re plced t the corners of n equilterl tringle with sides d. Two of the point chrges re identicl nd hve chrge q. If zero net work is required to plce the three chrges t the corners of the tringle, wht must the vlue of the third chrge e? trting from seprtion of severl meters, two protons re imed directly towrd ech other y cyclotron ccelertor with speeds of 1000 km>s, mesured reltive to the erth. Find the mximum electricl force tht these protons will exert on ech other. ection 23.2 Electric Potentil A smll prticle hs chrge mc nd mss 2.00 * 10-4 kg. It moves from point A, where the electric potentil is V A = 200 V, to point B, where the electric potentil is V B = 800 V. The electric force is the only force cting on the prticle. The prticle hs speed 5.00 m>s t point A. Wht is its speed t point B? Is it moving fster or slower t B thn t A? Explin.

94 780 CHAPTER 23 Electric Potentil A prticle with chrge of 4.20 nc is in uniform electric field E directed to the left. It is relesed from rest nd moves to the left; fter it hs moved 6.00 cm, its kinetic energy is found to e 1.50 * 10-6 J. () Wht work ws done y the electric force? () Wht is the potentil of the strting point with respect to the end point? (c) Wht is the mgnitude of E? A chrge of 28.0 nc is plced in uniform electric field tht is directed verticlly upwrd nd hs mgnitude of 4.00 * 10 4 V>m. Wht work is done y the electric force when the chrge moves () m to the right; () m upwrd; (c) 2.60 m t n ngle of 45.0 downwrd from the horizontl? Two sttionry point chrges 3.00 nc nd 2.00 nc re seprted y distnce of 50.0 cm. An electron is relesed from rest t point midwy etween the two chrges nd moves long the line connecting the two chrges. Wht is the speed of the electron when it is 10.0 cm from the 3.00-nC chrge? Point chrges q 1 = 2.00 mc nd q 2 = mc re plced t djcent corners of squre for which the length of ech side is 3.00 cm. Point is t the center of the squre, nd point is t the empty corner closest to q 2. Tke the electric potentil to e zero t distnce fr from oth chrges. () Wht is the electric potentil t point due to q 1 nd q 2? () Wht is the electric potentil t point? (c) A point chrge q 3 = mc moves from point to point. How much work is done on q 3 y the electric forces exerted y q 1 nd q 2? Is this work positive or negtive? Two chrges of equl mgnitude Q re held distnce d prt. Consider only points on the line pssing through oth chrges. () If the two chrges hve the sme sign, find the loction of ll points (if there re ny) t which (i) the potentil (reltive to infinity) is zero (is the electric field zero t these points?), nd (ii) the electric field is zero (is the potentil zero t these points?). () Repet prt () for two chrges hving opposite signs Two point chrges Figure E23.19 q 1 = 2.40 nc nd q 2 = nc re m prt. B Point A is midwy etween them; point B is m from q 1 nd m from q 2 (Fig. E23.19). Tke the electric potentil to e zero t infinity m m q A Find () the potentil t point A; 1 q 2 () the potentil t point B; (c) the work done y the electric field on chrge of 2.50 nc tht trvels from point B to point A A positive chrge q is locted t the point x = 0, y = -, nd negtive chrge -q is locted t the point x = 0, y =. () Derive n expression for the potentil V t points on the y-xis s function of the coordinte y. Tke V to e zero t n infinite distnce from the chrges. () Grph V t points on the y-xis s function of y over the rnge from y = -4 to y = 4. (c) how tht for y 7, the potentil t point on the positive y-xis is given y V = -11>4pP 0 22q>y 2. (d) Wht re the nswers to prts () nd (c) if the two chrges re interchnged so tht q is t y = nd -q is t y = -? A positive chrge q is fixed t the point x = 0, y = 0, nd negtive chrge -2q is fixed t the point x =, y = 0. () how the positions of the chrges in digrm. () Derive n expression for the potentil V t points on the x-xis s function of the coordinte x. Tke V to e zero t n infinite distnce from the chrges. (c) At which positions on the x-xis is V = 0? (d) Grph V t points on the x-xis s function of x in the rnge from x = -2 to x = 2. (e) Wht does the nswer to prt () ecome when x W? Explin why this result is otined m m Consider the rrngement of point chrges descried in Exercise () Derive n expression for the potentil V t points on the y-xis s function of the coordinte y. Tke V to e zero t n infinite distnce from the chrges. () At which positions on the y-xis is V = 0? (c) Grph V t points on the y-xis s function of y in the rnge from y = -2 to y = 2. (d) Wht does the nswer to prt () ecome when y 7? Explin why this result is otined () An electron is to e ccelerted from 3.00 * 10 6 m>s to 8.00 * 10 6 m>s. Through wht potentil difference must the electron pss to ccomplish this? () Through wht potentil difference must the electron pss if it is to e slowed from 8.00 * 10 6 m>s to hlt? At certin distnce from point chrge, the potentil nd electric-field mgnitude due to tht chrge re 4.98 V nd 12.0 V>m, respectively. (Tke the potentil to e zero t infinity.) () Wht is the distnce to the point chrge? () Wht is the mgnitude of the chrge? (c) Is the electric field directed towrd or wy from the point chrge? A uniform electric field hs mgnitude E nd is directed in the negtive x-direction. The potentil difference etween point (t x = 0.60 m) nd point (t x = 0.90 m) is 240 V. () Which point, or, is t the higher potentil? () Clculte the vlue of E. (c) A negtive point chrge q = mc is moved from to. Clculte the work done on the point chrge y the electric field For ech of the following rrngements of two point chrges, find ll the points long the line pssing through oth chrges for which the electric potentil V is zero (tke V = 0 infinitely fr from the chrges) nd for which the electric field E is zero: () chrges Q nd 2Q seprted y distnce d, nd () chrges -Q nd 2Q seprted y distnce d. (c) Are oth V nd E zero t the sme plces? Explin. ection 23.3 Clculting Electric Potentil A thin sphericl shell with rdius R 1 = 3.00 cm is concentric with lrger thin sphericl shell with rdius R 2 = 5.00 cm. Both shells re mde of insulting mteril. The smller shell hs chrge q 1 = 6.00 nc distriuted uniformly over its surfce, nd the lrger shell hs chrge q 2 = nc distriuted uniformly over its surfce. Tke the electric potentil to e zero t n infinite distnce from oth shells. () Wht is the electric potentil due to the two shells t the following distnce from their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00 cm? () Wht is the mgnitude of the potentil difference etween the surfces of the two shells? Which shell is t higher potentil: the inner shell or the outer shell? A totl electric chrge of 3.50 nc is distriuted uniformly over the surfce of metl sphere with rdius of 24.0 cm. If the potentil is zero t point t infinity, find the vlue of the potentil t the following distnces from the center of the sphere: () 48.0 cm; () 24.0 cm; (c) 12.0 cm A uniformly chrged, thin ring hs rdius 15.0 cm nd totl chrge 24.0 nc. An electron is plced on the ring s xis distnce 30.0 cm from the center of the ring nd is constrined to sty on the xis of the ring. The electron is then relesed from rest. () Descrie the susequent motion of the electron. () Find the speed of the electron when it reches the center of the ring An infinitely long line of chrge hs liner chrge density 5.00 * C>m. A proton (mss 1.67 * kg, chrge 1.60 * C) is 18.0 cm from the line nd moving directly towrd the line t 1.50 * 10 3 m>s. () Clculte the proton s initil kinetic energy. () How close does the proton get to the line of chrge?

95 Exercises A very long wire crries uniform liner chrge density l. Using voltmeter to mesure potentil difference, you find tht when one proe of the meter is plced 2.50 cm from the wire nd the other proe is 1.00 cm frther from the wire, the meter reds 575 V. () Wht is l? () If you now plce one proe t 3.50 cm from the wire nd the other proe 1.00 cm frther wy, will the voltmeter red 575 V? If not, will it red more or less thn 575 V? Why? (c) If you plce oth proes 3.50 cm from the wire ut 17.0 cm from ech other, wht will the voltmeter red? A very long insulting cylinder of chrge of rdius 2.50 cm crries uniform liner density of 15.0 nc>m. If you put one proe of voltmeter t the surfce, how fr from the surfce must the other proe e plced so tht the voltmeter reds 175 V? A very long insulting cylindricl shell of rdius 6.00 cm crries chrge of liner density 8.50 mc>m spred uniformly over its outer surfce. Wht would voltmeter red if it were connected etween () the surfce of the cylinder nd point 4.00 cm ove the surfce, nd () the surfce nd point 1.00 cm from the centrl xis of the cylinder? A ring of dimeter 8.00 cm is fixed in plce nd crries chrge of 5.00 mc uniformly spred over its circumference. () How much work does it tke to move tiny 3.00-mC chrged ll of mss 1.50 g from very fr wy to the center of the ring? () Is it necessry to tke pth long the xis of the ring? Why? (c) If the ll is slightly displced from the center of the ring, wht will it do nd wht is the mximum speed it will rech? A very smll sphere with positive chrge q = 8.00 mc is relesed from rest t point 1.50 cm from very long line of uniform liner chrge density l = 3.00 mc/m. Wht is the kinetic energy of the sphere when it is 4.50 cm from the line of chrge if the only force on it is the force exerted y the line of chrge? Chrge Q = 5.00 mc is distriuted uniformly over the volume of n insulting sphere tht hs rdius R = 12.0 cm. A smll sphere with chrge q = 3.00 mc nd mss 6.00 * 10-5 kg is projected towrd the center of the lrge sphere from n initil lrge distnce. The lrge sphere is held t fixed position nd the smll sphere cn e treted s point chrge. Wht minimum speed must the smll sphere hve in order to come within 8.00 cm of the surfce of the lrge sphere? BIO Axons. Neurons re the sic units of the nervous system. They contin long tuulr structures clled xons tht propgte electricl signls wy from the ends of the neurons. The xon contins solution of potssium Figure E nm K K 1 K 1 2 Axon 2 memrne K 1 2 K 1 1K 2 ions nd lrge negtive orgnic ions. The xon memrne prevents the lrge ions from leking out, ut the smller K ions re le to penetrte the memrne to some degree (Fig. E23.37). This leves n excess negtive chrge on the inner surfce of the xon memrne nd n excess positive chrge on the outer surfce, resulting in potentil difference cross the memrne tht prevents further ions from leking out. Mesurements show tht this potentil difference is typiclly out 70 mv. The thickness of the xon memrne itself vries from out 5 to 10 nm, so we ll use n verge of 7.5 nm. We cn model the memrne s lrge sheet hving equl nd opposite chrge densities on its fces. () Find the electric field inside the xon memrne, ssuming (not too relisticlly) tht it is filled with ir. Which wy does it point: into or out of the xon? K () Which is t higher potentil: the inside surfce or the outside surfce of the xon memrne? CP Two lrge, prllel conducting pltes crrying opposite chrges of equl mgnitude re seprted y 2.20 cm. () If the surfce chrge density for ech plte hs mgnitude 47.0 nc>m 2, wht is the mgnitude of E in the region etween the pltes? () Wht is the potentil difference etween the two pltes? (c) If the seprtion etween the pltes is douled while the surfce chrge density is kept constnt t the vlue in prt (), wht hppens to the mgnitude of the electric field nd to the potentil difference? Two lrge, prllel, metl pltes crry opposite chrges of equl mgnitude. They re seprted y 45.0 mm, nd the potentil difference etween them is 360 V. () Wht is the mgnitude of the electric field (ssumed to e uniform) in the region etween the pltes? () Wht is the mgnitude of the force this field exerts on prticle with chrge 2.40 nc? (c) Use the results of prt () to compute the work done y the field on the prticle s it moves from the higher-potentil plte to the lower. (d) Compre the result of prt (c) to the chnge of potentil energy of the sme chrge, computed from the electric potentil BIO Electricl ensitivity of hrks. Certin shrks cn detect n electric field s wek s 1.0 mv/m. To grsp how wek this field is, if you wnted to produce it etween two prllel metl pltes y connecting n ordinry 1.5-V AA ttery cross these pltes, how fr prt would the pltes hve to e? () how tht V for sphericl shell of rdius R, tht hs chrge q distriuted uniformly over its surfce, is the sme s V for solid conductor with rdius R nd chrge q. () You ru n inflted lloon on the crpet nd it cquires potentil tht is 1560 V lower thn its potentil efore it ecme chrged. If the chrge is uniformly distriuted over the surfce of the lloon nd if the rdius of the lloon is 15 cm, wht is the net chrge on the lloon? (c) In light of its 1200-V potentil difference reltive to you, do you think this lloon is dngerous? Explin () How much excess chrge must e plced on copper sphere 25.0 cm in dimeter so tht the potentil of its center, reltive to infinity, is 1.50 kv? () Wht is the potentil of the sphere s surfce reltive to infinity? The electric field t the surfce of chrged, solid, copper sphere with rdius m is 3800 N>C, directed towrd the center of the sphere. Wht is the potentil t the center of the sphere, if we tke the potentil to e zero infinitely fr from the sphere? ection 23.4 Equipotentil urfces nd ection 23.5 Potentil Grdient A very lrge plstic sheet crries uniform chrge density of nc>m 2 on one fce. () As you move wy from the sheet long line perpendiculr to it, does the potentil increse or decrese? How do you know, without doing ny clcultions? Does your nswer depend on where you choose the reference point for potentil? () Find the spcing etween equipotentil surfces tht differ from ech other y 1.00 V. Wht type of surfces re these? CALC In certin region of spce, the electric potentil is V(x, y, z) = Axy - Bx 2 Cy, where A, B, nd C re positive constnts. () Clculte the x-, y-, nd z-components of the electric field. () At which points is the electric field equl to zero? CALC In certin region of spce the electric potentil is given y V = Ax 2 y - Bxy 2, where A = 5.00 V/m 3 nd B = 8.00 V/m 3. Clculte the mgnitude nd direction of the electric field t the point in the region tht hs coordintes x = 2.00 m, y = m, nd z = 0.

96 782 CHAPTER 23 Electric Potentil (c) Use Eq. (23.23) nd the result from prt () to show tht the electric field t ny point etween the spheres hs mgnitude (d) Use Eq. (23.23) nd the result from prt () to find the electric field t point outside the lrger sphere t distnce r from the center, where r 7 r. (e) uppose the chrge on the outer sphere is not -q ut negtive chrge of different mgnitude, sy -Q. how tht the nswers for prts () nd (c) re the sme s efore ut the nswer for prt (d) is different A metl sphere with rdius r = 1.20 cm is supported on n insulting stnd t the center of hollow, metl, sphericl shell with rdius r = 9.60 cm. Chrge q is put on the inner sphere nd chrge -q on the outer sphericl shell. The mgnitude of q is chosen to mke the potentil difference etween the spheres 500 V, with the inner sphere t higher potentil. () Use the result of Exercise 23.47() to clculte q. () With the help of the result of Exercise 23.47(), sketch the equipotentil surfces tht correspond to 500, 400, 300, 200, 100, nd 0V. (c) In your sketch, show the electric field lines. Are the electric field lines nd equipotentil surfces mutully perpendiculr? Are the equipotentil surfces closer together when the mgnitude of E is lrgest? A very long cylinder of rdius 2.00 cm crries uniform chrge density of 1.50 nc>m. () Descrie the shpe of the equipotentil surfces for this cylinder. () Tking the reference level for the zero of potentil to e the surfce of the cylinder, find the rdius of equipotentil surfces hving potentils of 10.0 V, 20.0 V, nd 30.0 V. (c) Are the equipotentil surfces eqully spced? If not, do they get closer together or frther prt s r increses? PROBLEM V = E1r2 = q 4pP 0 1 r - 1 r V 11>r - 1>r CP A point chrge q 1 = 5.00 mc is held fixed in spce. From horizontl distnce of 6.00 cm, smll sphere with mss 4.00 * 10-3 kg nd chrge q 2 = 2.00 mc is fired towrd the fixed chrge with n initil speed of 40.0 m/s. Grvity cn e neglected. Wht is the ccelertion of the sphere t the instnt when its speed is 25.0 m/s? A point chrge q 1 = 4.00 nc is plced t the origin, nd second point chrge q 2 = nc is plced on the x-xis t x = 20.0 cm. A third point chrge q 3 = 2.00 nc is to e plced on the x-xis etween q 1 nd q 2. (Tke s zero the potentil energy of the three chrges when they re infinitely fr prt.) () Wht is the potentil energy of the system of the three chrges if q 3 is plced t x = 10.0 cm? () Where should q 3 e plced to mke the potentil energy of the system equl to zero? 1 r CALC A metl sphere with rdius r is supported on n A smll sphere with mss 5.00 * 10-7 kg nd chrge insulting stnd t the center of hollow, metl, sphericl shell 3.00 mc is relesed from rest distnce of m ove lrge with rdius r. There is chrge q on the inner sphere nd chrge horizontl insulting sheet of chrge tht hs uniform surfce -q on the outer sphericl shell. () Clculte the potentil V(r) for chrge density s = 8.00 pc/m 2. Using energy methods, clcu- (i) r 6 r ; (ii) r 6 r 6 r ; (iii) r 7 r. (Hint: The net potentil is the sum of the potentils due to the individul spheres.) Tke V to e zero when r is infinite. () how tht the potentil of the inner sphere with respect to the outer is lte the speed of the sphere when it is m ove the sheet of chrge? Determining the ize of the Nucleus. When rdium- 226 decys rdioctively, it emits n lph prticle (the nucleus of helium), nd the end product is rdon-222. We cn model this decy y thinking of the rdium-226 s consisting of n lph prticle emitted from the surfce of the sphericlly symmetric rdon- 222 nucleus, nd we cn tret the lph prticle s point chrge. The energy of the lph prticle hs een mesured in the lortory nd hs een found to e 4.79 MeV when the lph prticle is essentilly infinitely fr from the nucleus. ince rdon is much hevier thn the lph prticle, we cn ssume tht there is no pprecile recoil of the rdon fter the decy. The rdon nucleus contins 86 protons, while the lph prticle hs 2 protons nd the rdium nucleus hs 88 protons. () Wht ws the electric potentil energy of the lphrdon comintion just efore the decy, in MeV nd in joules? () Use your result from prt () to clculte the rdius of the rdon nucleus CP A proton nd n lph prticle re relesed from rest when they re nm prt. The lph prticle ( helium nucleus) hs essentilly four times the mss nd two times the chrge of proton. Find the mximum speed nd mximum ccelertion of ech of these prticles. When do these mxim occur: just following the relese of the prticles or fter very long time? A prticle with chrge 7.60 nc is in uniform electric field directed to the left. Another force, in ddition to the electric force, cts on the prticle so tht when it is relesed from rest, it moves to the right. After it hs moved 8.00 cm, the dditionl force hs done 6.50 * 10-5 J of work nd the prticle hs 4.35 * 10-5 J of kinetic energy. () Wht work ws done y the electric force? () Wht is the potentil of the strting point with respect to the end point? (c) Wht is the mgnitude of the electric field? CP In the Bohr model of the hydrogen tom, single electron revolves round single proton in circle of rdius r. Assume tht the proton remins t rest. () By equting the electric force to the electron mss times its ccelertion, derive n expression for the electron s speed. () Otin n expression for the electron s kinetic energy, nd show tht its mgnitude is just hlf tht of the electric potentil energy. (c) Otin n expression for the totl energy, nd evlute it using r = 5.29 * m. Give your numericl result in joules nd in electron volts CALC A vcuum tue diode consists of concentric cylindricl electrodes, the negtive cthode nd the positive node. Becuse of the ccumultion of chrge ner the cthode, the electric potentil etween the electrodes is not liner function of the position, even with plnr geometry, ut is given y V1x2 = Cx 4>3 where x is the distnce from the cthode nd C is constnt, chrcteristic of prticulr diode nd operting conditions. Assume tht the distnce etween the cthode nd node is 13.0 mm nd the potentil difference etween electrodes is 240 V. () Determine the vlue of C. () Otin formul for the electric field etween the electrodes s function of x. (c) Determine the force on n electron when the electron is hlfwy etween the electrodes.

97 Prolems Two oppositely chrged, Figure P23.58 identicl insulting spheres, ech 50.0 cm in dimeter nd crrying V uniform chrge of mgnitude 250 mc, re plced 1.00 m prt center to center (Fig. P23.58). () If voltmeter is connected etween the nerest points ( nd ) on their surfces, wht will it red? () Which point, or, is t the higher potentil? How cn you know this without ny clcultions? An Ionic Crystl. Figure P23.59 Figure P23.59 shows eight point 2q 1q chrges rrnged t the corners of cue with sides of length d. 2q The vlues of the chrges re q nd -q, s shown. This is 1q model of one cell of cuic ionic crystl. In sodium chloride d (NCl), for instnce, the positive ions re N nd the negtive ions re Cl - d 1q 2q. () Clculte 2q d the potentil energy U of this 1q rrngement. (Tke s zero the potentil energy of the eight chrges when they re infinitely fr prt.) () In prt (), you should hve found tht U 6 0. Explin the reltionship etween this result nd the oservtion tht such ionic crystls exist in nture () Clculte the potentil energy of system of two smll spheres, one crrying chrge of 2.00 mc nd the other chrge of mc, with their centers seprted y distnce of m. Assume zero potentil energy when the chrges re infinitely seprted. () uppose tht one of the spheres is held in plce nd the other sphere, which hs mss of 1.50 g, is shot wy from it. Wht minimum initil speed would the moving sphere need in order to escpe completely from the ttrction of the fixed sphere? (To escpe, the moving sphere would hve to rech velocity of zero when it ws infinitely distnt from the fixed sphere.) The Ion. The H 2 H 2 ion is composed of two protons, ech of chrge e = 1.60 * C, nd n electron of chrge -e nd mss 9.11 * kg. The seprtion etween the protons is 1.07 * m. The protons nd the electron my e treted s point chrges. () uppose the electron is locted t the point midwy etween the two protons. Wht is the potentil energy of the interction etween the electron nd the two protons? (Do not include the potentil energy due to the interction etween the two protons.) () uppose the electron in prt () hs velocity of mgnitude 1.50 * 10 6 m>s in direction long the perpendiculr isector of the line connecting the two protons. How fr from the point midwy etween the two protons cn the electron move? Becuse the msses of the protons re much greter thn the electron mss, the motions of the protons re very slow nd cn e ignored. Figure P23.62 (Note: A relistic description of the electron motion requires the use of quntum mechnics, not Newtonin mechnics.) CP A smll sphere with mss 1.50 g hngs y 30.0 thred etween two prllel verticl pltes 5.00 cm prt q (Fig. P23.62). The pltes re insulting nd hve uniform 5.00 cm surfce chrge densities s nd -s. The chrge on the sphere is q = 8.90 * 10-6 C. Wht potentil difference etween the pltes will cuse the thred to ssume n ngle of 30.0 with the verticl? CALC Coxil Cylinders. A long metl cylinder with rdius is supported on n insulting stnd on the xis of long, hollow, metl tue with rdius. The positive chrge per unit length on the inner cylinder is l, nd there is n equl negtive chrge per unit length on the outer cylinder. () Clculte the potentil V1r2 for (i) r 6 ; (ii) 6 r 6 ; (iii) r 7. (Hint: The net potentil is the sum of the potentils due to the individul conductors.) Tke V = 0 t r =. () how tht the potentil of the inner cylinder with respect to the outer is l V = ln 2pP 0 (c) Use Eq. (23.23) nd the result from prt () to show tht the electric field t ny point etween the cylinders hs mgnitude (d) Wht is the potentil difference etween the two cylinders if the outer cylinder hs no net chrge? A Geiger counter detects rdition such s lph prticles y using the fct tht the rdition ionizes the ir long its pth. A thin wire lies on the xis of hollow metl cylinder nd is insulted from it (Fig. P23.64). A lrge potentil difference is estlished etween the wire nd the outer cylinder, with the wire t higher potentil; this sets up strong electric field directed rdilly outwrd. When ionizing rdition enters the device, it ionizes few ir molecules. The free electrons produced re ccelerted y the electric field towrd the wire nd, on the wy there, ionize mny more ir molecules. Thus current pulse is produced tht cn e detected y pproprite electronic circuitry nd converted to n udile click. uppose the rdius of the centrl wire is 145 mm nd the rdius of the hollow cylinder is 1.80 cm. Wht potentil difference etween the wire nd the cylinder produces n electric field of 2.00 * 10 4 V>m t distnce of 1.20 cm from the xis of the wire? (The wire nd cylinder re oth very long in comprison to their rdii, so the results of Prolem pply.) Figure P23.64 V E1r2 = V 1 ln1>2 r Counter Free electron Rdition CP Deflection in CRT. Cthode-ry tues (CRTs) re often found in oscilloscopes nd computer monitors. In Fig. P23.65 n electron with n initil speed of 6.50 * 10 6 m>s is projected long the xis midwy etween the deflection pltes of cthodery tue. The potentil difference etween the two pltes is 22.0 V nd the lower plte is the one t higher potentil. () Wht is the force (mgnitude nd direction) on the electron when it is etween the pltes? () Wht is the ccelertion of the electron (mgnitude

98 784 CHAPTER 23 Electric Potentil nd direction) when cted on Figure P23.65 y the force in prt ()? (c) 2.0 cm How fr elow the xis hs the electron moved when it reches v 0 the end of the pltes? (d) At 6.0 cm 12.0 cm wht ngle with the xis is it moving s it leves the pltes? (e) How fr elow the xis will it strike the fluorescent screen? CP Deflecting Pltes of n Oscilloscope. The verticl deflecting pltes of typicl clssroom oscilloscope re pir of prllel squre metl pltes crrying equl ut opposite chrges. Typicl dimensions re out 3.0 cm on side, with seprtion of out 5.0 mm. The potentil difference etween the pltes is 25.0 V. The pltes re close enough tht we cn ignore fringing t the ends. Under these conditions: () how much chrge is on ech plte, nd () how strong is the electric field etween the pltes? (c) If n electron is ejected t rest from the negtive plte, how fst is it moving when it reches the positive plte? Electrosttic precipittors use electric forces to remove pollutnt prticles from smoke, in prticulr in the smokestcks of col-urning power plnts. One form of precipittor consists of verticl, hollow, metl cylinder with thin wire, insulted from the cylinder, running long its xis (Fig. P23.67). A lrge potentil difference is estlished etween the wire nd the outer cylinder, with the wire t lower potentil. This sets up strong rdil electric field directed inwrd. Figure P kv Air flow 14.0 cm The field produces region of ionized ir ner the wire. moke enters the precipittor t the ottom, sh nd dust in it pick up electrons, nd the chrged pollutnts re ccelerted towrd the outer cylinder wll y the electric field. uppose the rdius of the centrl wire is 90.0 mm, the rdius of the cylinder is 14.0 cm, nd potentil difference of 50.0 kv is estlished etween the wire nd the cylinder. Also ssume tht the wire nd cylinder re oth very long in comprison to the cylinder rdius, so the results of Prolem pply. () Wht is the mgnitude of the electric field midwy etween the wire nd the cylinder wll? () Wht mgnitude of chrge must 30.0-mg sh prticle hve if the electric field computed in prt () is to exert force ten times the weight of the prticle? CALC A disk with rdius R hs uniform surfce chrge density s. () By regrding the disk s series of thin concentric rings, clculte the electric potentil V t point on the disk s xis distnce x from the center of the disk. Assume tht the potentil is zero t infinity. (Hint: Use the result of Exmple in ection 23.3.) () Clculte -0V>0x. how tht the result grees with the expression for E x clculted in Exmple (ection 21.5) CALC () From the expression for E otined in Prolem 22.42, find the expressions for the electric potentil V s function of r, oth inside nd outside the cylinder. Let V = 0 t the surfce of the cylinder. In ech cse, express your result in terms of the chrge per unit length l of the chrge distriution. () Grph V nd E s functions of r from r = 0 to r = 3R CALC A thin insulting rod is ent into semicirculr rc of rdius, nd totl electric chrge Q is distriuted uniformly long the rod. Clculte the potentil t the center of curvture of the rc if the potentil is ssumed to e zero t infinity CALC elf-energy of phere of Chrge. A solid sphere of rdius R contins totl chrge Q distriuted uniformly throughout its volume. Find the energy needed to ssemle this chrge y ringing infinitesiml chrges from fr wy. This energy is clled the self-energy of the chrge distriution. (Hint: After you hve ssemled chrge q in sphere of rdius r, how much energy would it tke to dd sphericl shell of thickness dr hving chrge dq? Then integrte to get the totl energy.) CALC () From the expression for E otined in Exmple 22.9 (ection 22.4), find the expression for the electric potentil V s function of r oth inside nd outside the uniformly chrged sphere. Assume tht V = 0 t infinity. () Grph V nd E s functions of r from r = 0 to r = 3R Chrge Q = 4.00 mc is distriuted uniformly over the volume of n insulting sphere tht hs rdius R = 5.00 cm. Wht is the potentil difference etween the center of the sphere nd the surfce of the sphere? An insulting sphericl shell with inner rdius 25.0 cm nd outer rdius 60.0 cm crries chrge of mc uniformly distriuted over its outer surfce (see Exercise 23.41). Point is t the center of the shell, point is on the inner surfce, nd point c is on the outer surfce. () Wht will voltmeter red if it is connected etween the following points: (i) nd ; (ii) nd c; (iii) c nd infinity; (iv) nd c? () Which is t higher potentil: (i) or ; (ii) or c; (iii) or c? (c) Which, if ny, of the nswers would chnge sign if the chrge were -150 mc? Exercise shows tht, outside sphericl shell with uniform surfce chrge, the potentil is the sme s if ll the chrge were concentrted into point chrge t the center of the sphere. () Use this result to show tht for two uniformly chrged insulting shells, the force they exert on ech other nd their mutul electricl energy re the sme s if ll the chrge were concentrted t their centers. (Hint: ee ection 13.6.) () Does this sme result hold for solid insulting spheres, with chrge distriuted uniformly throughout their volume? (c) Does this sme result hold for the force etween two chrged conducting shells? Between two chrged solid conductors? Explin CP Two plstic spheres, ech crrying chrge uniformly distriuted throughout its interior, re initilly plced in contct nd then relesed. One sphere is 60.0 cm in dimeter, hs mss 50.0 g, nd contins mc of chrge. The other sphere is 40.0 cm in dimeter, hs mss g, nd contins mc of chrge. Find the mximum ccelertion nd the mximum speed chieved y ech sphere (reltive to the fixed point of their initil loction in spce), ssuming tht no other forces re cting on them. (Hint: The uniformly distriuted chrges ehve s though they were concentrted t the centers of the two spheres.) CALC Use the electric field clculted in Prolem to clculte the potentil difference etween the solid conducting sphere nd the thin insulting shell CALC Consider solid conducting sphere inside hollow conducting sphere, with rdii nd chrges specified in Prolem Tke V = 0 s r q. Use the electric field clculted in Prolem to clculte the potentil V t the following vlues of r: () r = c (t the outer surfce of the hollow sphere); () r = (t the inner surfce of the hollow sphere); (c) r = (t the surfce of the solid sphere); (d) r = 0 (t the center of the solid sphere) CALC Electric chrge is distriuted uniformly long thin rod of length, with totl chrge Q. Tke the potentil to e zero t

99 Chllenge Prolems 785 infinity. Find the potentil t the Figure P23.79 following points (Fig. P23.79): R () point P, distnce x to the right of the rod, nd () point R, y Q P distnce y ove the right-hnd x end of the rod. (c) In prts () nd (), wht does your result reduce to s x or y ecomes much lrger thn? () If sphericl rindrop of rdius mm crries chrge of pc uniformly distriuted over its volume, wht is the potentil t its surfce? (Tke the potentil to e zero t n infinite distnce from the rindrop.) () Two identicl rindrops, ech with rdius nd chrge specified in prt (), collide nd merge into one lrger rindrop. Wht is the rdius of this lrger drop, nd wht is the potentil t its surfce, if its chrge is uniformly distriuted over its volume? Two metl spheres of different sizes re chrged such tht the electric potentil is the sme t the surfce of ech. phere A hs rdius three times tht of sphere B. Let Q A nd Q B e the chrges on the two spheres, nd let E A nd E B e the electric-field mgnitudes t the surfces of the two spheres. Wht re () the rtio Q B >Q A nd () the rtio E B >E A? An lph prticle with kinetic energy 11.0 MeV mkes hed-on collision with led nucleus t rest. Wht is the distnce of closest pproch of the two prticles? (Assume tht the led nucleus remins sttionry nd tht it my e treted s point chrge. The tomic numer of led is 82. The lph prticle is helium nucleus, with tomic numer 2.) A metl sphere with rdius R 1 hs chrge Q 1. Tke the electric potentil to e zero t n infinite distnce from the sphere. () Wht re the electric field nd electric potentil t the surfce of the sphere? This sphere is now connected y long, thin conducting wire to nother sphere of rdius R 2 tht is severl meters from the first sphere. Before the connection is mde, this second sphere is unchrged. After electrosttic equilirium hs een reched, wht re () the totl chrge on ech sphere; (c) the electric potentil t the surfce of ech sphere; (d) the electric field t the surfce of ech sphere? Assume tht the mount of chrge on the wire is much less thn the chrge on ech sphere CALC Use the chrge distriution nd electric field clculted in Prolem () how tht for r Ú R the potentil is identicl to tht produced y point chrge Q. (Tke the potentil to e zero t infinity.) () Otin n expression for the electric potentil vlid in the region r R CP Nucler Fusion in the un. The source of the sun s energy is sequence of nucler rections tht occur in its core. The first of these rections involves the collision of two protons, which fuse together to form hevier nucleus nd relese energy. For this process, clled nucler fusion, to occur, the two protons must first pproch until their surfces re essentilly in contct. () Assume oth protons re moving with the sme speed nd they collide hed-on. If the rdius of the proton is 1.2 * m, wht is the minimum speed tht will llow fusion to occur? The chrge distriution within proton is sphericlly symmetric, so the electric field nd potentil outside proton re the sme s if it were point chrge. The mss of the proton is 1.67 * kg. () Another nucler fusion rection tht occurs in the sun s core involves collision etween two helium nuclei, ech of which hs 2.99 times the mss of the proton, chrge 2e, nd rdius 1.7 * m. Assuming the sme collision geometry s in prt (), wht minimum speed is required for this fusion rection to tke plce if the nuclei must pproch center-to-center distnce of out 3.5 * m? As for the proton, the chrge of the helium nucleus is uniformly distriuted throughout its volume. (c) In ection 18.3 it ws shown tht the verge trnsltionl kinetic energy of prticle with mss m in gs t solute temperture T 3 is 2 kt, where k is the Boltzmnn constnt (given in Appendix F). For two protons with kinetic energy equl to this verge vlue to e le to undergo the process descried in prt (), wht solute temperture is required? Wht solute temperture is required for two verge helium nuclei to e le to undergo the process descried in prt ()? (At these tempertures, toms re completely ionized, so nuclei nd electrons move seprtely.) (d) The temperture in the sun s core is out 1.5 * 10 7 K. How does this compre to the tempertures clculted in prt (c)? How cn the rections descried in prts () nd () occur t ll in the interior of the sun? (Hint: ee the discussion of the distriution of moleculr speeds in ection 18.5.) CALC The electric potentil V in region of spce is given y where A is constnt. () Derive n expression for the electric field E t ny point in this region. () The work done y the field when 1.50-mC test chrge moves from the point (x, y, z) = (0, 0, m) to the origin is mesured to e 6.00 * 10-5 J. Determine A. (c) Determine the electric field t the point (0, 0, m). (d) how tht in every plne prllel to the xz-plne the equipotentil contours re circles. (e) Wht is the rdius of the equipotentil contour corresponding to V = 1280 V nd y = 2.00 m? Nucler Fission. The Figure P23.87 unstle nucleus of urnium- 236 cn e regrded s uniformly chrged sphere of chrge Q 5192e BEFORE Q = 92e nd rdius R = 7.4 * m. In nucler fission, this cn divide into two smller nuclei, ech with hlf the chrge nd hlf the volume of the originl urnium-236 Q 5146e AFTER Q 5146e nucleus. This is one of the rections tht occurred in the nucler wepon tht exploded over Hiroshim, Jpn, in August () Find the rdii of the two dughter nuclei of chrge 46e. () In simple model for the fission process, immeditely fter the urnium-236 nucleus hs undergone fission, the dughter nuclei re t rest nd just touching, s shown in Fig. P Clculte the kinetic energy tht ech of the dughter nuclei will hve when they re very fr prt. (c) In this model the sum of the kinetic energies of the two dughter nuclei, clculted in prt (), is the energy relesed y the fission of one urnium- 236 nucleus. Clculte the energy relesed y the fission of 10.0 kg of urnium-236. The tomic mss of urnium-236 is 236 u, where 1 u = 1 tomic mss unit = 1.66 * kg. Express your nswer oth in joules nd in kilotons of TNT (1 kiloton of TNT releses 4.18 * J when it explodes). (d) In terms of this model, discuss why n tomic om could just s well e clled n electric om. CHALLENGE PROBLEM V1x, y, z2 = A1x 2-3y 2 z CP CALC In certin region, chrge distriution exists tht is sphericlly symmetric ut nonuniform. Tht is, the

100 786 CHAPTER 23 Electric Potentil volume chrge density r1r2 depends on the distnce r from the where r is the density of the oil. (Ignore the uoynt force of the center of the distriution ut not on the sphericl polr ngles u nd f. The electric potentil V(r) due to this chrge distriution is ir.) By djusting V AB to keep given drop t rest, the chrge on tht drop cn e determined, provided its rdius is known. () Millikn s oil drops were much too smll to mesure their rdii r 0 2 c1-3 r 2 V(r) = c 18P 0 2 r 3 directly. Insted, Millikn determined r y cutting off the electric d for r field nd mesuring the terminl speed v t of the drop s it fell. (We discussed the concept of terminl speed in ection 5.3.) The viscous force F on sphere of rdius r moving with speed v through 0 for r Ú where r 0 is constnt hving units of nd is constnt hving units of meters. () Derive expressions for E fluid with viscosity h is given y tokes s lw: F = 6phrv. When for the regions the drop is flling t v t, the viscous force just lnces the weight r nd r Ú. [Hint: Use Eq. (23.23).] Explin why E hs only rdil component. () Derive n expression for r1r2 in ech of w = mg of the drop. how tht the mgnitude of the chrge on the drop is the two regions r nd r Ú. [Hint: Use Guss s lw for two sphericl shells, one of rdius r nd the other of rdius r dr. The q = 18p d h 3 v 3 t chrge contined in the infinitesiml sphericl shell of rdius dr is V AB A 2rg dq = 4pr 2 r1r2 dr. ] (c) how tht the net chrge contined in the volume of sphere of rdius greter thn or equl to is zero. [Hint: Integrte the expressions derived in prt () for r(r) over sphericl volume of rdius greter thn or equl to. ] Is this result consistent with the electric field for r 7 tht you clculted in Within the limits of their experimentl error, every one of the thousnds of drops tht Millikn nd his coworkers mesured hd chrge equl to some smll integer multiple of sic chrge e. Tht is, they found drops with chrges of 2e, 5e, nd so on, ut none with vlues such s 0.76e or 2.49e. A drop with chrge prt ()? -e hs cquired one extr electron; if its chrge is -2e, it hs CP In experiments in which tomic nuclei collide, cquired two extr electrons, nd so on. (c) A chrged oil drop in hed-on collisions like tht descried in Prolem do hppen, ut ner misses re more common. uppose the lph prticle in Prolem ws not imed t the center of the led nucleus, Millikn oil-drop pprtus is oserved to fll 1.00 mm t constnt speed in 39.3 s if V AB = 0. The sme drop cn e held t rest etween two pltes seprted y 1.00 mm if V AB = 9.16 V. How ut hd n initil nonzero ngulr momentum (with respect to the mny excess electrons hs the drop cquired, nd wht is the sttionry led nucleus) of mgnitude L = p 0, where p 0 is the rdius of the drop? The viscosity of ir is 1.81 * 10-5 N # s>m 2, mgnitude of the initil momentum of the lph prticle nd = 1.00 * m. Wht is the distnce of closest pproch? nd the density of the oil is 824 kg>m CP Two point chrges re moving to the right long the Repet for = 1.00 * m nd = 1.00 * m. x-xis. Point chrge 1 hs chrge q = 2.00 mc, mss m 1 = CALC A hollow, thin-wlled insulting cylinder of 6.00 * 10-5 kg, nd speed v 1. Point chrge 2 is to the right of q 1 rdius R nd length L (like the crdord tue in roll of toilet pper) hs chrge Q uniformly distriuted over its surfce. () Clculte nd hs chrge q mss m 2 = 3.00 * = mc, kg, nd speed v 2. At prticulr instnt, the chrges re seprted y dis- the electric potentil t ll points long the xis of the tue. tnce of 9.00 mm nd hve speeds v 1 = 400 m>s nd Tke the origin to e t the center of the tue, nd tke the potentil v 2 = 1300 m>s. The only forces on the prticles re the forces to e zero t infinity. () how tht if L V R, the result of prt () they exert on ech other. () Determine the speed v cm of the reduces to the potentil on the xis of ring of chrge of rdius R. (ee Exmple in ection 23.3.) (c) Use the result of prt () to find the electric field t ll points long the xis of the tue The Millikn Oil-Drop Experiment. The chrge of n electron ws first mesured y the Americn physicist Roert center of mss of the system. () The reltive energy E rel of the system is defined s the totl energy minus the kinetic energy contriuted y the motion of the center of mss: E rel = E m 1 m 2 2vcm 2 Millikn during In his experiment, oil is spryed in very fine drops (round 10-4 where E = 1 2 is the totl energy of mm in dimeter) into the spce 1v m 2v 2 2 q 1 q 2 >4pP 0 r the system nd r is the distnce etween the chrges. how tht etween two prllel horizontl pltes seprted y distnce d. A E rel = 1 2 where is potentil difference V AB is mintined etween the prllel pltes, q 1 q 2 >4pP 0 r, m = m 1 m 2 >(m 1 m 2 ) clled the reduced mss of the system nd v = v 2 - v 1 is the reltive speed of the moving prticles. (c) For the numericl vlues cusing downwrd electric field etween them. ome of the oil drops cquire negtive chrge ecuse of frictionl effects or given ove, clculte the numericl vlue of E rel. (d) Bsed on the ecuse of ioniztion of the surrounding ir y x rys or rdioctivity. The drops re oserved through microscope. () how tht escpe from one nother? Explin. (e) If the prticles do escpe, result of prt (c), for the conditions given ove, will the prticles n oil drop of rdius r t rest etween the pltes will remin t rest wht will e their finl reltive speed when r q? If the prticles if the mgnitude of its chrge is do not escpe, wht will e their distnce of mximum seprtion? q = 4p rr 3 gd Tht is, wht will e the vlue of r when v = 0? (f) Repet prts (c)(e) for v nd v when the seprtion is V 1 = 400 m>s 2 = 1800 m>s AB mm.

101 Answers 787 Answers Chpter Opening Question? A lrge, constnt potentil difference V is mintined etween the welding tool () nd the metl pieces to e welded (). From Exmple 23.9 (ection 23.3) the electric field etween two conductors seprted y distnce d hs mgnitude E = V /d. Hence d must e smll in order for the field mgnitude E to e lrge enough to ionize the gs etween the conductors nd (see ection 23.3) nd produce n rc through this gs. Test Your Understnding Questions 23.1 Answers: () (i), () (ii) The three chrges q 1, q 2, nd re ll positive, so ll three of the terms in the sum in Eq. (23.11) q 1 q 2 /r 12, q 1 q 3 /r 13, nd q 2 q 3 /r 23 re positive. Hence the totl electric potentil energy U is positive. This mens tht it would tke positive work to ring the three chrges from infinity to the positions shown in Fig , nd hence negtive work to move the three chrges from these positions ck to infinity Answer: no If V = 0 t certin point, E does not hve to e zero t tht point. An exmple is point c in Figs nd 23.13, for which there is n electric field in the x-direction (see Exmple 21.9 in ection 21.5) even though V = 0 (see Exmple 23.4). This isn t surprising result ecuse V nd E re quite different quntities: V is the net mount of work required to ring unit chrge from infinity to the point in question, wheres E is the electric force tht cts on unit chrge when it rrives t tht point Answer: no If E 0 t certin point, V does not hve to e zero t tht point. An exmple is point O t the center of the q 3 chrged ring in Figs nd From Exmple 21.9 (ection 21.5), the electric field is zero t O ecuse the electric-field contriutions from different prts of the ring completely cncel. From Exmple 23.11, however, the potentil t O is not zero: This point corresponds to x = 0, so V = (1/4pP 0 )(Q/). This vlue of V corresponds to the work tht would hve to e done to move unit positive test chrge long pth from infinity to point O; it is nonzero ecuse the chrged ring repels the test chrge, so positive work must e done to move the test chrge towrd the ring Answer: no If the positive chrges in Fig were replced y negtive chrges, nd vice vers, the equipotentil surfces would e the sme ut the sign of the potentil would e reversed. For exmple, the surfces in Fig with potentil V = 30 V nd V = -50 V would hve potentil V = -30 V nd V = 50 V, respectively Answer: (iii) From Eqs. (23.19), the components of the electric field re E x =-0V/0x = B Dy, E y =-0V/0y = 3Cy 2 Dx, nd E z = -0V/0z = 0. The vlue of A hs no effect, which mens tht we cn dd constnt to the electric potentil t ll points without chnging E or the potentil difference etween two points. The potentil does not depend on z, so the z-component of E is zero. Note tht t the origin the electric field is not zero ecuse it hs nonzero x-component: E x = B, E y = 0, E z = 0. Bridging Prolem Answer: qq 8pP 0 ln L L -

102 24 CAPACITANCE AND DIELECTRIC LEARNING GOAL By studying this chpter, you will lern: The nture of cpcitors, nd how to clculte quntity tht mesures their ility to store chrge. How to nlyze cpcitors connected in network. How to clculte the mount of energy stored in cpcitor. Wht dielectrics re, nd how they mke cpcitors more effective.? The energy used in cmer s flsh unit is stored in cpcitor, which consists of two closely spced conductors tht crry opposite chrges. If the mount of chrge on the conductors is douled, y wht fctor does the stored energy increse? When you set n old-fshioned spring mousetrp or pull ck the string of n rcher s ow, you re storing mechnicl energy s elstic potentil energy. A cpcitor is device tht stores electric potentil energy nd electric chrge. To mke cpcitor, just insulte two conductors from ech other. To store energy in this device, trnsfer chrge from one conductor to the other so tht one hs negtive chrge nd the other hs n equl mount of positive chrge. Work must e done to move the chrges through the resulting potentil difference etween the conductors, nd the work done is stored s electric potentil energy. Cpcitors hve tremendous numer of prcticl pplictions in devices such s electronic flsh units for photogrphy, pulsed lsers, ir g sensors for crs, nd rdio nd television receivers. We ll encounter mny of these pplictions in lter chpters (prticulrly Chpter 31, in which we ll see the crucil role plyed y cpcitors in the lternting-current circuits tht pervde our technologicl society). In this chpter, however, our emphsis is on the fundmentl properties of cpcitors. For prticulr cpcitor, the rtio of the chrge on ech conductor to the potentil difference etween the conductors is constnt, clled the cpcitnce. The cpcitnce depends on the sizes nd shpes of the conductors nd on the insulting mteril (if ny) etween them. Compred to the cse in which there is only vcuum etween the conductors, the cpcitnce increses when n insulting mteril ( dielectric) is present. This hppens ecuse redistriution of chrge, clled polriztion, tkes plce within the insulting mteril. tudying polriztion will give us dded insight into the electricl properties of mtter. Cpcitors lso give us new wy to think out electric potentil energy. The energy stored in chrged cpcitor is relted to the electric field in the spce etween the conductors. We will see tht electric potentil energy cn e regrded s eing stored in the field itself. The ide tht the electric field is itself storehouse of energy is t the hert of the theory of electromgnetic wves nd our modern understnding of the nture of light, to e discussed in Chpter

103 24.1 Cpcitors nd Cpcitnce Cpcitors nd Cpcitnce Any two conductors seprted y n insultor (or vcuum) form cpcitor (Fig. 24.1). In most prcticl pplictions, ech conductor initilly hs zero net chrge nd electrons re trnsferred from one conductor to the other; this is clled chrging the cpcitor. Then the two conductors hve chrges with equl mgnitude nd opposite sign, nd the net chrge on the cpcitor s whole remins zero. We will ssume throughout this chpter tht this is the cse. When we sy tht cpcitor hs chrge Q, or tht chrge Q is stored on the cpcitor, we men tht the conductor t higher potentil hs chrge Q nd the conductor t lower potentil hs chrge -Q (ssuming tht Q is positive). Keep this in mind in the following discussion nd exmples. In circuit digrms cpcitor is represented y either of these symols: 24.1 Any two conductors nd insulted from ech other form cpcitor. Conductor 1Q E In either symol the verticl lines (stright or curved) represent the conductors nd the horizontl lines represent wires connected to either conductor. One common wy to chrge cpcitor is to connect these two wires to opposite terminls of ttery. Once the chrges Q nd -Q re estlished on the conductors, the ttery is disconnected. This gives fixed potentil difference V etween the conductors (tht is, the potentil of the positively chrged conductor with respect to the negtively chrged conductor ) tht is just equl to the voltge of the ttery. The electric field t ny point in the region etween the conductors is proportionl to the mgnitude Q of chrge on ech conductor. It follows tht the potentil difference V etween the conductors is lso proportionl to Q. If we doule the mgnitude of chrge on ech conductor, the chrge density t ech point doules, the electric field t ech point doules, nd the potentil difference etween conductors doules; however, the rtio of chrge to potentil difference does not chnge. This rtio is clled the cpcitnce C of the cpcitor: 2Q Conductor ActivPhysics : Electric Potentil: Qulittive Introduction ActivPhysics nd : Electric Potentil, Field, nd Force C = Q V (definition of cpcitnce) (24.1) The I unit of cpcitnce is clled one frd (1 F), in honor of the 19th-century English physicist Michel Frdy. From Eq. (24.1), one frd is equl to one coulom per volt (1 C>V): CAUTION Cpcitnce vs. couloms Don t confuse the symol C for cpcitnce (which is lwys in itlics) with the revition C for couloms (which is never itlicized). The greter the cpcitnce C of cpcitor, the greter the mgnitude Q of chrge on either conductor for given potentil difference V nd hence the greter the mount of stored energy. (Rememer tht potentil is potentil energy per unit chrge.) Thus cpcitnce is mesure of the ility of cpcitor to store energy. We will see tht the vlue of the cpcitnce depends only on the shpes nd sizes of the conductors nd on the nture of the insulting mteril etween them. (The ove remrks out cpcitnce eing independent of Q nd V do not pply to certin specil types of insulting mterils. We won t discuss these mterils in this ook, however.) Clculting Cpcitnce: Cpcitors in Vcuum We cn clculte the cpcitnce C of given cpcitor y finding the potentil difference etween the conductors for given mgnitude of chrge Q nd V 1 F = 1 frd = 1 C>V = 1 coulom>volt

104 790 CHAPTER 24 Cpcitnce nd Dielectrics 24.2 A chrged prllel-plte cpcitor. () Arrngement of the cpcitor pltes Q Q Potentil difference 5 V Wire Wire () ide view of the electric field E Plte, re A When the seprtion of the pltes is smll compred to their size, the fringing of the field is slight. d Plte, re A E then using Eq. (24.1). For now we ll consider only cpcitors in vcuum; tht is, we ll ssume tht the conductors tht mke up the cpcitor re seprted y empty spce. The simplest form of cpcitor consists of two prllel conducting pltes, ech with re A, seprted y distnce d tht is smll in comprison with their dimensions (Fig. 24.2). When the pltes re chrged, the electric field is lmost completely loclized in the region etween the pltes (Fig. 24.2). As we discussed in Exmple 22.8 (ection 22.4), the field etween such pltes is essentilly uniform, nd the chrges on the pltes re uniformly distriuted over their opposing surfces. We cll this rrngement prllel-plte cpcitor. We worked out the electric-field mgnitude E for this rrngement in Exmple (ection 21.5) using the principle of superposition of electric fields nd gin in Exmple 22.8 (ection 22.4) using Guss s lw. It would e good ide to review those exmples. We found tht E = s>p 0, where s is the mgnitude (solute vlue) of the surfce chrge density on ech plte. This is equl to the mgnitude of the totl chrge Q on ech plte divided y the re A of the plte, or s = Q>A, so the field mgnitude E cn e expressed s E = s P 0 = Q P 0 A The field is uniform nd the distnce etween the pltes is d, so the potentil difference (voltge) etween the two pltes is V = Ed = 1 P 0 Qd A From this we see tht the cpcitnce C of prllel-plte cpcitor in vcuum is C = Q V =P 0 A d (cpcitnce of prllel-plte cpcitor in vcuum) (24.2) 24.3 Inside condenser microphone is cpcitor with one rigid plte nd one flexile plte. The two pltes re kept t constnt potentil difference V. ound wves cuse the flexile plte to move ck nd forth, vrying the cpcitnce C nd cusing chrge to flow to nd from the cpcitor in ccordnce with the reltionship C = Q/V. Thus sound wve is converted to chrge flow tht cn e mplified nd recorded digitlly. The cpcitnce depends only on the geometry of the cpcitor; it is directly proportionl to the re A of ech plte nd inversely proportionl to their seprtion d. The quntities A nd d re constnts for given cpcitor, nd P 0 is universl constnt. Thus in vcuum the cpcitnce C is constnt independent of the chrge on the cpcitor or the potentil difference etween the pltes. If one of the cpcitor pltes is flexile, the cpcitnce C chnges s the plte seprtion d chnges. This is the operting principle of condenser microphone (Fig. 24.3). When mtter is present etween the pltes, its properties ffect the cpcitnce. We will return to this topic in ection Menwhile, we remrk tht if the spce contins ir t tmospheric pressure insted of vcuum, the cpcitnce differs from the prediction of Eq. (24.2) y less thn 0.06%. In Eq. (24.2), if A is in squre meters nd d in meters, C is in frds. The units of re C 2 >N # m 2, so we see tht P 0 1 F = 1 C 2 >N # m = 1 C 2 >J Becuse 1 V = 1 J>C (energy per unit chrge), this is consistent with our definition 1 F = 1 C>V. Finlly, the units of P cn e expressed s 1 C 2 >N # m 2 0 = 1 F>m, so P 0 = 8.85 * F>m This reltionship is useful in cpcitnce clcultions, nd it lso helps us to verify tht Eq. (24.2) is dimensionlly consistent. One frd is very lrge cpcitnce, s the following exmple shows. In mny pplictions the most convenient units of cpcitnce re the microfrd

105 24.1 Cpcitors nd Cpcitnce mf = 10-6 F2 nd the picofrd 11 pf = F2. For exmple, the flsh unit in point-nd-shoot cmer uses cpcitor of few hundred microfrds (Fig. 24.4), while cpcitnces in rdio tuning circuit re typiclly from 10 to 100 picofrds. For ny cpcitor in vcuum, the cpcitnce C depends only on the shpes, dimensions, nd seprtion of the conductors tht mke up the cpcitor. If the conductor shpes re more complex thn those of the prllel-plte cpcitor, the expression for cpcitnce is more complicted thn in Eq. (24.2). In the following exmples we show how to clculte C for two other conductor geometries A commercil cpcitor is leled with the vlue of its cpcitnce. For these cpcitors, C = 2200 mf, 1000 mf, nd 470 mf. Exmple 24.1 ize of 1-F cpcitor The prllel pltes of 1.0-F cpcitor re 1.0 mm prt. Wht is their re? OLUTION IDENTIFY nd ET UP: This prolem uses the reltionship mong the cpcitnce C, plte seprtion d, nd plte re A (our trget vrile) for prllel-plte cpcitor. We solve Eq. (24.2) for A. EXECUTE: From Eq. (24.2), A = Cd 11.0 F)11.0 * 10-3 m2 = = 1.1 * 10 8 m 2 P * F>m EVALUATE: This corresponds to squre out 10 km (out 6 miles) on side! The volume of such cpcitor would e t lest Ad = 1.1 * 10 5 m 3, equivlent to tht of cue out 50 m on side. In fct, it s possile to mke 1-F cpcitors few centimeters on side. The trick is to hve n pproprite sustnce etween the pltes rther thn vcuum, so tht (mong other things) the plte seprtion d cn gretly reduced. We ll explore this further in ection Exmple 24.2 Properties of prllel-plte cpcitor The pltes of prllel-plte cpcitor in vcuum re 5.00 mm () The chrge on the cpcitor is prt nd 2.00 m 2 in re. A 10.0-kV potentil difference is pplied Q = CV cross the cpcitor. Compute () the cpcitnce; () the chrge = * 10-9 C>V * 10 4 V2 on ech plte; nd (c) the mgnitude of the electric field etween = 3.54 * 10-5 C = 35.4 mc the pltes. The plte t higher potentil hs chrge 35.4 mc, nd the other plte hs chrge mc. OLUTION (c) The electric-field mgnitude is IDENTIFY nd ET UP: We re given the plte re A, the plte spcing d, nd the potentil difference V = 1.00 * 10 4 V for E = s = Q P this prllel-plte cpcitor. Our trget vriles re the cpcitnce C, the chrge Q on ech plte, nd the electric-field mgni- = 2.00 * 10 6 N>C 0 P 0 A = 3.54 * 10-5 C * C 2 >N # m m 2 2 tude E. We use Eq. (24.2) to clculte C nd then use Eq. (24.1) nd V to find Q. We use E = Q>P 0 A to find E. EVALUATE: We cn lso find E y reclling tht the electric field is equl in mgnitude to the potentil grdient [Eq. (23.22)]. The EXECUTE: () From Eq. (24.2), field etween the pltes is uniform, so A C =P 0 d = * m F>m * 10-3 m = 3.54 * 10-9 F = mf E = V d = 1.00 * 104 V 5.00 * 10-3 m = 2.00 * 106 V>m (Rememer tht 1 N>C = 1 V>m.)

106 792 CHAPTER 24 Cpcitnce nd Dielectrics Exmple 24.3 A sphericl cpcitor Two concentric sphericl conducting shells re seprted y vcuum (Fig. 24.5). The inner shell hs totl chrge Q nd outer rdius r, nd the outer shell hs chrge -Q nd inner rdius r. Find the cpcitnce of this sphericl cpcitor. OLUTION IDENTIFY nd ET UP: By definition, the cpcitnce C is the mgnitude Q of the chrge on either sphere divided y the potentil difference V etween the spheres. We first find V, nd then use Eq. (24.1) to find the cpcitnce C = Q>V. EXECUTE: Using Gussin surfce such s tht shown in Fig. 24.5, we found in Exmple 22.5 (ection 22.4) tht the chrge on conducting sphere produces zero field inside the sphere, so the outer sphere mkes no contriution to the field etween the spheres. Therefore the electric field nd the electric potentil 24.5 A sphericl cpcitor. r r Inner shell, chrge 1Q r Gussin surfce Outer shell, chrge 2Q etween the shells re the sme s those outside chrged conducting sphere with chrge Q. We considered tht prolem in Exmple 23.8 (ection 23.3), so the sme result pplies here: The potentil t ny point etween the spheres is V = Q>4pP 0 r. Hence the potentil of the inner (positive) conductor t r = r with respect to tht of the outer (negtive) conductor t r = r is Q Q V = V - V = - 4pP 0 r 4pP 0 r = Q 1-1 = Q r - r 4pP 0 r r 4pP 0 r r The cpcitnce is then C = Q r r = 4pP V 0 r - r As n exmple, if r = 9.5 cm nd r = 10.5 cm, m) m2 C = 4p18.85 * F>m m = 1.1 * F = 110 pf EVALUATE: We cn relte our expression for C to tht for prllelplte cpcitor. The quntity 4pr r is intermedite etween the res 4pr 2 nd 4pr 2 of the two spheres; in fct, it s the geometric men of these two res, which we cn denote y A gm. The distnce etween spheres is d = r - r, so we cn write C = 4pP 0 r r >1r - r 2 =P 0 A gm >d. This hs the sme form s for prllel pltes: C =P 0 A>d. If the distnce etween spheres is very smll in comprison to their rdii, their cpcitnce is the sme s tht of prllel pltes with the sme re nd spcing. Exmple 24.4 A cylindricl cpcitor Two long, coxil cylindricl conductors re seprted y vcuum (ection 23.3) to find the potentil difference V etween the (Fig. 24.6). The inner cylinder hs rdius r nd liner chrge density l. The outer cylinder hs inner rdius r nd liner chrge the liner chrge density. We then find the corresponding cpci- cylinders, nd find the chrge Q on length L of the cylinders from density -l. Find the cpcitnce per unit length for this cpcitor. tnce C using Eq. (24.1). Our trget vrile is this cpcitnce divided y L. OLUTION EXECUTE: As in Exmple 24.3, the potentil V etween the cylinders is not ffected y the presence of the chrged outer cylinder. IDENTIFY nd ET UP: As in Exmple 24.3, we use the definition of cpcitnce, C = Q>V. We use the result of Exmple Hence our result in Exmple for the potentil outside chrged conducting cylinder lso holds in this exmple for potentil in the spce etween the cylinders: 24.6 A long cylindricl cpcitor. The liner chrge density l is ssumed to e positive in this figure. The mgnitude of chrge in l V = ln r 0 length L of either cylinder is ll. 2pP 0 r 2l Here r 0 is the ritrry, finite rdius t which V = 0. We tke 1l r 0 = r, the rdius of the inner surfce of the outer cylinder. Then the potentil t the outer surfce of the inner cylinder (t which r r = r ) is just the potentil V of the inner (positive) cylinder r with respect to the outer (negtive) cylinder : l V = ln r 2pP 0 r L If l is positive s in Fig. 24.6, then V is positive s well: The inner cylinder is t higher potentil thn the outer.

107 24.2 Cpcitors in eries nd Prllel 793 The totl chrge Q in length L is Q = ll, so from Eq. (24.1) the cpcitnce C of length L is ustituting P 0 = 8.85 * F>m = 8.85 pf>m, we get C = Q C 55.6 pf>m = = L ln1r >r 2 V ll l 2pP 0 ln r r The cpcitnce per unit length is C L = 2pP 0 ln1r >r 2 = 2pP 0L ln1r >r 2 EVALUATE: The cpcitnce of coxil cylinders is determined entirely y their dimensions, just s for prllel-plte nd sphericl cpcitors. Ordinry coxil cles re mde like this ut with n insulting mteril insted of vcuum etween the conductors. A typicl cle used for connecting television to cle TV feed hs cpcitnce per unit length of 69 pf>m. Test Your Understnding of ection 24.1 A cpcitor hs vcuum in the spce etween the conductors. If you doule the mount of chrge on ech conductor, wht hppens to the cpcitnce? (i) It increses; (ii) it decreses; (iii) it remins the sme; (iv) the nswer depends on the size or shpe of the conductors An ssortment of commercilly ville cpcitors Cpcitors in eries nd Prllel Cpcitors re mnufctured with certin stndrd cpcitnces nd working voltges (Fig. 24.7). However, these stndrd vlues my not e the ones you ctully need in prticulr ppliction. You cn otin the vlues you need y comining cpcitors; mny comintions re possile, ut the simplest comintions re series connection nd prllel connection. Cpcitors in eries Figure 24.8 is schemtic digrm of series connection. Two cpcitors re connected in series (one fter the other) y conducting wires etween points nd. Both cpcitors re initilly unchrged. When constnt positive potentil difference V is pplied etween points nd, the cpcitors ecome chrged; the figure shows tht the chrge on ll conducting pltes hs the sme mgnitude. To see why, note first tht the top plte of C 1 cquires positive chrge Q. The electric field of this positive chrge pulls negtive chrge up to the ottom plte of C 1 until ll of the field lines tht egin on the top plte end on the ottom plte. This requires tht the ottom plte hve chrge -Q. These negtive chrges hd to come from the top plte of C 2, which ecomes positively chrged with chrge Q. This positive chrge then pulls negtive chrge -Q from the connection t point onto the ottom plte of C 2. The totl chrge on the lower plte of C 1 nd the upper plte of C 2 together must lwys e zero ecuse these pltes ren t connected to nything except ech other. Thus in series connection the mgnitude of chrge on ll pltes is the sme. Referring to Fig. 24.8, we cn write the potentil differences etween points nd c, c nd, nd nd s nd so (24.3) Following common convention, we use the symols V 1, V 2, nd V to denote the potentil differences V c (cross the first cpcitor), V c (cross the second cpcitor), nd (cross the entire comintion of cpcitors), respectively. V V c = V 1 = Q V C c = V 2 = Q 1 C 2 V = V = V 1 V 2 = Q 1 1 C 1 C 2 V Q = 1 C 1 1 C A series connection of two cpcitors. () Two cpcitors in series Cpcitors in series: The cpcitors hve the sme chrge Q. Their potentil differences dd: V c 1 V c 5 V. V 5 V 1Q 2Q C 1 V c 5 V 1 Chrge is 1Q the sme V s for the C eq 5 individul 2Q cpcitors. 5 c 1Q C V 2Q 2 c 5 V 2 () The equivlent single cpcitor Equivlent cpcitnce is less thn the individul cpcitnces: Q V 1 1 C eq C C 2

108 794 CHAPTER 24 Cpcitnce nd Dielectrics Appliction Touch creens nd Cpcitnce The touch screen on moile phone, n MP3 plyer, or (s shown here) medicl device uses the physics of cpcitors. Behind the screen re two prllel lyers, one ehind the other, of thin strips of trnsprent conductor such s indium tin oxide. A voltge is mintined etween the two lyers. The strips in one lyer re oriented perpendiculr to those in the other lyer; the points where two strips overlp ct s grid of cpcitors. When you ring your finger ( conductor) up to point on the screen, your finger nd the front conducting lyer ct like second cpcitor in series t tht point. The circuitry ttched to the conducting lyers detects the loction of the cpcitnce chnge, nd so detects where you touched the screen. The equivlent cpcitnce C eq of the series comintion is defined s the cpcitnce of single cpcitor for which the chrge Q is the sme s for the comintion, when the potentil difference V is the sme. In other words, the comintion cn e replced y n equivlent cpcitor of cpcitnce C eq. For such cpcitor, shown in Fig. 24.8, C eq = Q V or 1 C eq = V Q Comining Eqs. (24.3) nd (24.4), we find 1 C eq = 1 C 1 1 C 2 (24.4) We cn extend this nlysis to ny numer of cpcitors in series. We find the following result for the reciprocl of the equivlent cpcitnce: 1 C eq = 1 C 1 1 C 2 1 C 3 Á (cpcitors in series) (24.5) The reciprocl of the equivlent cpcitnce of series comintion equls the sum of the reciprocls of the individul cpcitnces. In series connection the equivlent cpcitnce is lwys less thn ny individul cpcitnce. CAUTION Cpcitors in series The mgnitude of chrge is the sme on ll pltes of ll the cpcitors in series comintion; however, the potentil differences of the individul cpcitors re not the sme unless their individul cpcitnces re the sme. The potentil differences of the individul cpcitors dd to give the totl potentil difference cross the series comintion: V totl = V 1 V 2 V 3 Á A prllel connection of two cpcitors. () Two cpcitors in prllel Cpcitors in prllel: The cpcitors hve the sme potentil V. The chrge on ech cpcitor depends on its cpcitnce: Q 1 5 C 1 V, Q 2 5 C 2 V. V 5 V C 1 Q 1 C 2 Q 2 () The equivlent single cpcitor Chrge is the sum of the individul chrges: 1Q V C eq Q 5 Q 1 1 Q 2 2Q Equivlent cpcitnce: C eq 5 C 1 1 C 2 Cpcitors in Prllel The rrngement shown in Fig is clled prllel connection. Two cpcitors re connected in prllel etween points nd. In this cse the upper pltes of the two cpcitors re connected y conducting wires to form n equipotentil surfce, nd the lower pltes form nother. Hence in prllel connection the potentil difference for ll individul cpcitors is the sme nd is equl to V = V. The chrges Q 1 nd Q 2 re not necessrily equl, however, since chrges cn rech ech cpcitor independently from the source (such s ttery) of the voltge V. The chrges re The totl chrge Q of the comintion, nd thus the totl chrge on the equivlent cpcitor, is so Q 1 = C 1 V nd Q 2 = C 2 V Q = Q 1 Q 2 = 1C 1 C 2 2V Q V = C 1 C 2 (24.6) The prllel comintion is equivlent to single cpcitor with the sme totl chrge Q = Q 1 Q 2 nd potentil difference V s the comintion (Fig. 24.9). The equivlent cpcitnce of the comintion, C eq, is the sme s the cpcitnce Q>V of this single equivlent cpcitor. o from Eq. (24.6), C eq = C 1 C 2

109 24.2 Cpcitors in eries nd Prllel 795 In the sme wy we cn show tht for ny numer of cpcitors in prllel, C eq = C 1 C 2 C 3 Á (cpcitors in prllel) (24.7) The equivlent cpcitnce of prllel comintion equls the sum of the individul cpcitnces. In prllel connection the equivlent cpcitnce is lwys greter thn ny individul cpcitnce. CAUTION Cpcitors in prllel The potentil differences re the sme for ll the cpcitors in prllel comintion; however, the chrges on individul cpcitors re not the sme unless their individul cpcitnces re the sme. The chrges on the individul cpcitors dd to give the totl chrge on the prllel comintion: Q 2 Q 3 Á Q totl = Q 1. [Compre these sttements to those in the Cution prgrph following Eq. (24.5).] Prolem-olving trtegy 24.1 Equivlent Cpcitnce IDENTIFY the relevnt concepts: The concept of equivlent cpcitnce is useful whenever two or more cpcitors re connected. ET UP the prolem using the following steps: 1. Mke drwing of the cpcitor rrngement. 2. Identify ll groups of cpcitors tht re connected in series or in prllel. 3. Keep in mind tht when we sy cpcitor hs chrge Q, we men tht the plte t higher potentil hs chrge Q nd the other plte hs chrge -Q. EXECUTE the solution s follows: 1. Use Eq. (24.5) to find the equivlent cpcitnce of cpcitors connected in series, s in Fig uch cpcitors ech hve the sme chrge if they were unchrged efore they were connected; tht chrge is the sme s tht on the equivlent cpcitor. The potentil difference cross the comintion is the sum of the potentil differences cross the individul cpcitors. 2. Use Eq. (24.7) to find the equivlent cpcitnce of cpcitors connected in prllel, s in Fig uch cpcitors ll hve the sme potentil difference cross them; tht potentil difference is the sme s tht cross the equivlent cpcitor. The totl chrge on the comintion is the sum of the chrges on the individul cpcitors. 3. After replcing ll the series or prllel groups you initilly identified, you my find tht more such groups revel themselves. Replce those groups using the sme procedure s ove until no more replcements re possile. If you then need to find the chrge or potentil difference for n individul originl cpcitor, you my hve to retrce your steps. EVALUATE your nswer: Check whether your result mkes sense. If the cpcitors re connected in series, the equivlent cpcitnce C eq must e smller thn ny of the individul cpcitnces. If the cpcitors re connected in prllel, C eq must e greter thn ny of the individul cpcitnces. Exmple 24.5 Cpcitors in series nd in prllel In Figs nd 24.9, let C 1 = 6.0 mf, C 2 = 3.0 mf, nd V = 18 V. Find the equivlent cpcitnce nd the chrge nd potentil difference for ech cpcitor when the cpcitors re connected () in series (see Fig. 24.8) nd () in prllel (see Fig. 24.9). OLUTION IDENTIFY nd ET UP: In oth prts of this exmple trget vrile is the equivlent cpcitnce C eq, which is given y Eq. (24.5) for the series comintion in prt () nd y Eq. (24.7) for the prllel comintion in prt (). In ech prt we find the chrge nd potentil difference using the definition of cpcitnce, Eq. (24.1), nd the rules outlined in Prolem-olving trtegy EXECUTE: () From Eq. (24.5) for series comintion, 1 C eq = 1 C 1 1 C 2 = mf mf C eq = 2.0 mf The chrge Q on ech cpcitor in series is the sme s tht on the equivlent cpcitor: Q = C eq V = 12.0 mf2118 V2 = 36 mc The potentil difference cross ech cpcitor is inversely proportionl to its cpcitnce: V c = V 1 = Q C 1 = V c = V 2 = Q C 2 = 36 mc 6.0 mf = 6.0 V 36 mc 3.0 mf = 12.0 V () From Eq. (24.7) for prllel comintion, C eq = C 1 C 2 = 6.0 mf 3.0 mf = 9.0 mf The potentil difference cross ech of the cpcitors is the sme s tht cross the equivlent cpcitor, 18 V. The chrge on ech cpcitor is directly proportionl to its cpcitnce: Q 1 = C 1 V = 16.0 mf2118 V2 = 108 mc Q 2 = C 2 V = 13.0 mf2118 V2 = 54 mc EVALUATE: As expected, the equivlent cpcitnce C eq for the series comintion in prt () is less thn either or C 2, while C 1 Continued

110 796 CHAPTER 24 Cpcitnce nd Dielectrics tht for the prllel comintion in prt () is greter thn either C 1 equivlent cpcitor: V c V c = V = 18 V. By contrst, for or C 2. For two cpcitors in series, s in prt (), the chrge is the two cpcitors in prllel, s in prt (), ech cpcitor hs the sme on either cpcitor nd the lrger potentil difference sme potentil difference nd the lrger chrge ppers on the ppers cross the cpcitor with the smller cpcitnce. Furthermore, the sum of the potentil differences cross the individul chrge Q 1 Q 2 on the prllel comintion is equl to the chrge cpcitor with the lrger cpcitnce. Cn you show tht the totl cpcitors in series equls the potentil difference cross the Q = C eq V on the equivlent cpcitor? Exmple 24.6 A cpcitor network Find the equivlent cpcitnce of the five-cpcitor network shown in Fig OLUTION IDENTIFY nd ET UP: These cpcitors re neither ll in series nor ll in prllel. We cn, however, identify portions of the rrngement tht re either in series or prllel. We comine these s descried in Prolem-olving trtegy 24.1 to find the net equivlent cpcitnce, using Eq. (24.5) for series connections nd Eq. (24.7) for prllel connections. EXECUTE: The cption of Fig outlines our procedure. We first use Eq. (24.5) to replce the 12-mF nd 6-mF series comintion y its equivlent cpcitnce C : 1 C = 1 12 mf 1 6 mf C =4 mf This gives us the equivlent comintion of Fig Now we see three cpcitors in prllel, nd we use Eq. (24.7) to replce them with their equivlent cpcitnce C: C =3 mf 11 mf 4 mf = 18 mf This gives us the equivlent comintion of Fig c, which hs two cpcitors in series. We use Eq. (24.5) to replce them with their equivlent cpcitnce C eq, which is our trget vrile (Fig d): 1 C eq = 1 18 mf 1 9 mf C eq = 6 mf EVALUATE: If the potentil difference cross the entire network in Fig is V = 9.0 V, the net chrge on the network is Q = C eq V = 16 mf219.0 V2 = 54 mc. Cn you find the chrge on, nd the voltge cross, ech of the five individul cpcitors? () A cpcitor network etween points nd. () The 12-mF nd 6-mF cpcitors in series in () re replced y n equivlent 4-mF cpcitor. (c) The 3-mF, 11-mF, nd 4-mF cpcitors in prllel in () re replced y n equivlent 18-mF cpcitor. (d) Finlly, the 18-mF nd 9-mF cpcitors in series in (c) re replced y n equivlent 6-mF cpcitor. () () (c) (d) 12 mf 3 mf 11 mf 3 mf 11 mf 4 mf 18 mf 6 mf Replce these series cpcitors 9 mf 9 mf y n equivlent cpcitor replce these prllel cpcitors y n equivlent cpcitor... 9 mf... replce these series cpcitors y n equivlent cpcitor. 6 mf Test Your Understnding of ection 24.2 You wnt to connect 4-mF cpcitor nd n 8-mF cpcitor. () With which type of connection will the 4-mF cpcitor hve greter potentil difference cross it thn the 8-mF cpcitor? (i) series; (ii) prllel; (iii) either series or prllel; (iv) neither series nor prllel. () With which type of connection will the 4-mF cpcitor hve greter chrge thn the 8-mF cpcitor? (i) series; (ii) prllel; (iii) either series or prllel; (iv) neither series nor prllel Energy torge in Cpcitors nd Electric-Field Energy Mny of the most importnt pplictions of cpcitors depend on their ility to store energy. The electric potentil energy stored in chrged cpcitor is just equl to the mount of work required to chrge it tht is, to seprte opposite chrges nd plce them on different conductors. When the cpcitor is dischrged, this stored energy is recovered s work done y electricl forces.

111 24.3 Energy torge in Cpcitors nd Electric-Field Energy 797 We cn clculte the potentil energy U of chrged cpcitor y clculting the work W required to chrge it. uppose tht when we re done chrging the cpcitor, the finl chrge is Q nd the finl potentil difference is V. From Eq. (24.1) these quntities re relted y V = Q C Let q nd v e the chrge nd potentil difference, respectively, t n intermedite stge during the chrging process; then v = q>c. At this stge the work dw required to trnsfer n dditionl element of chrge dq is dw = v dq = qdq C The totl work W needed to increse the cpcitor chrge q from zero to finl vlue Q is W W = dw = 1 Q qdq = Q2 (work to chrge cpcitor) (24.8) L 0 C L0 2C This is lso equl to the totl work done y the electric field on the chrge when the cpcitor dischrges. Then q decreses from n initil vlue Q to zero s the elements of chrge dq fll through potentil differences v tht vry from V down to zero. If we define the potentil energy of n unchrged cpcitor to e zero, then W in Eq. (24.8) is equl to the potentil energy U of the chrged cpcitor. The finl stored chrge is Q = CV, so we cn express U (which is equl to W) s U = Q2 2C = 1 2 CV 2 = 1 2 QV (potentil energy stored in cpcitor) (24.9) When Q is in couloms, C in frds (couloms per volt), nd V in volts (joules per coulom), U is in joules. The lst form of Eq. (24.9), U = 1 2 QV, shows tht the totl work W required to chrge the cpcitor is equl to the totl chrge Q multiplied y the verge 1 potentil difference 2 V during the chrging process. The expression U = 1 2 1Q2 >C2 in Eq. (24.9) shows tht chrged cpcitor is the electricl nlog of stretched spring with elstic potentil energy U = 1 2 kx 2. The chrge Q is nlogous to the elongtion x, nd the reciprocl of the cpcitnce, 1>C, is nlogous to the force constnt k. The energy supplied to cpcitor in the chrging process is nlogous to the work we do on spring when we stretch it. Equtions (24.8) nd (24.9) tell us tht cpcitnce mesures the ility of cpcitor to store oth energy nd chrge. If cpcitor is chrged y connecting it to ttery or other source tht provides fixed potentil difference V, then incresing the vlue of C gives greter chrge Q = CV nd greter mount of stored energy U = 1 2 CV 2. If insted the gol is to trnsfer given quntity of chrge Q from one conductor to nother, Eq. (24.8) shows tht the work W required is inversely proportionl to C; the greter the cpcitnce, the esier it is to give cpcitor fixed mount of chrge. Applictions of Cpcitors: Energy torge Most prcticl pplictions of cpcitors tke dvntge of their ility to store nd relese energy. In electronic flsh units used y photogrphers, the? energy stored in cpcitor (see Fig. 24.4) is relesed y depressing the cmer s shutter utton. This provides conducting pth from one cpcitor plte to the other through the flsh tue. Once this pth is estlished, the stored energy is rpidly converted into rief ut intense flsh of light. An extreme exmple of the sme principle is the Z mchine t ndi Ntionl Lortories in New Mexico,

112 798 CHAPTER 24 Cpcitnce nd Dielectrics The Z mchine uses lrge numer of cpcitors in prllel to give tremendous equivlent cpcitnce C (see ection 24.2). Hence lrge mount of energy U = 1 2 CV 2 cn e stored with even modest potentil difference V. The rcs shown here re produced when the cpcitors dischrge their energy into trget, which is no lrger thn spool of thred. This hets the trget to temperture higher thn 2 * 10 9 K. which is used in experiments in controlled nucler fusion (Fig ). A nk of chrged cpcitors releses more thn million joules of energy in just few illionths of second. For tht rief spce of time, the power output of the Z mchine is 2.9 * W, or out 80 times the power output of ll the electric power plnts on erth comined! In other pplictions, the energy is relesed more slowly. prings in the suspension of n utomoile help smooth out the ride y soring the energy from sudden jolts nd relesing tht energy grdully; in n nlogous wy, cpcitor in n electronic circuit cn smooth out unwnted vritions in voltge due to power surges. We ll discuss these circuits in detil in Chpter 26. Electric-Field Energy We cn chrge cpcitor y moving electrons directly from one plte to nother. This requires doing work ginst the electric field etween the pltes. Thus we cn think of the energy s eing stored in the field in the region etween the pltes. To develop this reltionship, let s find the energy per unit volume in the spce etween the pltes of prllel-plte cpcitor with plte re A nd seprtion d. We cll this the energy density, denoted y u. From Eq. (24.9) the totl stored 1 potentil energy is 2 CV 2 nd the volume etween the pltes is just Ad; hence the energy density is 1 2 u = Energy density = CV 2 (24.10) Ad From Eq. (24.2) the cpcitnce C is given y C =P 0 A>d. The potentil difference V is relted to the electric-field mgnitude E y V = Ed. If we use these expressions in Eq. (24.10), the geometric fctors A nd d cncel, nd we find u = 1 2 P 0E 2 (electric energy density in vcuum) (24.11) Although we hve derived this reltionship only for prllel-plte cpcitor, it turns out to e vlid for ny cpcitor in vcuum nd indeed for ny electric field configurtion in vcuum. This result hs n interesting impliction. We think of vcuum s spce with no mtter in it, ut vcuum cn nevertheless hve electric fields nd therefore energy. Thus empty spce need not e truly empty fter ll. We will use this ide nd Eq. (24.11) in Chpter 32 in connection with the energy trnsported y electromgnetic wves. CAUTION Electric-field energy is electric potentil energy It s common misconception tht electric-field energy is new kind of energy, different from the electric potentil energy descried efore. This is not the cse; it is simply different wy of interpreting electric potentil energy. We cn regrd the energy of given system of chrges s eing shred property of ll the chrges, or we cn think of the energy s eing property of the electric field tht the chrges crete. Either interprettion leds to the sme vlue of the potentil energy. Exmple 24.7 Trnsferring chrge nd energy etween cpcitors We connect cpcitor C 1 = 8.0 mf to power supply, chrge it to potentil difference V 0 = 120 V, nd disconnect the power supply (Fig ). witch is open. () Wht is the chrge Q 0 on C 1? () Wht is the energy stored in C 1? (c) Cpcitor C 2 = 4.0 mf is initilly unchrged. We close switch. After chrge no longer flows, wht is the potentil difference cross ech cpcitor, nd wht is the chrge on ech cpcitor? (d) Wht is the finl energy of the system? When the switch is closed, the chrged cpcitor C 1 is connected to n unchrged cpcitor C 2. The center prt of the switch is n insulting hndle; chrge cn flow only etween the two upper terminls nd etween the two lower terminls. C mf Q 0 V V C mf

113 24.3 Energy torge in Cpcitors nd Electric-Field Energy 799 OLUTION IDENTIFY nd ET UP: In prts () nd () we find the chrge Q 1 Q 2 = Q 0 V 0 nd stored energy U initil for the single chrged cpcitor C 1 using Eqs. (24.1) nd (24.9), respectively. After we close switch, one wire connects the upper pltes of the two cpcitors nd nother wire connects the lower pltes; the cpcitors re now connected etween the pltes is the sme for oth cpcitors ecuse they re connected in prllel, so the chrges re Q 1 = C 1 V nd Q 2 = C 2 V. We now hve three independent equtions relting the three unknowns Q 1, Q 2, nd V. olving these, we find in prllel. In prt (c) we use the chrcter of the prllel connection to determine how Q is shred etween the two cpcitors. In mc Q 0 V = = prt (d) we gin use Eq. (24.9) to find the energy stored in cpcitors C 1 nd C 2 ; the energy of the system is the sum of these vlues. Q 1 = 640 mc Q 2 = 320 mc C 1 C mf 4.0 mf = 80 V EXECUTE: () The initil chrge Q 0 on C 1 is Q 0 = C 1 V 0 = 18.0 mf21120 V2 = 960 mc (d) The finl energy of the system is U finl = 1 2 Q 1V 1 2 Q 2V = 1 2 Q 0V () The energy initilly stored in C 1 is = * 10-6 C2180 V2 = J U initil = 1 2 Q 0V 0 = * 10-6 C21120 V2 = J EVALUATE: The finl energy is less thn the initil energy; the difference (c) When we close the switch, the positive chrge Q 0 is distriuted over the upper pltes of oth cpcitors nd the negtive chrge -Q 0 is distriuted over the lower pltes. Let Q 1 nd Q 2 e the mgnitudes of the finl chrges on the cpcitors. Conservtion ws converted to energy of some other form. The conduc- tors ecome little wrmer ecuse of their resistnce, nd some energy is rdited s electromgnetic wves. We ll study the ehvior of cpcitors in more detil in Chpters 26 nd 31. Exmple 24.8 Electric-field energy () Wht is the mgnitude of the electric field required to store 1.00 J of electric potentil energy in volume of 1.00 m 3 in vcuum? () If the field mgnitude is 10 times lrger thn tht, how much energy is stored per cuic meter? OLUTION IDENTIFY nd ET UP: We use the reltionship etween the electricfield mgnitude E nd the energy density u. In prt () we use the given informtion to find u; then we use Eq. (24.11) to find the corresponding vlue of E. In prt (), Eq. (24.11) tells us how u vries with E. EXECUTE: () The desired energy density is u = 1.00 J>m 3. Then from Eq. (24.11), 2u J>m 3 2 E = = A P 0 B 8.85 * C 2 >N # m 2 = 4.75 * 10 5 N>C = 4.75 * 10 5 V>m () Eqution (24.11) shows tht u is proportionl to E 2. If E increses y fctor of 10, u increses y fctor of 10 2 = 100, so the energy density ecomes u = 100 J>m 3. EVALUATE: Dry ir cn sustin n electric field of out 3 * 10 6 V>m without experiencing dielectric rekdown, which we will discuss in ection There we will see tht field mgnitudes in prcticl insultors cn e s gret s this or even lrger. Exmple 24.9 Two wys to clculte energy stored in cpcitor The sphericl cpcitor descried in Exmple 24.3 (ection 24.1) hs chrges Q nd -Q on its inner nd outer conductors. Find the electric potentil energy stored in the cpcitor () y using the cpcitnce C found in Exmple 24.3 nd () y integrting the electric-field energy density u. OLUTION IDENTIFY nd ET UP: We cn determine the energy U stored in cpcitor in two wys: in terms of the work done to put the chrges on the two conductors, nd in terms of the energy in the electric field etween the conductors. The descriptions re equivlent, so they must give us the sme result. In Exmple 24.3 we found the cpcitnce C nd the field mgnitude E in the spce etween the conductors. (The electric field is zero inside the inner sphere nd is lso zero outside the inner surfce of the outer sphere, ecuse Gussin surfce with rdius r 6 r or r 7 r encloses zero net chrge. Hence the energy density is nonzero only in the spce etween the spheres, r 6 r 6 r.) In prt () we use Eq. (24.9) to find U. In prt () we use Eq. (24.11) to find u, which we integrte over the volume etween the spheres to find U. EXECUTE: () From Exmple 24.3, the sphericl cpcitor hs cpcitnce r r C = 4pP 0 r - r where r nd r re the rdii of the inner nd outer conducting spheres, respectively. From Eq. (24.9) the energy stored in this cpcitor is U = Q2 2C = Q2 8pP 0 r - r r r Continued

114 800 CHAPTER 24 Cpcitnce nd Dielectrics () The electric field in the region r 6 r 6 r etween the two conducting spheres hs mgnitude E = Q>4pP 0 r 2. The energy density in this region is u = 1 2 P 0E 2 = 1 2 P Q 2 0 4pP 0 r 2 = 32p 2 P 0 r 4 The energy density is not uniform; it decreses rpidly with incresing distnce from the center of the cpcitor. To find the totl electric-field energy, we integrte u (the energy per unit volume) over the region r 6 r 6 r. We divide this region into sphericl shells of rdius r, surfce re 4pr 2, thickness dr, nd volume dv = 4pr 2 dr. Then Q 2 r U = udv = L L 32p 2 P 0 r 4 4pr 2 dr = Q2 8pP 0 L r r r dr r 2 = = Q2 8pP 0 r - r r r Q pP 0 r r EVALUATE: Electric potentil energy cn e regrded s eing ssocited with either the chrges, s in prt (), or the field, s in prt (); the clculted mount of stored energy is the sme in either cse. Q 2 Test Your Understnding of ection 24.3 You wnt to connect 4-mF cpcitor nd n 8-mF cpcitor. With which type of connection will the 4-mF cpcitor hve greter mount of stored energy thn the 8-mF cpcitor? (i) series; (ii) prllel; (iii) either series or prllel; (iv) neither series nor prllel Dielectrics A common type of cpcitor uses dielectric sheets to seprte the conductors. Conductor (metl foil) Conductor (metl foil) Dielectric (plstic sheets) Most cpcitors hve nonconducting mteril, or dielectric, etween their conducting pltes. A common type of cpcitor uses long strips of metl foil for the pltes, seprted y strips of plstic sheet such s Mylr. A sndwich of these mterils is rolled up, forming unit tht cn provide cpcitnce of severl microfrds in compct pckge (Fig ). Plcing solid dielectric etween the pltes of cpcitor serves three functions. First, it solves the mechnicl prolem of mintining two lrge metl sheets t very smll seprtion without ctul contct. econd, using dielectric increses the mximum possile potentil difference etween the cpcitor pltes. As we descried in ection 23.3, ny insulting mteril, when sujected to sufficiently lrge electric field, experiences prtil ioniztion tht permits conduction through it. This is clled dielectric rekdown. Mny dielectric mterils cn tolerte stronger electric fields without rekdown thn cn ir. Thus using dielectric llows cpcitor to sustin higher potentil difference V nd so store greter mounts of chrge nd energy. Third, the cpcitnce of cpcitor of given dimensions is greter when there is dielectric mteril etween the pltes thn when there is vcuum. We cn demonstrte this effect with the id of sensitive electrometer, device tht mesures the potentil difference etween two conductors without letting ny pprecile chrge flow from one to the other. Figure shows n electrometer connected cross chrged cpcitor, with mgnitude of chrge Q on ech plte nd potentil difference V 0. When we insert n unchrged sheet of dielectric, such s glss, prffin, or polystyrene, etween the pltes, experiment shows tht the potentil difference decreses to smller vlue V (Fig ). When we remove the dielectric, the potentil difference returns to its originl vlue V 0, showing tht the originl chrges on the pltes hve not chnged. The originl cpcitnce C 0 is given y C 0 = Q>V 0, nd the cpcitnce C with the dielectric present is C = Q>V. The chrge Q is the sme in oth cses, nd V is less thn V 0, so we conclude tht the cpcitnce C with the dielectric present is greter thn C 0. When the spce etween pltes is completely filled y the dielectric, the rtio of C to C 0 (equl to the rtio of V 0 to V) is clled the dielectric constnt of the mteril, K: K = C C 0 (definition of dielectric constnt) (24.12)

115 24.4 Dielectrics 801 When the chrge is constnt, Q = C 0 V 0 = CV nd C>C 0 = V 0 >V. In this cse, Eq. (24.12) cn e rewritten s V = V 0 1when Q is constnt2 (24.13) K With the dielectric present, the potentil difference for given chrge Q is reduced y fctor K. The dielectric constnt K is pure numer. Becuse C is lwys greter thn C 0, K is lwys greter thn unity. ome representtive vlues of K re given in Tle For vcuum, K = 1 y definition. For ir t ordinry tempertures nd pressures, K is out ; this is so nerly equl to 1 tht for most purposes n ir cpcitor is equivlent to one in vcuum. Note tht while wter hs very lrge vlue of K, it is usully not very prcticl dielectric for use in cpcitors. The reson is tht while pure wter is very poor conductor, it is lso n excellent ionic solvent. Any ions tht re dissolved in the wter will cuse chrge to flow etween the cpcitor pltes, so the cpcitor dischrges. Tle 24.1 Vlues of Dielectric Constnt K t 20 C Mteril K Mteril Vcuum 1 Polyvinyl chloride 3.18 Air (1 tm) Plexigls 3.40 Air (100 tm) Glss 510 Teflon 2.1 Neoprene 6.70 Polyethylene 2.25 Germnium 16 Benzene 2.28 Glycerin 42.5 Mic 36 Wter 80.4 Mylr 3.1 trontium titnte 310 No rel dielectric is perfect insultor. Hence there is lwys some lekge current etween the chrged pltes of cpcitor with dielectric. We tcitly ignored this effect in ection 24.2 when we derived expressions for the equivlent cpcitnces of cpcitors in series, Eq. (24.5), nd in prllel, Eq. (24.7). But if lekge current flows for long enough time to sustntilly chnge the chrges from the vlues we used to derive Eqs. (24.5) nd (24.7), those equtions my no longer e ccurte. Induced Chrge nd Polriztion When dielectric mteril is inserted etween the pltes while the chrge is kept constnt, the potentil difference etween the pltes decreses y fctor K. Therefore the electric field etween the pltes must decrese y the sme fctor. If is the vcuum vlue nd E is the vlue with the dielectric, then E 0 E = E 0 1when Q is constnt2 (24.14) K ince the electric-field mgnitude is smller when the dielectric is present, the surfce chrge density (which cuses the field) must e smller s well. The surfce chrge on the conducting pltes does not chnge, ut n induced chrge of the opposite sign ppers on ech surfce of the dielectric (Fig ). The dielectric ws originlly electriclly neutrl nd is still neutrl; the induced surfce chrges rise s result of redistriution of positive nd negtive chrge within the dielectric mteril, phenomenon clled polriztion. We first encountered polriztion in ection 21.2, nd we suggest tht you rered the discussion of Fig We will ssume tht the induced surfce chrge is directly proportionl to the electric-field mgnitude E in the mteril; this is indeed the cse for mny common dielectrics. (This direct proportionlity is nlogous to K Effect of dielectric etween the pltes of prllel-plte cpcitor. () With given chrge, the potentil difference is V 0. () With the sme chrge ut with dielectric etween the pltes, the potentil difference V is smller thn V 0. () Vcuum () () s Q Dielectric Q Vcuum E 0 2Q V 0 V 2Q 2s Electrometer (mesures potentil difference cross pltes) Adding the dielectric reduces the potentil difference cross the cpcitor Electric field lines with () vcuum etween the pltes nd () dielectric etween the pltes. () s 2s i 2s i Dielectric 2s s i s i s 2s s 2s For given chrge density s, the induced chrges on the dielectric s surfces reduce the electric field etween the pltes. E Induced chrges

116 802 CHAPTER 24 Cpcitnce nd Dielectrics Hooke s lw for spring.) In tht cse, K is constnt for ny prticulr mteril. When the electric field is very strong or if the dielectric is mde of certin crystlline mterils, the reltionship etween induced chrge nd the electric field cn e more complex; we won t consider such cses here. We cn derive reltionship etween this induced surfce chrge nd the chrge on the pltes. Let s denote the mgnitude of the chrge per unit re induced on the surfces of the dielectric (the induced surfce chrge density) y s i. The mgnitude of the surfce chrge density on the cpcitor pltes is s, s usul. Then the net surfce chrge on ech side of the cpcitor hs mgnitude 1s - s i 2, s shown in Fig As we found in Exmple (ection 21.5) nd in Exmple 22.8 (ection 22.4), the field etween the pltes is relted to the net surfce chrge density y E = s net >P 0. Without nd with the dielectric, respectively, we hve E 0 = s E = s - s i (24.15) P 0 P 0 Using these expressions in Eq. (24.14) nd rerrnging the result, we find s i = s1-1 (induced surfce chrge density) K (24.16) This eqution shows tht when K is very lrge, s i is nerly s lrge s s. In this cse, s i nerly cncels s, nd the field nd potentil difference re much smller thn their vlues in vcuum. The product KP 0 is clled the permittivity of the dielectric, denoted y P: P=KP 0 1definition of permittivity2 (24.17) In terms of P we cn express the electric field within the dielectric s E = s P (24.18) The cpcitnce when the dielectric is present is given y C = KC 0 = KP 0 A d =PA d (prllel-plte cpcitor, dielectric etween pltes) (24.19) We cn repet the derivtion of Eq. (24.11) for the energy density u in n electric field for the cse in which dielectric is present. The result is u = 1 2 KP 0E 2 = 1 2 PE 2 (electric energy density in dielectric) (24.20) In empty spce, where K = 1, P=P 0 nd Eqs. (24.19) nd (24.20) reduce to Eqs. (24.2) nd (24.11), respectively, for prllel-plte cpcitor in vcuum. For this reson, P 0 is sometimes clled the permittivity of free spce or the permittivity of vcuum. Becuse K is pure numer, P nd P hve the sme units, C 2 >N # 0 m 2 or F>m. Eqution (24.19) shows tht extremely high cpcitnces cn e otined with pltes tht hve lrge surfce re A nd re seprted y smll distnce d y dielectric with lrge vlue of K. In n electrolytic doule-lyer cpcitor, tiny cron grnules dhere to ech plte: The vlue of A is the comined surfce re of the grnules, which cn e tremendous. The pltes with grnules ttched re seprted y very thin dielectric sheet. A cpcitor of this kind cn hve cpcitnce of 5000 frds yet fit in the plm of your hnd (compre Exmple 24.1 in ection 24.1). everl prcticl devices mke use of the wy in which cpcitor responds to chnge in dielectric constnt. One exmple is n electric stud finder, used y

117 24.4 Dielectrics 803 home repir workers to locte metl studs hidden ehind wll s surfce. It consists of metl plte with ssocited circuitry. The plte cts s one hlf of cpcitor, with the wll cting s the other hlf. If the stud finder moves over metl stud, the effective dielectric constnt for the cpcitor chnges, chnging the cpcitnce nd triggering signl. Prolem-olving trtegy 24.2 Dielectrics IDENTIFY the relevnt concepts: The reltionships in this section re useful whenever there is n electric field in dielectric, such s dielectric etween chrged cpcitor pltes. Typiclly you must relte the potentil difference V etween the pltes, the electric field mgnitude E in the cpcitor, the chrge density s on the cpcitor pltes, nd the induced chrge density s i on the surfces of the cpcitor. ET UP the prolem using the following steps: 1. Mke drwing of the sitution. 2. Identify the trget vriles, nd choose which equtions from this section will help you solve for those vriles. EXECUTE the solution s follows: 1. In prolems such s the next exmple, it is esy to get lost in lizzrd of formuls. Ask yourself t ech step wht kind of quntity ech symol represents. For exmple, distinguish clerly etween chrges nd chrge densities, nd etween electric fields nd electric potentil differences. 2. Check for consistency of units. Distnces must e in meters. A microfrd is 10-6 frd, nd so on. Don t confuse the numericl vlue of P 0 with the vlue of 1>4pP 0. Electric-field mgnitude cn e expressed in oth N>C nd V>m. The units of P 0 re C 2 >N # m 2 or F>m. EVALUATE your nswer: With dielectric present, () the cpcitnce is greter thn without dielectric; () for given chrge on the cpcitor, the electric field nd potentil difference re less thn without dielectric; nd (c) the mgnitude of the induced surfce chrge density s i on the dielectric is less thn tht of the chrge density s on the cpcitor pltes. Exmple A cpcitor with nd without dielectric uppose the prllel pltes in Fig ech hve n re of 2000 cm 2 ( 2.00 * 10-1 m 2 ) nd re 1.00 cm ( 1.00 * 10-2 m) prt. We connect the cpcitor to power supply, chrge it to potentil difference V 0 = 3.00 kv, nd disconnect the power supply. We then insert sheet of insulting plstic mteril etween the pltes, completely filling the spce etween them. We find tht the potentil difference decreses to 1.00 kv while the chrge on ech cpcitor plte remins constnt. Find () the originl cpcitnce C 0 ; () the mgnitude of chrge Q on ech plte; (c) the cpcitnce C fter the dielectric is inserted; (d) the dielectric constnt K of the dielectric; (e) the permittivity P of the dielectric; (f ) the mgnitude of the induced chrge Q i on ech fce of the dielectric; (g) the originl electric field E 0 etween the pltes; nd (h) the electric field E fter the dielectric is inserted. OLUTION IDENTIFY nd ET UP: This prolem uses most of the reltionships we hve discussed for cpcitors nd dielectrics. (Energy reltionships re treted in Exmple ) Most of the trget vriles cn e otined in severl wys. The methods used elow re smple; we encourge you to think of others nd compre your results. EXECUTE: () With vcuum etween the pltes, we use Eq. (24.19) with K = 1: A C 0 =P 0 d = * F>m * 10-1 m * 10-2 m = 1.77 * F = 177 pf () From the definition of cpcitnce, Eq. (24.1), Q = C 0 V 0 = * F * 10 3 V2 = 5.31 * 10-7 C = mc (c) When the dielectric is inserted, Q is unchnged ut the potentil difference decreses to V = 1.00 kv. Hence from Eq. (24.1), the new cpcitnce is C = Q V = 5.31 * 10-7 C 1.00 * 10 3 V = 5.31 * F = 531 pf (d) From Eq. (24.12), the dielectric constnt is K = C = 5.31 * F 531 pf C * = F 177 pf = 3.00 Alterntively, from Eq. (24.13), K = V 0 V = 3000 V 1000 V = 3.00 (e) Using K from prt (d) in Eq. (24.17), the permittivity is P=KP 0 = * C 2 >N # m 2 2 = 2.66 * C 2 >N # m 2 (f) Multiplying oth sides of Eq. (24.16) y the plte re A gives the induced chrge Q i = s i A in terms of the chrge Q = sa on ech plte: Q i = Q1-1 K = * 10-7 C = 3.54 * 10-7 C Continued

118 804 CHAPTER 24 Cpcitnce nd Dielectrics (g) ince the electric field etween the pltes is uniform, its mgnitude is the potentil difference divided y the plte seprtion: E 0 = V 0 d = (h) After the dielectric is inserted, E = V d = 1000 V 1.00 * 10-2 m = 1.00 * 105 V>m or, from Eq. (24.18), E = s P = Q PA = 5.31 * 10-7 C * C 2 >N # m * 10-1 m 2 2 = 1.00 * 10 5 V>m or, from Eq. (24.15), 3000 V 1.00 * 10-2 m = 3.00 * 105 V>m E = s - s i P * 10-7 C = * C 2 >N # m * 10-1 m 2 2 = 1.00 * 10 5 V>m or, from Eq. (24.14), = Q - Q i P 0 A E = E 0 K = 3.00 * 105 V>m = 1.00 * 10 5 V>m 3.00 EVALUATE: Inserting the dielectric incresed the cpcitnce y fctor of K = 3.00 nd reduced the electric field etween the pltes y fctor of 1>K = 1>3.00. It did so y developing induced chrges on the fces of the dielectric of mgnitude Q11-1>K2 = Q11-1>3.002 = 0.667Q. Exmple Energy storge with nd without dielectric Find the energy stored in the electric field of the cpcitor in Exmple nd the energy density, oth efore nd fter the dielectric sheet is inserted. OLUTION IDENTIFY nd ET UP: We now consider the ides of energy stored in cpcitor nd of electric-field energy density. We use Eq. (24.9) to find the stored energy nd Eq. (24.20) to find the energy density. EXECUTE: From Eq. (24.9), the stored energies U 0 nd U without nd with the dielectric in plce re U 0 = 1 2 C 0V 0 2 = * F V2 2 = 7.97 * 10-4 J U = 1 2 CV 2 = * F V2 2 = 2.66 * 10-4 J The finl energy is one-third of the originl energy. Eqution (24.20) gives the energy densities without nd with the dielectric: u 0 = 1 2 P 0E 2 0 = * C 2 >N # m * 10 5 N>C2 2 = J>m 3 u = 1 2 PE 2 = * C 2 >N # m * 10 5 N>C2 2 = J>m 3 The energy density with the dielectric is one-third of the originl energy density. EVALUATE: We cn check our nswer for u 0 y noting tht the volume etween the pltes is V = m m2 = m 3. ince the electric field etween the pltes is uniform, u 0 is uniform s well nd the energy density is just the stored energy divided y the volume: u 0 = U 0 V = 7.97 * 10-4 J m 3 = J>m 3 This grees with our erlier nswer. You cn use the sme pproch to check our result for u. In generl, when dielectric is inserted into cpcitor while the chrge on ech plte remins the sme, the permittivity P increses y fctor of K (the dielectric constnt), nd the electric field E nd the energy density u = 1 2 PE 2 decrese y fctor of 1>K. Where does the energy go? The nswer lies in the fringing field t the edges of rel prllel-plte cpcitor. As Fig shows, tht field tends to pull the dielectric into the spce etween the pltes, doing work on it s it does so. We could ttch spring to the left end of the dielectric in Fig nd use this force to stretch the spring. Becuse work is done y the field, the field energy density decreses The fringing field t the edges of the cpcitor exerts forces F nd F -i i on the negtive nd positive induced surfce chrges of dielectric, pulling the dielectric into the cpcitor. Dielectric F 2i F 1i E Dielectric Brekdown We mentioned erlier tht when dielectric is sujected to sufficiently strong electric field, dielectric rekdown tkes plce nd the dielectric ecomes conductor. This occurs when the electric field is so strong tht electrons re ripped loose from their molecules nd crsh into other molecules, lierting even more

119 24.5 Moleculr Model of Induced Chrge 805 electrons. This vlnche of moving chrge forms sprk or rc dischrge. Lightning is drmtic exmple of dielectric rekdown in ir. Becuse of dielectric rekdown, cpcitors lwys hve mximum voltge rtings. When cpcitor is sujected to excessive voltge, n rc my form through lyer of dielectric, urning or melting hole in it. This rc cretes conducting pth ( short circuit) etween the conductors. If conducting pth remins fter the rc is extinguished, the device is rendered permnently useless s cpcitor. The mximum electric-field mgnitude tht mteril cn withstnd without the occurrence of rekdown is clled its dielectric strength. This quntity is ffected significntly y temperture, trce impurities, smll irregulrities in the metl electrodes, nd other fctors tht re difficult to control. For this reson we cn give only pproximte figures for dielectric strengths. The dielectric strength of dry ir is out 3 * 10 6 V>m. Tle 24.2 lists the dielectric strengths of few common insulting mterils. Note tht the vlues re ll sustntilly greter thn the vlue for ir. For exmple, lyer of polycronte 0.01 mm thick (out the smllest prcticl thickness) hs 10 times the dielectric strength of ir nd cn withstnd mximum voltge of out 13 * 10 7 V>m211 *10-5 m2 = 300 V. Appliction Dielectric Cell Memrne The memrne of living cell ehves like dielectric etween the pltes of cpcitor. The memrne is mde of two sheets of lipid molecules, with their wter-insolule ends in the middle nd their wter-solule ends (shown in red) on the surfces of the memrne. The conductive fluids on either side of the memrne (wter with negtive ions inside the cell, wter with positive ions outside) ct s chrged cpcitor pltes, nd the nonconducting memrne cts s dielectric with K of out 10. The potentil difference V cross the memrne is out 0.07 V nd the memrne thickness d is out 7 * 10-9 m, so the electric field E V/d in the memrne is out 10 7 V/m close to the dielectric strength of the memrne. If the memrne were mde of ir, V nd E would e lrger y fctor of K L 10 nd dielectric rekdown would occur. Tle 24.2 Dielectric Constnt nd Dielectric trength of ome Insulting Mterils Mteril Dielectric Constnt, K Polycronte 2.8 Polyester 3.3 Polypropylene 2.2 Polystyrene 2.6 Dielectric trength, E m (V/m) 3 * * * * 10 7 Pyrex glss * 10 7 Test Your Understnding of ection 24.4 The spce etween the pltes of n isolted prllel-plte cpcitor is filled y sl of dielectric with dielectric constnt K. The two pltes of the cpcitor hve chrges Q nd -Q. You pull out the dielectric sl. If the chrges do not chnge, how does the energy in the cpcitor chnge when you remove the sl? (i) It increses; (ii) it decreses; (iii) it remins the sme Moleculr Model of Induced Chrge In ection 24.4 we discussed induced surfce chrges on dielectric in n electric field. Now let s look t how these surfce chrges cn rise. If the mteril were conductor, the nswer would e simple. Conductors contin chrge tht is free to move, nd when n electric field is present, some of the chrge redistriutes itself on the surfce so tht there is no electric field inside the conductor. But n idel dielectric hs no chrges tht re free to move, so how cn surfce chrge occur? To understnd this, we hve to look gin t rerrngement of chrge t the moleculr level. ome molecules, such s H 2 O nd N 2 O, hve equl mounts of positive nd negtive chrges ut lopsided distriution, with excess positive chrge concentrted on one side of the molecule nd negtive chrge on the other. As we descried in ection 21.7, such n rrngement is clled n electric dipole, nd the molecule is clled polr molecule. When no electric field is present in gs or liquid with polr molecules, the molecules re oriented rndomly (Fig ). When they re plced in n electric field, however, they tend to orient themselves s in Fig , s result of the electric-field torques descried in ection Becuse of therml gittion, the lignment of the molecules with E is not perfect Polr molecules () without nd () with n pplied electric field E. () () In the sence of n electric field, polr molecules orient rndomly. E When n electric field is pplied, the molecules tend to lign with it.

120 806 CHAPTER 24 Cpcitnce nd Dielectrics Nonpolr molecules () without nd () with n pplied electric field E. PhET: Moleculr Motors PhET: Opticl Tweezers nd Applictions PhET: tretching DNA () In the sence of n electric field, nonpolr molecules re not electric dipoles. () E An electric field cuses the molecules positive nd negtive chrges to seprte slightly, mking the molecule effectively polr Polriztion of dielectric in n electric field E gives rise to thin lyers of ound chrges on the surfces, creting surfce chrge densities s i nd -s i. The sizes of the molecules re gretly exggerted for clrity. 2s i 2s i d s i s i E Even molecule tht is not ordinrily polr ecomes dipole when it is plced in n electric field ecuse the field pushes the positive chrges in the molecules in the direction of the field nd pushes the negtive chrges in the opposite direction. This cuses redistriution of chrge within the molecule (Fig ). uch dipoles re clled induced dipoles. With either polr or nonpolr molecules, the redistriution of chrge cused y the field leds to the formtion of lyer of chrge on ech surfce of the dielectric mteril (Fig ). These lyers re the surfce chrges descried in ection 24.4; their surfce chrge density is denoted y s i. The chrges re not free to move indefinitely, s they would e in conductor, ecuse ech chrge is ound to molecule. They re in fct clled ound chrges to distinguish them from the free chrges tht re dded to nd removed from the conducting cpcitor pltes. In the interior of the mteril the net chrge per unit volume remins zero. As we hve seen, this redistriution of chrge is clled polriztion, nd we sy tht the mteril is polrized. The four prts of Fig show the ehvior of sl of dielectric when it is inserted in the field etween pir of oppositely chrged cpcitor pltes. Figure shows the originl field. Figure is the sitution fter the dielectric hs een inserted ut efore ny rerrngement of chrges hs occurred. Figure 24.20c shows y thinner rrows the dditionl field set up in the dielectric y its induced surfce chrges. This field is opposite to the originl field, ut it is not gret enough to cncel the originl field completely ecuse the chrges in the dielectric re not free to move indefinitely. The resultnt field () Electric field of mgnitude E 0 etween two chrged pltes. () Introduction of dielectric of dielectric constnt K. (c) The induced surfce chrges nd their field. (d) Resultnt field of mgnitude E 0 /K. () No dielectric s E 2s 0 s 2s () Dielectric just inserted (c) Induced chrges (d) Resultnt field crete electric field s 2s 2s i s i 2s i s i 2s i s i 2s i s i s 2s Originl electric field Weker field in dielectric due to induced (ound) chrges

121 24.6 Guss s Lw in Dielectrics 807 in the dielectric, shown in Fig d, is therefore decresed in mgnitude. In the field-line representtion, some of the field lines leving the positive plte go through the dielectric, while others terminte on the induced chrges on the fces of the dielectric. As we discussed in ection 21.2, polriztion is lso the reson chrged ody, such s n electrified plstic rod, cn exert force on n unchrged ody such s it of pper or pith ll. Figure shows n unchrged dielectric sphere B in the rdil field of positively chrged ody A. The induced positive chrges on B experience force towrd the right, while the force on the induced negtive chrges is towrd the left. The negtive chrges re closer to A, nd thus re in stronger field, thn re the positive chrges. The force towrd the left is stronger thn tht towrd the right, nd B is ttrcted towrd A, even though its net chrge is zero. The ttrction occurs whether the sign of A s chrge is positive or negtive (see Fig. 21.8). Furthermore, the effect is not limited to dielectrics; n unchrged conducting ody would e ttrcted in the sme wy A neutrl sphere B in the rdil electric field of positively chrged sphere A is ttrcted to the chrge ecuse of polriztion. A B E Test Your Understnding of ection 24.5 A prllel-plte cpcitor hs chrges Q nd -Q on its two pltes. A dielectric sl with K = 3 is then inserted into the spce etween the pltes s shown in Fig Rnk the following electric-field mgnitudes in order from lrgest to smllest. (i) the field efore the sl is inserted; (ii) the resultnt field fter the sl is inserted; (iii) the field due to the ound chrges Guss s Lw in Dielectrics We cn extend the nlysis of ection 24.4 to reformulte Guss s lw in form tht is prticulrly useful for dielectrics. Figure is close-up view of the left cpcitor plte nd left surfce of the dielectric in Fig Let s pply Guss s lw to the rectngulr ox shown in cross section y the purple line; the surfce re of the left nd right sides is A. The left side is emedded in the conductor tht forms the left cpcitor plte, nd so the electric field everywhere on tht surfce is zero. The right side is emedded in the dielectric, where the electric field hs mgnitude E, nd E = 0 everywhere on the other four sides. The totl chrge enclosed, including oth the chrge on the cpcitor plte nd the induced chrge on the dielectric surfce, is Q encl = 1s - s i 2A, so Guss s lw gives EA = 1s - s i2a (24.21) P 0 This eqution is not very illuminting s it stnds ecuse it reltes two unknown quntities: E inside the dielectric nd the induced surfce chrge density s i. But now we cn use Eq. (24.16), developed for this sme sitution, to simplify this eqution y eliminting s i. Eqution (24.16) is Comining this with Eq. (24.21), we get s i = s1-1 K or s - s i = s K EA = sa KP 0 or KEA = sa P 0 (24.22) Eqution (24.22) sys tht the flux of KE, not E, through the Gussin surfce in Fig is equl to the enclosed free chrge sa divided y P 0. It turns out tht for ny Gussin surfce, whenever the induced chrge is proportionl to the electric field in the mteril, we cn rewrite Guss s lw s Guss s lw with dielectric. This figure shows close-up of the lefthnd cpcitor plte in Fig The Gussin surfce is rectngulr ox tht lies hlf in the conductor nd hlf in the dielectric. ide view Perspective view E 5 0 Conductor Dielectric Conductor A σ σi A E Gussin surfce Dielectric C KE # da = Q encl-free P 0 (Guss s lw in dielectric) (24.23)

122 808 CHAPTER 24 Cpcitnce nd Dielectrics where Q encl-free is the totl free chrge (not ound chrge) enclosed y the Gussin surfce. The significnce of these results is tht the right sides contin only the free chrge on the conductor, not the ound (induced) chrge. In fct, lthough we hve not proved it, Eq. (24.23) remins vlid even when different prts of the Gussin surfce re emedded in dielectrics hving different vlues of K, provided tht the vlue of K in ech dielectric is independent of the electric field (usully the cse for electric fields tht re not too strong) nd tht we use the pproprite vlue of K for ech point on the Gussin surfce. Exmple A sphericl cpcitor with dielectric Use Guss s lw to find the cpcitnce of the sphericl cpcitor of Exmple 24.3 (ection 24.1) if the volume etween the shells is filled with n insulting oil with dielectric constnt K. OLUTION IDENTIFY nd ET UP: The sphericl symmetry of the prolem is not chnged y the presence of the dielectric, so s in Exmple 24.3, we use concentric sphericl Gussin surfce of rdius r etween the shells. ince dielectric is present, we use Guss s lw in the form of Eq. (24.23). EXECUTE: From Eq. (24.23), # C KE da = KE da = KE da = 1KE214pr 2 2 = Q C C P 0 Q E = 4pKP 0 r 2 = Q 4pPr 2 where P=KP 0. Compred to the cse in which there is vcuum etween the shells, the electric field is reduced y fctor of 1>K. The potentil difference V etween the shells is reduced y the sme fctor, nd so the cpcitnce C = Q>V is incresed y fctor of K, just s for prllel-plte cpcitor when dielectric is inserted. Using the result of Exmple 24.3, we find tht the cpcitnce with the dielectric is C = 4pKP 0r r r - r = 4pPr r r - r EVALUATE: If the dielectric fills the volume etween the two conductors, the cpcitnce is just K times the vlue with no dielectric. The result is more complicted if the dielectric only prtilly fills this volume (see Chllenge Prolem 24.78). Test Your Understnding of ection 24.6 A single point chrge q is imedded in dielectric of dielectric constnt K. At point inside the dielectric distnce r from the point chrge, wht is the mgnitude of the electric field? (i) q>4pp 0 r 2 ; (ii) Kq>4pP (iii) q>4pkp 0 r 2 0 r 2 ; ; (iv) none of these.

123 CHAPTER 24 UMMARY Cpcitors nd cpcitnce: A cpcitor is ny pir of conductors seprted y n insulting mteril. When the cpcitor is chrged, there re chrges of equl mgnitude Q nd opposite sign on the two conductors, nd the potentil V of the positively chrged conductor with respect to the negtively chrged conductor is proportionl to Q. The cpcitnce C is defined s the rtio of Q to V. The I unit of cpcitnce is the frd (F): 1 F = 1 C>V. A prllel-plte cpcitor consists of two prllel conducting pltes, ech with re A, seprted y distnce d. If they re seprted y vcuum, the cpcitnce depends only on A nd d. For other geometries, the cpcitnce cn e found y using the definition C = Q>V. (ee Exmples ) C = Q V C = Q V =P 0 A d (24.1) (24.2) 1Q 2Q Wire Plte, re A d Potentil Wire Plte, re A difference 5 V Cpcitors in series nd prllel: When cpcitors with cpcitnces C 1, C 2, C 3, Á re connected in series, the reciprocl of the equivlent cpcitnce C eq equls the sum of the reciprocls of the individul cpcitnces. When cpcitors re connected in prllel, the equivlent cpcitnce C eq equls the sum of the individul cpcitnces. (ee Exmples 24.5 nd 24.6.) 1 = C eq C 1 C 2 C Á 3 (cpcitors in series) C eq = C 1 C 2 C 3 Á (cpcitors in prllel) (24.5) (24.7) 1Q 2Q 1Q 2Q C 1 V 5 V c C 2 V c 5 V 1 V c 5 V 2 V 5 V C 1 Q 1 C 2 Q 2 Energy in cpcitor: The energy U required to chrge cpcitor C to potentil difference V nd chrge Q is equl to the energy stored in the cpcitor. This energy cn e thought of s residing in the electric field etween the conductors; the energy density u (energy per unit volume) is proportionl to the squre of the electric-field mgnitude. (ee Exmples ) U = Q2 2C = 1 2 CV 2 = 1 2 QV u = 1 2 P 0 E 2 (24.9) (24.11) 1Q V E 2Q Dielectrics: When the spce etween the conductors is A C = KC 0 = KP 0 filled with dielectric mteril, the cpcitnce d =PA d increses y fctor K, clled the dielectric constnt (prllel-plte cpcitor of the mteril. The quntity P=KP 0 is clled the permittivity of the dielectric. For fixed mount of chrge filled with dielectric) (24.19) on the cpcitor pltes, induced chrges on the surfce of the dielectric decrese the electric field nd potentil u = 1 2 KP 0 E 2 = 1 2 PE 2 (24.20) difference etween the pltes y the sme fctor K. The surfce chrge results from polriztion, microscopic Q # encl-free (24.23) rerrngement of chrge in the dielectric. (ee Exmple C KE da = P ) Under sufficiently strong fields, dielectrics ecome conductors, sitution clled dielectric rekdown. The mximum field tht mteril cn withstnd without rekdown is clled its dielectric strength. In dielectric, the expression for the energy density is the sme s in vcuum ut with P 0 replced y P=KP. (ee Exmple ) Guss s lw in dielectric hs lmost the sme form s in vcuum, with two key differences: E is replced y KE nd Q encl is replced y Q encl-free, which includes only the free chrge (not ound chrge) enclosed y the Gussin surfce. (ee Exmple ) Dielectric etween pltes s 2s 2s i s i 2s i s i s 2s 809

124 810 CHAPTER 24 Cpcitnce nd Dielectrics BRIDGING PROBLEM Electric-Field Energy nd Cpcitnce of Conducting phere A solid conducting sphere of rdius R crries chrge Q. Clculte the electric-field energy density t point distnce r from the center of the sphere for () r R nd () r R. (c) Clculte the totl electric-field energy ssocited with the chrged sphere. (d) How much work is required to ssemle the chrge Q on the sphere? (e) Use the result of prt (c) to find the cpcitnce of the sphere. (You cn think of the second conductor s hollow conducting shell of infinite rdius.) OLUTION GUIDE ee MsteringPhysics study re for Video Tutor solution IDENTIFY nd ET UP 1. You know the electric field for this sitution t ll vlues of r from Exmple 22.5 (ection 22.4). You ll use this to find the electric-field energy density u nd the totl electric-field energy U. You cn then find the cpcitnce from the reltionship U = Q 2 >2C. 2. To find U, consider sphericl shell of rdius r nd thickness dr tht hs volume dv = 4pr 2 dr. The energy stored in this volume is udv, nd the totl energy is the integrl of udvfrom r 0 to r q. et up this integrl. EXECUTE 3. Find u for r R nd for r R. 4. ustitute your results from step 3 into the expression from step 2. Then clculte the integrl to find the totl electric-field energy U. 5. Use your understnding of the energy stored in chrge distriution to find the work required to ssemle the chrge Q. 6. Find the cpcitnce of the sphere. EVALUATE 7. Where is the electric-field energy density gretest? Where is it lest? 8. How would the results e ffected if the solid sphere were replced y hollow conducting sphere of the sme rdius R? 9. You cn find the potentil difference etween the sphere nd infinity from C = Q>V. Does this gree with the result of Exmple 23.8 (ection 23.3)? Prolems For instructor-ssigned homework, go to : Prolems of incresing difficulty. CP: Cumultive prolems incorporting mteril from erlier chpters. CALC: Prolems requiring clculus. BIO: Biosciences prolems. DICUION QUETION Q24.1 Eqution (24.2) shows tht the cpcitnce of prllelplte cpcitor ecomes lrger s the plte seprtion d decreses. However, there is prcticl limit to how smll d cn e mde, which plces limits on how lrge C cn e. Explin wht sets the limit on d. (Hint: Wht hppens to the mgnitude of the electric field s d 0? ) Q24.2 uppose severl different prllel-plte cpcitors re chrged up y constnt-voltge source. Thinking of the ctul movement nd position of the chrges on n tomic level, why does it mke sense tht the cpcitnces re proportionl to the surfce res of the pltes? Why does it mke sense tht the cpcitnces re inversely proportionl to the distnce etween the pltes? Q24.3 uppose the two pltes of cpcitor hve different res. When the cpcitor is chrged y connecting it to ttery, do the chrges on the two pltes hve equl mgnitude, or my they e different? Explin your resoning. Q24.4 At the Fermi Ntionl Accelertor Lortory (Fermil) in Illinois, protons re ccelerted round ring 2 km in rdius to speeds tht pproch tht of light. The energy for this is stored in cpcitors the size of house. When these cpcitors re eing chrged, they mke very loud creking sound. Wht is the origin of this sound? Q24.5 In the prllel-plte cpcitor of Fig. 24.2, suppose the pltes re pulled prt so tht the seprtion d is much lrger thn the size of the pltes. () Is it still ccurte to sy tht the electric field etween the pltes is uniform? Why or why not? () In the sit- ution shown in Fig. 24.2, the potentil difference etween the pltes is V = Qd>P 0 A. If the pltes re pulled prt s descried ove, is V more or less thn this formul would indicte? Explin your resoning. (c) With the pltes pulled prt s descried ove, is the cpcitnce more thn, less thn, or the sme s tht given y Eq. (24.2)? Explin your resoning. Q24.6 A prllel-plte cpcitor is chrged y eing connected to ttery nd is kept connected to the ttery. The seprtion etween the pltes is then douled. How does the electric field chnge? The chrge on the pltes? The totl energy? Explin your resoning. Q24.7 A prllel-plte cpcitor is chrged y eing connected to ttery nd is then disconnected from the ttery. The seprtion etween the pltes is then douled. How does the electric field chnge? The potentil difference? The totl energy? Explin your resoning. Q24.8 Two prllel-plte cpcitors, identicl except tht one hs twice the plte seprtion of the other, re chrged y the sme voltge source. Which cpcitor hs stronger electric field etween the pltes? Which cpcitor hs greter chrge? Which hs greter energy density? Explin your resoning. Q24.9 The chrged pltes of cpcitor ttrct ech other, so to pull the pltes frther prt requires work y some externl force. Wht ecomes of the energy dded y this work? Explin your resoning. Q24.10 The two pltes of cpcitor re given chrges ;Q. The cpcitor is then disconnected from the chrging device so tht the

125 Exercises 811 chrges on the pltes cn t chnge, nd the cpcitor is immersed in tnk of oil. Does the electric field etween the pltes increse, decrese, or sty the sme? Explin your resoning. How cn this field e mesured? Q24.11 As shown in Tle 24.1, wter hs very lrge dielectric constnt K = Why do you think wter is not commonly used s dielectric in cpcitors? Q24.12 Is dielectric strength the sme thing s dielectric constnt? Explin ny differences etween the two quntities. Is there simple reltionship etween dielectric strength nd dielectric constnt (see Tle 24.2)? Q24.13 A cpcitor mde of luminum foil strips seprted y Mylr film ws sujected to excessive voltge, nd the resulting dielectric rekdown melted holes in the Mylr. After this, the cpcitnce ws found to e out the sme s efore, ut the rekdown voltge ws much less. Why? Q24.14 uppose you ring sl of dielectric close to the gp etween the pltes of chrged cpcitor, prepring to slide it etween the pltes. Wht force will you feel? Wht does this force tell you out the energy stored etween the pltes once the dielectric is in plce, compred to efore the dielectric is in plce? Q24.15 The freshness of fish cn e mesured y plcing fish etween the pltes of cpcitor nd mesuring the cpcitnce. How does this work? (Hint: As time psses, the fish dries out. ee Tle 24.1.) Q24.16 Electrolytic cpcitors use s their dielectric n extremely thin lyer of nonconducting oxide etween metl plte nd conducting solution. Discuss the dvntge of such cpcitor over one constructed using solid dielectric etween the metl pltes. Q24.17 In terms of the dielectric constnt K, wht hppens to the electric flux through the Gussin surfce shown in Fig when the dielectric is inserted into the previously empty spce etween the pltes? Explin. Q24.18 A prllel-plte cpcitor is connected to power supply tht mintins fixed potentil difference etween the pltes. () If sheet of dielectric is then slid etween the pltes, wht hppens to (i) the electric field etween the pltes, (ii) the mgnitude of chrge on ech plte, nd (iii) the energy stored in the cpcitor? () Now suppose tht efore the dielectric is inserted, the chrged cpcitor is disconnected from the power supply. In this cse, wht hppens to (i) the electric field etween the pltes, (ii) the mgnitude of chrge on ech plte, nd (iii) the energy stored in the cpcitor? Explin ny differences etween the two situtions. Q24.19 Liquid dielectrics tht hve polr molecules (such s wter) lwys hve dielectric constnts tht decrese with incresing temperture. Why? Q24.20 A conductor is n extreme cse of dielectric, since if n electric field is pplied to conductor, chrges re free to move within the conductor to set up induced chrges. Wht is the dielectric constnt of perfect conductor? Is it K = 0, K q, or something in etween? Explin your resoning. EXERCIE ection 24.1 Cpcitors nd Cpcitnce The pltes of prllel-plte cpcitor re 2.50 mm prt, nd ech crries chrge of mgnitude 80.0 nc. The pltes re in vcuum. The electric field etween the pltes hs mgnitude of 4.00 * 10 6 V>m. () Wht is the potentil difference etween the pltes? () Wht is the re of ech plte? (c) Wht is the cpcitnce? The pltes of prllel-plte cpcitor re 3.28 mm prt, nd ech hs n re of 12.2 cm 2. Ech plte crries chrge of mgnitude 4.35 * 10-8 C. The pltes re in vcuum. () Wht is the cpcitnce? () Wht is the potentil difference etween the pltes? (c) Wht is the mgnitude of the electric field etween the pltes? A prllel-plte ir cpcitor of cpcitnce 245 pf hs chrge of mgnitude mc on ech plte. The pltes re mm prt. () Wht is the potentil difference etween the pltes? () Wht is the re of ech plte? (c) Wht is the electricfield mgnitude etween the pltes? (d) Wht is the surfce chrge density on ech plte? Cpcitnce of n Oscilloscope. Oscilloscopes hve prllel metl pltes inside them to deflect the electron em. These pltes re clled the deflecting pltes. Typiclly, they re squres 3.0 cm on side nd seprted y 5.0 mm, with vcuum in etween. Wht is the cpcitnce of these deflecting pltes nd hence of the oscilloscope? (Note: This cpcitnce cn sometimes hve n effect on the circuit you re trying to study nd must e tken into considertion in your clcultions.) A 10.0-mF prllel-plte cpcitor with circulr pltes is connected to 12.0-V ttery. () Wht is the chrge on ech plte? () How much chrge would e on the pltes if their seprtion were douled while the cpcitor remined connected to the ttery? (c) How much chrge would e on the pltes if the cpcitor were connected to the 12.0-V ttery fter the rdius of ech plte ws douled without chnging their seprtion? A 10.0-mF prllel-plte cpcitor is connected to 12.0-V ttery. After the cpcitor is fully chrged, the ttery is disconnected without loss of ny of the chrge on the pltes. () A voltmeter is connected cross the two pltes without dischrging them. Wht does it red? () Wht would the voltmeter red if (i) the plte seprtion were douled; (ii) the rdius of ech plte were douled ut their seprtion ws unchnged? How fr prt would prllel pennies hve to e to mke 1.00-pF cpcitor? Does your nswer suggest tht you re justified in treting these pennies s infinite sheets? Explin A 5.00-pF, prllel-plte, ir-filled cpcitor with circulr pltes is to e used in circuit in which it will e sujected to potentils of up to 1.00 * 10 2 V. The electric field etween the pltes is to e no greter thn 1.00 * 10 4 N>C. As udding electricl engineer for Live-Wire Electronics, your tsks re to () design the cpcitor y finding wht its physicl dimensions nd seprtion must e; () find the mximum chrge these pltes cn hold A prllel-plte ir cpcitor is to store chrge of mgnitude pc on ech plte when the potentil difference etween the pltes is 42.0 V. () If the re of ech plte is 6.80 cm 2, wht is the seprtion etween the pltes? () If the seprtion etween the two pltes is doule the vlue clculted in prt (), wht potentil difference is required for the cpcitor to store chrge of mgnitude pc on ech plte? A cylindricl cpcitor consists of solid inner conducting core with rdius cm, surrounded y n outer hollow conducting tue. The two conductors re seprted y ir, nd the length of the cylinder is 12.0 cm. The cpcitnce is 36.7 pf. () Clculte the inner rdius of the hollow tue. () When the cpcitor is chrged to 125 V, wht is the chrge per unit length l on the cpcitor? A cpcitor is mde from two hollow, coxil, iron cylinders, one inside the other. The inner cylinder is negtively chrged

126 812 CHAPTER 24 Cpcitnce nd Dielectrics nd the outer is positively chrged; the mgnitude of the chrge on ech is 10.0 pc. The inner cylinder hs rdius 0.50 mm, the outer one hs rdius 5.00 mm, nd the length of ech cylinder is 18.0 cm. () Wht is the cpcitnce? () Wht pplied potentil difference is necessry to produce these chrges on the cylinders? A cylindricl cpcitor hs n inner conductor of rdius 1.5 mm nd n outer conductor of rdius 3.5 mm. The two conductors re seprted y vcuum, nd the entire cpcitor is 2.8 m long. () Wht is the cpcitnce per unit length? () The potentil of the inner conductor is 350 mv higher thn tht of the outer conductor. Find the chrge (mgnitude nd sign) on oth conductors A sphericl cpcitor contins chrge of 3.30 nc when connected to potentil difference of 220 V. If its pltes re seprted y vcuum nd the inner rdius of the outer shell is 4.00 cm, clculte: () the cpcitnce; () the rdius of the inner sphere; (c) the electric field just outside the surfce of the inner sphere A sphericl cpcitor is formed from two concentric, sphericl, conducting shells seprted y vcuum. The inner sphere hs rdius 15.0 cm nd the cpcitnce is 116 pf. () Wht is the rdius of the outer sphere? () If the potentil difference etween the two spheres is 220 V, wht is the mgnitude of chrge on ech sphere? ection 24.2 Cpcitors in eries nd Prllel BIO Electric Eels. Electric eels nd electric fish generte lrge potentil differences tht re used to stun enemies nd prey. These potentils re produced y cells tht ech cn generte 0.10 V. We cn plusily model such cells s chrged cpcitors. () How should these cells e connected (in series or in prllel) to produce totl potentil of more thn 0.10 V? () Using Figure E24.16 the connection in prt (), how mny cells must e connected together to produce the 500-V surge of the electric eel? 15 pf For the system of cpcitors shown in Fig. E24.16, find the equivlent cpcitnce () etween nd c, nd () etween nd c. 9.0 pf 11 pf In Fig. E24.17, ech cpcitor hs C = 4.00 mf nd V = 28.0 V. Clculte () the chrge on ech cpcitor; c () the potentil difference cross ech cpcitor; (c) the potentil difference etween points Figure E24.17 nd d In Fig. 24.8, let C 1 C 2 C 1 = 3.00 mf, C 2 = 5.00 mf, nd V = 52.0 V. Clculte () the chrge on ech cpcitor C 3 nd () the potentil difference cross ech cpcitor. d In Fig. 24.9, let C 1 = 3.00 mf, C 2 = 5.00 mf, nd V = 52.0 V. Clculte () C 4 the chrge on ech cpcitor nd () the potentil difference cross ech cpcitor In Fig. E24.20, C 1 = 6.00 mf, C 2 = 3.00 mf, nd C 3 = 5.00 mf. The cpcitor network is connected to n pplied potentil V. After the chrges on the cpcitors hve reched their C 2 finl vlues, the chrge on is 40.0 mc. () Wht re the chrges on cpcitors C 1 nd C 3? () Wht is the pplied voltge V? For the system of cpcitors shown in Fig. E24.21, potentil difference of 25 V is mintined cross. () Wht is the equivlent cpcitnce of this system etween nd? () How much chrge is stored y this system? (c) How much chrge does the 6.5-nF cpcitor store? (d) Wht is the potentil difference cross the 7.5-nF cpcitor? Figure E24.22 shows system of four cpcitors, where the potentil difference cross is 50.0 V. () Find the equivlent cpcitnce of this system etween nd. () How much chrge is stored y this comintion of cpcitors? (c) How much chrge is Figure E24.20 stored in ech of the 10.0-mF 10.0 mf 9.0 mf nd the 9.0-mF cpcitors? uppose the 3-mF cpcitor in Fig were removed nd replced y different one, nd tht this 8.0 mf chnged the equivlent cpcitnce etween points nd to 8 mf. Wht would e the cpcitnce of the replcement cpcitor? ection 24.3 Energy torge in Cpcitors nd Electric-Field Energy A prllel-plte ir cpcitor hs cpcitnce of 920 pf. The chrge on ech plte is 2.55 mc. () Wht is the potentil difference etween the pltes? () If the chrge is kept constnt, wht will e the potentil difference etween the pltes if the seprtion is douled? (c) How much work is required to doule the seprtion? A 5.80-mF, prllel-plte, ir cpcitor hs plte seprtion of 5.00 mm nd is chrged to potentil difference of 400 V. Clculte the energy density in the region etween the pltes, in units of J>m An ir cpcitor is mde from two flt prllel pltes 1.50 mm prt. The mgnitude of chrge on ech plte is mc when the potentil difference is 200 V. () Wht is the cpcitnce? () Wht is the re of ech plte? (c) Wht mximum voltge cn e pplied without dielectric rekdown? (Dielectric rekdown for ir occurs t n electric-field strength of 3.0 * 10 6 V>m. ) (d) When the chrge is mc, wht totl energy is stored? A prllel-plte vcuum cpcitor with plte re A nd seprtion x hs chrges Q nd -Q on its pltes. The cpcitor is disconnected from the source of chrge, so the chrge on ech plte remins fixed. () Wht is the totl energy stored in the cpcitor? () The pltes re pulled prt n dditionl distnce dx. Wht is the chnge in the stored energy? (c) If F is the force with C 1 C 2 C 3 Figure E nf 18.0 nf 30.0 nf 10.0 nf 6.5 nf Figure E mf d

127 Exercises 813 which the pltes ttrct ech other, then the chnge in the stored energy must equl the work dw = Fdx done in pulling the pltes prt. Find n expression for F. (d) Explin why F is not equl to QE, where E is the electric field etween the pltes A prllel-plte vcuum cpcitor hs 8.38 J of energy stored in it. The seprtion etween the pltes is 2.30 mm. If the seprtion is decresed to 1.15 mm, wht is the energy stored () if the cpcitor is disconnected from the potentil source so the chrge on the pltes remins constnt, nd () if the cpcitor remins connected to the potentil source so the potentil difference etween the pltes remins constnt? You hve two identicl cpcitors nd n externl potentil source. () Compre the totl energy stored in the cpcitors when they re connected to the pplied potentil in series nd in prllel. () Compre the mximum mount of chrge stored in ech cse. (c) Energy storge in cpcitor cn e limited y the mximum electric field etween the pltes. Wht is the rtio of the electric field for the series nd prllel comintions? For the cpcitor network shown in Fig. E24.30, the Figure E24.30 potentil difference cross is 150 nf 120 nf 36 V. Find () the totl chrge stored in this network; () the chrge on ech cpcitor; (c) the totl energy stored in the network; (d) the energy stored in ech cpcitor; (e) the potentil differences cross ech cpcitor For the cpcitor network shown in Fig. E24.31, the potentil difference cross Figure E24.31 is 220 V. Find () the totl chrge stored in this network; () the chrge on ech cpcitor; (c) the totl energy stored in the network; (d) the energy stored in ech cpcitor; (e) the potentil difference cross ech cpcitor. 35 nf 75 nf A m-long cylindricl cpcitor consists of solid conducting core with rdius of 1.20 mm nd n outer hollow conducting tue with n inner rdius of 2.00 mm. The two conductors re seprted y ir nd chrged to potentil difference of 6.00 V. Clculte () the chrge per length for the cpcitor; () the totl chrge on the cpcitor; (c) the cpcitnce; (d) the energy stored in the cpcitor when fully chrged A cylindricl ir cpcitor of length 15.0 m stores 3.20 * 10-9 J of energy when the potentil difference etween the two conductors is 4.00 V. () Clculte the mgnitude of the chrge on ech conductor. () Clculte the rtio of the rdii of the inner nd outer conductors A cpcitor is formed from two concentric sphericl conducting shells seprted y vcuum. The inner sphere hs rdius 12.5 cm, nd the outer sphere hs rdius 14.8 cm. A potentil difference of 120 V is pplied to the cpcitor. () Wht is the energy density t r = 12.6 cm, just outside the inner sphere? () Wht is the energy density t r = 14.7 cm, just inside the outer sphere? (c) For prllel-plte cpcitor the energy density is uniform in the region etween the pltes, except ner the edges of the pltes. Is this lso true for sphericl cpcitor? ection 24.4 Dielectrics A 12.5-mF cpcitor is connected to power supply tht keeps constnt potentil difference of 24.0 V cross the pltes. A piece of mteril hving dielectric constnt of 3.75 is plced etween the pltes, completely filling the spce etween them. () How much energy is stored in the cpcitor efore nd fter the dielectric is inserted? () By how much did the energy chnge during the insertion? Did it increse or decrese? A prllel-plte cpcitor hs cpcitnce C 0 = 5.00 pf when there is ir etween the pltes. The seprtion etween the pltes is 1.50 mm. () Wht is the mximum mgnitude of chrge Q tht cn e plced on ech plte if the electric field in the region etween the pltes is not to exceed 3.00 * 10 4 V>m? () A dielectric with K = 2.70 is inserted etween the pltes of the cpcitor, completely filling the volume etween the pltes. Now wht is the mximum mgnitude of chrge on ech plte if the electric field etween the pltes is not to exceed 3.00 * 10 4 V>m? Two prllel pltes hve equl nd opposite chrges. When the spce etween the pltes is evcuted, the electric field is E = 3.20 * 10 5 V>m. When the spce is filled with dielectric, the electric field is E = 2.50 * 10 5 V>m. () Wht is the chrge density on ech surfce of the dielectric? () Wht is the dielectric constnt? A udding electronics hoyist wnts to mke simple 1.0-nF cpcitor for tuning her crystl rdio, using two sheets of luminum foil s pltes, with few sheets of pper etween them s dielectric. The pper hs dielectric constnt of 3.0, nd the thickness of one sheet of it is 0.20 mm. () If the sheets of pper mesure 22 * 28 cm nd she cuts the luminum foil to the sme dimensions, how mny sheets of pper should she use etween her pltes to get the proper cpcitnce? () uppose for convenience she wnts to use single sheet of posterord, with the sme dielectric constnt ut thickness of 12.0 mm, insted of the pper. Wht re of luminum foil will she need for her pltes to get her 1.0 nf of cpcitnce? (c) uppose she goes high-tech nd finds sheet of Teflon of the sme thickness s the posterord to use s dielectric. Will she need lrger or smller re of Teflon thn of posterord? Explin The dielectric to e used in prllel-plte cpcitor hs dielectric constnt of 3.60 nd dielectric strength of 1.60 * 10 7 V>m. The cpcitor is to hve cpcitnce of 1.25 * 10-9 F nd must e le to withstnd mximum potentil difference of 5500 V. Wht is the minimum re the pltes of the cpcitor my hve? BIO Potentil in Humn Cells. ome cell wlls in the humn ody hve lyer of negtive chrge on the inside surfce nd lyer of positive chrge of equl mgnitude on the outside surfce. uppose tht the chrge density on either surfce is 0.50 * 10-3 C>m 2, the cell wll is 5.0 nm thick, nd the cellwll mteril is ir. () Find the mgnitude of E in the wll etween the two lyers of chrge. () Find the potentil difference etween the inside nd the outside of the cell. Which is t the higher potentil? (c) A typicl cell in the humn ody hs volume of m 3. Estimte the totl electric-field energy stored in the wll of cell of this size. (Hint: Assume tht the cell is sphericl, nd clculte the volume of the cell wll.) (d) In relity, the cell wll is mde up, not of ir, ut of tissue with dielectric constnt of 5.4. Repet prts () nd () in this cse A cpcitor hs prllel pltes of re 12 cm 2 seprted y 2.0 mm. The spce etween the pltes is filled with polystyrene (see Tle 24.2). () Find the permittivity of polystyrene. () Find the mximum permissile voltge cross the cpcitor to void dielectric rekdown. (c) When the voltge equls the vlue found in prt (), find the surfce chrge density on ech plte nd the induced surfce chrge density on the surfce of the dielectric.

128 814 CHAPTER 24 Cpcitnce nd Dielectrics A constnt potentil difference of 12 V is mintined plte? (c) Wht is the electric field etween the pltes? (d) Wht is etween the terminls of 0.25-mF, prllel-plte, ir cpcitor. the energy stored in the cpcitor? (e) If the ttery is disconnected () A sheet of Mylr is inserted etween the pltes of the cpcitor, nd then the pltes re pulled prt to seprtion of 7.4 mm, wht completely filling the spce etween the pltes. When this is done, re the nswers to prts ()(d)? how much dditionl chrge flows onto the positive plte of the uppose the ttery in Prolem remins connected while the pltes re pulled prt. Wht re the nswers then cpcitor (see Tle 24.1)? () Wht is the totl induced chrge on either fce of the Mylr sheet? (c) Wht effect does the Mylr to prts ()(d) fter the pltes hve een pulled prt? sheet hve on the electric field etween the pltes? Explin how BIO Cell Memrnes. you cn reconcile this with the increse in chrge on the pltes, Cell memrnes (the wlled enclosure round cell) re typiclly 7.5 nm Figure P24.50 which cts to increse the electric field When 360-nF ir cpcitor 11 nf = 10-9 F2 is connected to power supply, the energy stored in the cpcitor is tilly permele to llow chrged out 7.5 nm thick. They re pr- Outside xon 1.85 * 10-5 Axon memrne J. While the cpcitor is kept connected to the power mteril to pss in nd out, s supply, sl of dielectric is inserted tht completely fills the spce needed. Equl ut opposite chrge Inside xon etween the pltes. This increses the stored energy y densities uild up on the inside nd 2.32 * 10-5 J. () Wht is the potentil difference etween the cpcitor pltes? () Wht is the dielectric constnt of the sl? A prllel-plte cpcitor hs cpcitnce C = 12.5 pf when the volume etween the pltes is filled with ir. The pltes re circulr, with rdius 3.00 cm. The cpcitor is connected to ttery, nd chrge of mgnitude 25.0 pc goes onto ech plte. With the cpcitor still connected to the ttery, sl of dielectric is inserted etween the pltes, completely filling the spce etween the pltes. After the dielectric hs een inserted, the chrge on ech plte hs mgnitude 45.0 pc. () Wht is the dielectric constnt K of the dielectric? () Wht is the potentil difference etween the pltes efore nd fter the dielectric hs een inserted? (c) Wht is the electric field t point midwy etween the pltes efore nd fter the dielectric hs een inserted? ection 24.6 Guss s Lw in Dielectrics A prllel-plte cpcitor hs the volume etween its pltes filled with plstic with dielectric constnt K. The mgnitude of the chrge on ech plte is Q. Ech plte hs re A, nd the distnce etween the pltes is d. () Use Guss s lw s stted in Eq. (24.23) to clculte the mgnitude of the electric field in the dielectric. () Use the electric field determined in prt () to clculte the potentil difference etween the two pltes. (c) Use the result of prt () to determine the cpcitnce of the cpcitor. Compre your result to Eq. (24.12) A prllel-plte cpcitor hs pltes with re m 2 seprted y 1.00 mm of Teflon. () Clculte the chrge on the pltes when they re chrged to potentil difference of 12.0 V. () Use Guss s lw (Eq ) to clculte the electric field inside the Teflon. (c) Use Guss s lw to clculte the electric field if the voltge source is disconnected nd the Teflon is removed. PROBLEM Electronic flsh units for cmers contin cpcitor for storing the energy used to produce the flsh. In one such unit, the 1 flsh lsts for with n verge light power output of 2.70 * s W. () If the conversion of electricl energy to light is 95% efficient (the rest of the energy goes to therml energy), how much energy must e stored in the cpcitor for one flsh? () The cpcitor hs potentil difference etween its pltes of 125 V when the stored energy equls the vlue clculted in prt (). Wht is the cpcitnce? A prllel-plte ir cpcitor is mde y using two pltes 16 cm squre, spced 3.7 mm prt. It is connected to 12-V ttery. () Wht is the cpcitnce? () Wht is the chrge on ech outside fces of such memrne, nd these chrges prevent dditionl chrges from pssing through the cell wll. We cn model cell memrne s prllel-plte cpcitor, with the memrne itself contining proteins emedded in n orgnic mteril to give the memrne dielectric constnt of out 10. (ee Fig. P24.50.) () Wht is the cpcitnce per squre centimeter of such cell wll? () In its norml resting stte, cell hs potentil difference of 85 mv cross its memrne. Wht is the electric field inside this memrne? A cpcitor is mde from two hollow, coxil copper cylinders, one inside the other. There is ir in the spce etween the cylinders. The inner cylinder hs net positive chrge nd the outer cylinder hs net negtive chrge. The inner cylinder hs rdius 2.50 mm, the outer cylinder hs rdius 3.10 mm, nd the length of ech cylinder is 36.0 cm. If the potentil difference etween the surfces of the two cylinders is 80.0 V, wht is the mgnitude of the electric field t point etween the two cylinders tht is distnce of 2.80 mm from their common xis nd midwy etween the ends of the cylinders? In one type of computer keyord, ech key holds smll metl plte tht serves s one plte of prllel-plte, irfilled cpcitor. When the key is depressed, the plte seprtion decreses nd the cpcitnce increses. Electronic circuitry detects the chnge in cpcitnce nd thus detects tht the key hs een pressed. In one prticulr keyord, the re of ech metl plte is 42.0 mm 2, nd the seprtion etween the pltes is mm efore the key is depressed. () Clculte the cpcitnce efore the key is depressed. () If the circuitry cn detect chnge in cpcitnce of pf, how fr must the key e depressed efore the circuitry detects its depression? A 20.0-mF cpcitor is chrged to potentil difference of 800 V. The terminls of the chrged cpcitor re then connected to those of n unchrged 10.0-mF cpcitor. Compute () the originl chrge of the system, () the finl potentil difference cross ech cpcitor, (c) the finl energy of the system, nd (d) the decrese in energy when the cpcitors re connected In Fig. 24.9, let C 1 = 9.0 mf, C 2 = 4.0 mf, nd V = 36 V. uppose the chrged cpcitors re disconnected from the source nd from ech other, nd then reconnected to ech other with pltes of opposite sign together. By how much does the energy of the system decrese? For the cpcitor network shown in Fig. P24.55, the potentil difference cross is 12.0 V. Find () the totl energy stored in this network nd () the energy stored in the 4.80-mF cpcitor.

129 Prolems 815 Figure P mf 4.80 mf 6.20 mf 3.50 mf 11.8 mf everl 0.25-mF cpcitors re ville. The voltge cross ech is not to exceed 600 V. You need to mke cpcitor with cpcitnce 0.25 mf to e connected cross potentil difference of 960 V. () how in digrm how n equivlent cpcitor with the desired properties cn e otined. () No dielectric is perfect insultor tht would not permit the flow of ny chrge through its volume. uppose tht the dielectric in one of the cpcitors in your digrm is modertely good conductor. Wht will hppen in this cse when your comintion of cpcitors is connected cross the 960-V potentil difference? In Fig. P24.57, C 1 = C 5 = 8.4 mf nd C 2 = C 3 = Figure P24.57 C 4 = 4.2 mf. The pplied potentil is V = 220 V. () Wht is 1 C 3 C the equivlent cpcitnce of the network etween points nd? () Clculte the chrge C 2 C on ech cpcitor nd the 5 C 4 potentil difference cross ech cpcitor You re working on n electronics project requiring vriety of cpcitors, ut you hve only lrge supply of 100-nF cpcitors ville. how how you cn connect these cpcitors to produce ech of the following equivlent cpcitnces: () 50 nf; () 450 nf; (c) 25 nf; (d) 75 nf In Fig. E24.20, C 1 = 3.00 mf nd V = 150 V. The chrge on cpcitor C 1 is 150 mc nd the chrge on C 3 is 450 mc. Wht re the vlues of the cpcitnces of C 2 nd C 3? The cpcitors in Fig. P24.60 re initilly unchrged Figure P24.60 nd re connected, s in the digrm, with switch open. The d 3.00 mf 6.00 mf pplied potentil difference is V = 210 V. () Wht is the potentil difference V cd? () Wht is the potentil difference c cross ech cpcitor fter 6.00 mf 3.00 mf switch is closed? (c) How much chrge flowed through the switch when it ws closed? Three cpcitors hving cpcitnces of 8.4, 8.4, nd 4.2 mf re connected in series cross 36-V potentil difference. () Wht is the chrge on the 4.2-mF cpcitor? () Wht is the totl energy stored in ll three cpcitors? (c) The cpcitors re disconnected from the potentil difference without llowing them to dischrge. They re then reconnected in prllel with ech other, with the positively chrged pltes connected together. Wht is the voltge cross ech cpcitor in the prllel comintion? (d) Wht is the totl energy now stored in the cpcitors? Cpcitnce of Thundercloud. The chrge center of thundercloud, drifting 3.0 km ove the erth s surfce, contins 20 C of negtive chrge. Assuming the chrge center hs rdius of 1.0 km, nd modeling the chrge center nd the erth s surfce s prllel pltes, clculte: () the cpcitnce of the system; () the potentil difference etween chrge center nd ground; (c) the verge strength of the electric field etween cloud nd ground; (d) the electricl energy stored in the system In Fig. P24.63, ech Figure P24.63 cpcitnce C 1 is 6.9 mf, nd ech cpcitnce C 2 is 4.6 mf. () Compute the equivlent cpcitnce C 1 C 1 C 1 c of the network etween points nd. () Compute the chrge on ech of the three cpcitors nerest nd when V = 420 V. (c) With 420 V cross nd, C 1 C 1 d C 1 compute V cd Ech comintion of cpcitors etween points nd in Fig. P24.64 is first connected cross 120-V ttery, Figure P24.64 () chrging the comintion mf to 120 V. These comintions mf mf ignl re then connected to mke the device circuits shown. When the switch is thrown, surge of chrge for the dischrging cpcitors flows () to trigger the signl device. How much chrge flows through the signl device in ech cse? 10.0 mf A prllel-plte cpcitor with only ir etween the 20.0 mf pltes is chrged y connecting 30.0 mf it to ttery. The cpcitor is ignl then disconnected from the ttery, device without ny of the chrge leving the pltes. () A voltmeter reds 45.0 V when plced cross the cpcitor. When dielectric is inserted etween the pltes, completely filling the spce, the voltmeter reds 11.5 V. Wht is the dielectric constnt of this mteril? () Wht will the voltmeter red if the dielectric is now pulled prtwy out so it fills only onethird of the spce etween the pltes? An ir cpcitor is mde y Figure P24.66 using two flt pltes, ech with re A, seprted y distnce d. Then metl sl hving thickness (less thn d) nd the sme shpe nd size s the pltes is inserted etween them, prllel to the pltes nd not touching either plte (Fig. P24.66). () Wht is d the cpcitnce of this rrngement? () Express the cpcitnce s multiple of the cpcitnce C 0 when the metl sl is not present. (c) Discuss wht hppens to the cpcitnce in the limits 0 nd d Cpcitnce of the Erth. Consider sphericl cpcitor with one conductor eing solid conducting sphere of rdius R nd the other conductor eing t infinity. () Use Eq. (24.1) nd wht you know out the potentil t the surfce of conducting sphere with chrge Q to derive n expression for the cpcitnce of the chrged sphere. () Use your result in prt () to clculte the cpcitnce of the erth. The erth is good conductor nd hs rdius of 6380 km. Compre your results to the cpcitnce of typicl cpcitors used in electronic circuits, which rnges from 10 pf to 100 pf A potentil difference V = 48.0 V is pplied cross the cpcitor network of Fig. E If C 1 = C 2 = 4.00 mf nd C 2 C 2 C 1

130 816 CHAPTER 24 Cpcitnce nd Dielectrics C 4 = 8.00 mf, wht must the cpcitnce C 3 e if the network is w nd length L (Fig. P24.74). The height of the fuel etween the to store 2.90 * 10-3 J of electricl energy? pltes is h. You cn ignore ny fringing effects. () Derive n Erth-Ionosphere Cpcitnce. The erth cn e considered expression for K eff s function of h. () Wht is the effective s single-conductor cpcitor (see Prolem 24.67). It cn dielectric constnt for tnk full, 2 full, nd 4 full if the fuel is lso e considered in comintion with chrged lyer of the gsoline 1K = 1.952? (c) Repet prt () for methnol tmosphere, the ionosphere, s sphericl cpcitor with two 1K = (d) For which fuel is this fuel guge more prcticl? pltes, the surfce of the erth eing the negtive plte. The ionosphere is t level of out 70 km, nd the potentil difference pltes A, B, nd C, ech Three squre metl Figure P24.75 etween erth nd ionosphere is out 350,000 V. Clculte: () 12.0 cm on side nd 1.50 mm the cpcitnce of this system; () the totl chrge on the cpcitor; thick, re rrnged s in Fig. Pper Metl (c) the energy stored in the system. P The pltes re seprted y sheets of pper B A CALC The inner cylinder of long, cylindricl cpcitor hs rdius r nd liner chrge density l. It is surrounded y 0.45 mm thick nd with dielectric constnt 4.2. The outer C coxil cylindricl conducting shell with inner rdius r nd liner chrge density -l (see Fig. 24.6). () Wht is the energy density in the region etween the conductors t distnce r from the xis? () Integrte the energy density clculted in prt () over the volume etween the conductors in length L of the cpcitor to otin the totl electric-field energy per unit length. (c) Use Eq. (24.9) nd the cpcitnce per unit length clculted in Exmple 24.4 pltes re connected together nd connected to point. The inner plte is connected to point. () Copy the digrm nd show y plus nd minus signs the chrge distriution on the pltes when point is mintined t positive potentil reltive to point. () Wht is the cpcitnce etween points nd? (ection 24.1) to clculte U>L. Does your result gree with tht otined in prt ()? CP A cpcitor hs potentil difference of CHALLENGE PROBLEM 2.25 * 10 3 V etween its pltes. A short luminum wire with initil temperture CP The prllel-plte ir cpcitor in Fig. P24.76 consists 23.0 o C is connected etween the pltes of the cpcitor nd ll the energy stored in the cpcitor goes into heting the wire. The wire hs mss 12.0 g. If no het is lost to the surroundings nd the finl temperture of the wire is 34.2 o C, wht is the of two horizontl conducting pltes of equl re A. The ot- tom plte rests on fixed support, nd the top plte is suspended y four springs with spring constnt k, positioned t ech of the four corners of the top plte s shown in the figure. When cpcitnce of the cpcitor? unchrged, the pltes re seprted y distnce z 0. A ttery is A prllel-plte cpcitor is mde from two pltes etween them. This cuses the plte seprtion to decrese to z. connected to the pltes nd produces potentil difference V Figure P cm on ech side nd 4.50 mm prt. Hlf of the Neglect ny fringing effects. () how tht the electrosttic force etween the chrged pltes hs mgnitude P 0 AV 2 >2z 2. (Hint: ee spce etween these pltes Plexigls Air Exercise ) () Otin n expression tht reltes the plte seprtion z to the potentil difference V. The resulting eqution will contins only ir, ut the other hlf is filled with Plexigls of e cuic in z. (c) Given the vlues A = m 2, z 0 = 1.20 mm, dielectric constnt 3.40 (Fig. P24.72). An 18.0-V ttery is connected cross the pltes. () Wht is the cpcitnce of this comintion? (Hint: Cn you think of this cpcitor s equivlent to two cpcitors in prllel?) () How much energy is stored in the cpcitor? (c) If we remove the Plexigls ut chnge nothing else, how much energy will e stored in k = 25.0 N>m, nd V = 120 V, find the two vlues of z for which the top plte will e in equilirium. (Hint: You cn solve the cuic eqution y plugging tril vlue of z into the eqution nd then djusting your guess until the eqution is stisfied to three significnt figures. Locting the roots of the cuic eqution grphiclly cn help you pick strting vlues of z for this tril-nd-error procedure. One root of the cuic eqution hs nonphysicl negtive the cpcitor? A prllel-plte cpcitor hs squre pltes tht re vlue.) (d) For ech of the two vlues of z found in prt (c), is the 8.00 cm on ech side nd 3.80 mm prt. The spce etween the pltes is completely filled with two squre sls of dielectric, ech 8.00 cm on side nd 1.90 mm thick. One sl is pyrex glss nd the other is polystyrene. If the potentil difference etween the pltes is equilirium stle or unstle? For stle equilirium smll displcement of the oject will give rise to net force tending to return the oject to the equilirium position. For unstle equilirium smll displcement gives rise to net force tht tkes the 86.0 V, how much electricl energy is stored in the cpcitor? oject frther wy from equilirium A fuel guge uses cpcitor to determine the height Figure P24.74 Figure P24.76 of the fuel in tnk. The effective dielectric constnt K eff chnges V k k from vlue of 1 when the tnk is Bttery z k k V empty to vlue of K, the dielectric A constnt of the fuel, when the Air tnk is full. The pproprite electronic L A circuitry cn determine the effective dielectric constnt of the Two squre conducting pltes with sides of length L re w comined ir nd fuel etween seprted y distnce D. A dielectric sl with constnt K with h Fuel the cpcitor pltes. Ech of the dimensions L * L * D is inserted distnce x into the spce two rectngulr pltes hs width etween the pltes, s shown in Fig. P () Find the cpcitnce

131 Answers 817 C of this system. () uppose tht the cpcitor is connected to ttery tht mintins constnt potentil difference V etween the pltes. If the dielectric sl is inserted n dditionl distnce dx into the spce etween the pltes, show tht the chnge in stored energy is du = 1K - 12P 0V 2 L dx 2D Figure P24.77 Dielectric sl, constnt K (c) uppose tht efore the sl is moved y dx, the pltes re disconnected from the ttery, so tht the D chrges on the pltes remin constnt. Determine the mgnitude of the chrge on ech plte, nd then show tht when the sl is moved dx frther into the spce etween the pltes, the stored energy chnges y n mount tht is the negtive of the expression for du given in prt (). (d) If F is the force exerted on the sl y the chrges on the pltes, then du should equl the work done ginst this force to move the sl distnce dx. Thus du = -Fdx. how tht pplying this expression to the result of prt () suggests tht the electric force on the sl pushes it out of the cpcitor, while the result of prt (c) suggests tht the force x L L pulls the sl into the cpcitor. (e) Figure shows tht the force in fct pulls the sl into the cpcitor. Explin why the result of prt () gives n incorrect nswer for the direction of this force, nd clculte the mgnitude of the force. (This method does not require knowledge of the nture of the fringing field.) An isolted sphericl cpcitor hs chrge Q on its inner conductor (rdius r ) nd chrge -Q on its Figure P24.78 outer conductor (rdius r ). Hlf of the r volume etween the two conductors is then filled with liquid dielectric of constnt K, s shown in cross section in r Fig. P () Find the cpcitnce of the hlf-filled cpcitor. () Find the mgnitude of E in the volume etween K the two conductors s function of the distnce r from the center of the cpcitor. Give nswers for oth the upper nd lower hlves of this volume. (c) Find the surfce density of free chrge on the upper nd lower hlves of the inner nd outer conductors. (d) Find the surfce density of ound chrge on the inner 1r = r 2 nd outer 1r = r 2 surfces of the dielectric. (e) Wht is the surfce density of ound chrge on the flt surfce of the dielectric? Explin. Answers Chpter Opening Question? Eqution (24.9) shows tht the energy stored in cpcitor with cpcitnce C nd chrge Q is U = Q 2 >2C. If the chrge Q is douled, the stored energy increses y fctor of 2 2 = 4. Note tht if the vlue of Q is too gret, the electric-field mgnitude inside the cpcitor will exceed the dielectric strength of the mteril etween the pltes nd dielectric rekdown will occur (see ection 24.4). This puts prcticl limit on the mount of energy tht cn e stored. Test Your Understnding Questions 24.1 Answer: (iii) The cpcitnce does not depend on the vlue of the chrge Q. Douling the vlue of Q cuses the potentil difference V to doule, so the cpcitnce C = Q>V remins the sme. These sttements re true no mtter wht the geometry of the cpcitor Answers: () (i), () (iv) In series connection the two cpcitors crry the sme chrge Q ut hve different potentil differences V = Q>C; the cpcitor with the smller cpcitnce C hs the greter potentil difference. In prllel connection the two cpcitors hve the sme potentil difference V ut crry different chrges Q = CV ; the cpcitor with the lrger cpcitnce C hs the greter chrge. Hence 4-mF cpcitor will hve greter potentil difference thn n 8-mF cpcitor if the two re connected in series. The 4-mF cpcitor cnnot crry more chrge thn the 8-mF cpcitor no mtter how they re connected: In series connection they will crry the sme chrge, nd in prllel connection the 8-mF cpcitor will crry more chrge Answer: (i) Cpcitors connected in series crry the sme chrge Q. To compre the mount of energy stored, we use the expression U = Q 2 >2C from Eq. (24.9); it shows tht the cpcitor with the smller cpcitnce 1C = 4 mf2 hs more stored energy in series comintion. By contrst, cpcitors in prllel hve the sme potentil difference V, so to compre them we use U = 1 2 CV 2 from Eq. (24.9). It shows tht in prllel comintion, the cpcitor with the lrger cpcitnce 1C = 8 mf2 hs more stored energy. (If we hd insted used U = 1 2 CV 2 to nlyze the series comintion, we would hve to ccount for the different potentil differences cross the two cpcitors. Likewise, using U = Q 2 >2C to study the prllel comintion would require us to ccount for the different chrges on the cpcitors.) 24.4 Answer: (i) Here Q remins the sme, so we use U = Q 2 >2C from Eq. (24.9) for the stored energy. Removing the dielectric lowers the cpcitnce y fctor of 1>K; since U is inversely proportionl to C, the stored energy increses y fctor of K. It tkes work to pull the dielectric sl out of the cpcitor ecuse the fringing field tries to pull the sl ck in (Fig ). The work tht you do goes into the energy stored in the cpcitor Answer: (i), (iii), (ii) Eqution (24.14) sys tht if E 0 is the initil electric-field mgnitude (efore the dielectric sl is inserted), then the resultnt field mgnitude fter the sl is inserted is E 0 >K = E 0 >3. The mgnitude of the resultnt field equls the difference etween the initil field mgnitude nd the mgnitude E i of the field due to the ound chrges (see Fig ). Hence E 0 - E i = E 0 >3 nd E i = 2E 0 > Answer: (iii) Eqution (24.23) shows tht this sitution is the sme s n isolted point chrge in vcuum ut with E replced y KE. Hence KE t the point of interest is equl to q>4pp 0 r 2, nd so E = q>4pkp 0 r 2. As in Exmple 24.12, filling the spce with dielectric reduces the electric field y fctor of 1>K. Bridging Prolem Answers: () 0 () Q 2 >32p 2 P (c) Q 2 0 r 4 >8pP 0 R (d) Q 2 >8pP 0 R (e) C = 4pP 0 R

132 25 CURRENT, REITANCE, AND ELECTROMOTIVE FORCE LEARNING GOAL By studying this chpter, you will lern: The mening of electric current, nd how chrges move in conductor. Wht is ment y the resistivity nd conductivity of sustnce. How to clculte the resistnce of conductor from its dimensions nd its resistivity. How n electromotive force (emf) mkes it possile for current to flow in circuit. How to do clcultions involving energy nd power in circuits.? In flshlight, is the mount of current tht flows out of the ul less thn, greter thn, or equl to the mount of current tht flows into the ul? In the pst four chpters we studied the interctions of electric chrges t rest; now we re redy to study chrges in motion. An electric current consists of chrges in motion from one region to nother. If the chrges follow conducting pth tht forms closed loop, the pth is clled n electric circuit. Fundmentlly, electric circuits re mens for conveying energy from one plce to nother. As chrged prticles move within circuit, electric potentil energy is trnsferred from source (such s ttery or genertor) to device in which tht energy is either stored or converted to nother form: into sound in stereo system or into het nd light in toster or light ul. Electric circuits re useful ecuse they llow energy to e trnsported without ny moving prts (other thn the moving chrged prticles themselves). They re t the hert of flshlights, computers, rdio nd television trnsmitters nd receivers, nd household nd industril power distriution systems. Your nervous system is specilized electric circuit tht crries vitl signls from one prt of your ody to nother. In Chpter 26 we will see how to nlyze electric circuits nd will exmine some prcticl pplictions of circuits. Before we cn do so, however, you must understnd the sic properties of electric currents. These properties re the suject of this chpter. We ll egin y descriing the nture of electric conductors nd considering how they re ffected y temperture. We ll lern why short, ft, cold copper wire is etter conductor thn long, skinny, hot steel wire. We ll study the properties of tteries nd see how they cuse current nd energy trnsfer in circuit. In this nlysis we will use the concepts of current, potentil difference (or voltge), resistnce, nd electromotive force. Finlly, we ll look t electric current in mteril from microscopic viewpoint. 818

133 25.1 Current Current A current is ny motion of chrge from one region to nother. In this section we ll discuss currents in conducting mterils. The vst mjority of technologicl pplictions of chrges in motion involve currents of this kind. In electrosttic situtions (discussed in Chpters 21 through 24) the electric field is zero everywhere within the conductor, nd there is no current. However, this does not men tht ll chrges within the conductor re t rest. In n ordinry metl such s copper or lumium, some of the electrons re free to move within the conducting mteril. These free electrons move rndomly in ll directions, somewht like the molecules of gs ut with much greter speeds, of the order of 10 6 m/s. The electrons nonetheless do not escpe from the conducting mteril, ecuse they re ttrcted to the positive ions of the mteril. The motion of the electrons is rndom, so there is no net flow of chrge in ny direction nd hence no current. Now consider wht hppens if constnt, stedy electric field E is estlished inside conductor. (We ll see lter how this cn e done.) A chrged prticle (such s free electron) inside the conducting mteril is then sujected to stedy force F qe. If the chrged prticle were moving in vcuum, this stedy force would cuse stedy ccelertion in the direction of F, nd fter time the chrged prticle would e moving in tht direction t high speed. But chrged prticle moving in conductor undergoes frequent collisions with the mssive, nerly sttionry ions of the mteril. In ech such collision the prticle s direction of motion undergoes rndom chnge. The net effect of the electric field E is tht in ddition to the rndom motion of the chrged prticles within the conductor, there is lso very slow net motion or drift of the moving chrged prticles s group in the direction of the electric force F qe (Fig. 25.1). This motion is descried in terms of the drift velocity v d of the prticles. As result, there is net current in the conductor. While the rndom motion of the electrons hs very fst verge speed of out 10 6 m/s, the drift speed is very slow, often on the order of 10-4 m/s. Given tht the electrons move so slowly, you my wonder why the light comes on immeditely when you turn on the switch of flshlight. The reson is tht the electric field is set up in the wire with speed pproching the speed of light, nd electrons strt to move ll long the wire t very nerly the sme time. The time tht it tkes ny individul electron to get from the switch to the light ul isn t relly relevnt. A good nlogy is group of soldiers stnding t ttention when the sergent orders them to strt mrching; the order reches the soldiers ers t the speed of sound, which is much fster thn their mrching speed, so ll the soldiers strt to mrch essentilly in unison. The Direction of Current Flow The drift of moving chrges through conductor cn e interpreted in terms of work nd energy. The electric field E does work on the moving chrges. The resulting kinetic energy is trnsferred to the mteril of the conductor y mens of collisions with the ions, which virte out their equilirium positions in the crystlline structure of the conductor. This energy trnsfer increses the verge virtionl energy of the ions nd therefore the temperture of the mteril. Thus much of the work done y the electric field goes into heting the conductor, not into mking the moving chrges move ever fster nd fster. This heting is sometimes useful, s in n electric toster, ut in mny situtions is simply n unvoidle y-product of current flow. In different current-crrying mterils, the chrges of the moving prticles my e positive or negtive. In metls the moving chrges re lwys (negtive) electrons, while in n ionized gs (plsm) or n ionic solution the moving chrges my include oth electrons nd positively chrged ions. In semiconductor 25.1 If there is no electric field inside conductor, n electron moves rndomly from point P 1 to point P in time If n electric field E 2 t. is present, the electric force F qe imposes smll drift (gretly exggerted here) tht tkes the electron to point P 2, distnce v d t from P 2 in the direction of the force. E Conductor without internl E field Pth of electron without E field. Electron moves rndomly. Pth of electron with E field. The motion is mostly P 1 rndom, ut... P 2 P 2 v d Dt... the E field results in net displcement long the wire. Conductor with internl E field F 5 qe An electron hs negtive chrge q, so the force on it due to the E field is in the direction opposite to E. E

134 820 CHAPTER 25 Current, Resistnce, nd Electromotive Force 25.2 The sme current cn e produced y () positive chrges moving in the direction of the electric field E or () the sme numer of negtive chrges moving t the sme speed in the direction opposite to E. () v d v d v d I v d v d v d v d v d A conventionl current is treted s flow of positive chrges, regrdless of whether the free chrges in the conductor re positive, negtive, or oth. () v d v d v d v d v d v d I v d v d v d v d In metllic conductor, the moving chrges re electrons ut the current still points in the direction positive chrges would flow The current I is the time rte of chrge trnsfer through the cross-sectionl re A. The rndom component of ech moving chrged prticle s motion verges to zero, nd the current is in the sme direction s E whether the moving chrges re positive (s shown here) or negtive (see Fig. 25.2). E E mteril such s germnium or silicon, conduction is prtly y electrons nd prtly y motion of vcncies, lso known s holes; these re sites of missing electrons nd ct like positive chrges. Figure 25.2 shows segments of two different current-crrying mterils. In Fig the moving chrges re positive, the electric force is in the sme direction s E, nd the drift velocity v d is from left to right. In Fig the chrges re negtive, the electric force is opposite to E, nd the drift velocity v d is from right to left. In oth cses there is net flow of positive chrge from left to right, nd positive chrges end up to the right of negtive ones. We define the current, denoted y I, to e in the direction in which there is flow of positive chrge. Thus we descrie currents s though they consisted entirely of positive chrge flow, even in cses in which we know tht the ctul current is due to electrons. Hence the current is to the right in oth Figs nd This choice or convention for the direction of current flow is clled conventionl current. While the direction of the conventionl current is not necessrily the sme s the direction in which chrged prticles re ctully moving, we ll find tht the sign of the moving chrges is of little importnce in nlyzing electric circuits. Figure 25.3 shows segment of conductor in which current is flowing. We consider the moving chrges to e positive, so they re moving in the sme direction s the current. We define the current through the cross-sectionl re A to e the net chrge flowing through the re per unit time. Thus, if net chrge dq flows through n re in time dt, the current I through the re is I = dq dt (definition of current) (25.1) CAUTION Current is not vector Although we refer to the direction of current, current s defined y Eq. (25.1) is not vector quntity. In current-crrying wire, the current is lwys long the length of the wire, regrdless of whether the wire is stright or curved. No single vector could descrie motion long curved pth, which is why current is not vector. We ll usully descrie the direction of current either in words (s in the current flows clockwise round the circuit ) or y choosing current to e positive if it flows in one direction long conductor nd negtive if it flows in the other direction. The I unit of current is the mpere; one mpere is defined to e one coulom per second 11 A = 1 C/s2. This unit is nmed in honor of the French scientist André Mrie Ampère ( ). When n ordinry flshlight (D-cell size) is turned on, the current in the flshlight is out 0.5 to 1 A; the current in the wires of cr engine s strter motor is round 200 A. Currents in rdio nd television circuits re usully expressed in millimperes 11 ma = 10-3 A2 or micromperes 11 ma = 10-6 A2, nd currents in computer circuits re expressed in nnomperes 11 na = 10-9 A2 or picomperes 11 pa = A2. r v d r v d r v d I v d dt A Current I 5 dq dt r v d r v d r v d E Current, Drift Velocity, nd Current Density We cn express current in terms of the drift velocity of the moving chrges. Let s consider gin the sitution of Fig of conductor with cross-sectionl re A nd n electric field E directed from left to right. To egin with, we ll ssume tht the free chrges in the conductor re positive; then the drift velocity is in the sme direction s the field. uppose there re n moving chrged prticles per unit volume. We cll n the concentrtion of prticles; its I unit is m -3. Assume tht ll the prticles move with the sme drift velocity with mgnitude v d. In time intervl dt, ech prticle moves distnce v d dt. The prticles tht flow out of the right end of the shded cylinder with length v d dt during dt re the prticles tht were within this cylinder t the eginning of the intervl dt. The volume of the cylinder is Av d dt, nd the numer of prticles within it is nav d dt. If ech

135 25.1 Current 821 prticle hs chrge q, the chrge dq tht flows out of the end of the cylinder during time dt is nd the current is dq = q1nav d dt2 = nqv d Adt I = dq dt = nqv d A The current per unit cross-sectionl re is clled the current density J: J = I A = nqv d The units of current density re mperes per squre meter 1A/m 2 2. If the moving chrges re negtive rther thn positive, s in Fig. 25.2, the drift velocity is opposite to E. But the current is still in the sme direction s E t ech point in the conductor. Hence the current I nd current density J don t depend on the sign of the chrge, nd so in the ove expressions for I nd J we replce the chrge q y its solute vlue ƒqƒ: I = dq dt = nƒqƒv d A (generl expression for current) (25.2) J = I A = nƒqƒv d (generl expression for current density) (25.3) The current in conductor is the product of the concentrtion of moving chrged prticles, the mgnitude of chrge of ech such prticle, the mgnitude of the drift velocity, nd the cross-sectionl re of the conductor. We cn lso define vector current density J tht includes the direction of the drift velocity: J nqvd (vector current density) (25.4) There re no solute vlue signs in Eq. (25.4). If q is positive, v is in the sme direction s E ; if q is negtive, v is opposite to E d. In either cse, J is in the sme direction s E. Eqution (25.3) gives the mgnitude J of the vector current density J. d CAUTION Current density vs. current Note tht current density J is vector, ut current I is not. The difference is tht the current density J descries how chrges flow t certin point, nd the vector s direction tells you out the direction of the flow t tht point. By contrst, the current I descries how chrges flow through n extended oject such s wire. For exmple, I hs the sme vlue t ll points in the circuit of Fig. 25.3, ut J does not: The current density is directed downwrd in the left-hnd side of the loop nd upwrd in the right-hnd side. The mgnitude of J cn lso vry round circuit. In Fig the current density mgnitude J = I/A is less in the ttery (which hs lrge cross-sectionl re A) thn in the wires (which hve smll cross-sectionl re) Prt of the electric circuit tht includes this light ul psses through eker with solution of sodium chloride. The current in the solution is crried y oth positive chrges (N ions) nd negtive chrges ( Cl - ions). In generl, conductor my contin severl different kinds of moving chrged prticles hving chrges q 1, q 2, Á, concentrtions n 1, n 2, Á, nd drift velocities with mgnitudes v d1, v d2, Á. An exmple is current flow in n ionic solution (Fig. 25.4). In sodium chloride solution, current cn e crried y oth positive sodium ions nd negtive chlorine ions; the totl current I is found y dding up the currents due to ech kind of chrged prticle, using Eq. (25.2). Likewise, the totl vector current density J is found y using Eq. (25.4) for ech kind of chrged prticle nd dding the results. We will see in ection 25.4 tht it is possile to hve current tht is stedy (tht is, one tht is constnt in time) only if the conducting mteril forms

136 822 CHAPTER 25 Current, Resistnce, nd Electromotive Force closed loop, clled complete circuit. In such stedy sitution, the totl chrge in every segment of the conductor is constnt. Hence the rte of flow of chrge out t one end of segment t ny instnt equls the rte of flow of chrge in t the other end of the segment, nd the current is the sme t ll cross sections of the circuit. We ll mke use of this oservtion when we nlyze electric circuits lter in this chpter. In mny simple circuits, such s flshlights or cordless electric drills, the direction of the current is lwys the sme; this is clled direct current. But home pplinces such s tosters, refrigertors, nd televisions use lternting current, in which the current continuously chnges direction. In this chpter we ll consider direct current only. Alternting current hs mny specil fetures worthy of detiled study, which we ll exmine in Chpter 31. Exmple 25.1 Current density nd drift velocity in wire An 18-guge copper wire (the size usully used for lmp cords), with dimeter of 1.02 mm, crries constnt current of 1.67 A to 200-W lmp. The free-electron density in the wire is 8.5 * per cuic meter. Find () the current density nd () the drift speed. OLUTION IDENTIFY nd ET UP: This prolem uses the reltionships mong current I, current density J, nd drift speed v d. We re given I nd the wire dimeter d, so we use Eq. (25.3) to find J. We use Eq. (25.3) gin to find from J nd the known electron density n. EXECUTE: () The cross-sectionl re is A = pd2 4 v d = p11.02 * 10-3 m2 2 4 = 8.17 * 10-7 m 2 The mgnitude of the current density is then J = I A = 1.67 A 8.17 * 10-7 m 2 = 2.04 * 106 A>m 2 () From Eq. (25.3) for the drift velocity mgnitude v d, we find v d = J 2.04 * 10 6 A>m 2 = nƒqƒ 18.5 * m -3 2ƒ * Cƒ = 1.5 * 10-4 m>s = 0.15 mm>s EVALUATE: At this speed n electron would require 6700 s (lmost 2 h) to trvel 1 m long this wire. The speeds of rndom motion of the electrons re roughly 10 6 m>s, round times the drift speed. Picture the electrons s ouncing round frnticlly, with very slow drift! Test Your Understnding of ection 25.1 uppose we replced the wire in Exmple 25.1 with 12-guge copper wire, which hs twice the dimeter of 18-guge wire. If the current remins the sme, wht effect would this hve on the mgnitude of the drift velocity v d? (i) none v d would e unchnged; (ii) v d would e twice s gret; (iii) v d would e four times greter; (iv) v d would e hlf s gret; (v) would e one-fourth s gret. v d 25.2 Resistivity The current density J in conductor depends on the electric field nd on the properties of the mteril. In generl, this dependence cn e quite complex. But for some mterils, especilly metls, t given temperture, J is nerly directly proportionl to E, nd the rtio of the mgnitudes of E nd J is constnt. This reltionship, clled Ohm s lw, ws discovered in 1826 y the Germn physicist Georg imon Ohm ( ). The word lw should ctully e in quottion mrks, since Ohm s lw, like the idel-gs eqution nd Hooke s lw, is n idelized model tht descries the ehvior of some mterils quite well ut is not generl description of ll mtter. In the following discussion we ll ssume tht Ohm s lw is vlid, even though there re mny situtions in which it is not. The sitution is comprle to our representtion of the ehvior of the sttic nd kinetic friction forces; we treted these friction forces s eing directly proportionl to the norml force, even though we knew tht this ws t est n pproximte description. E

137 25.2 Resistivity Tle 25.1 Resistivities t Room Temperture (20 C) ustnce R 1æ # m2 ustnce R 1æ # m2 Conductors emiconductors Metls ilver 1.47 * 10-8 Pure cron (grphite) Copper 1.72 * 10-8 Pure germnium 3.5 * Gold 2.44 * 10-8 Pure silicon 2300 Aluminum 2.75 * 10-8 Insultors Tungsten 5.25 * 10-8 Amer 5 * teel 20 * 10-8 Glss Led 22 * 10-8 Lucite Mercury 95 * 10-8 Mic Alloys Mngnin (Cu 84%, Mn 12%, Ni 4%) 44 * 10-8 Qurtz (fused) 75 * Constntn (Cu 60%, Ni 40%) 49 * 10-8 ulfur Nichrome 100 * 10-8 Teflon Wood We define the resistivity r of mteril s the rtio of the mgnitudes of electric field nd current density: r = E J (definition of resistivity) (25.5) The greter the resistivity, the greter the field needed to cuse given current density, or the smller the current density cused y given field. From Eq. (25.5) the units of r re 1V>m2>1A>m 2 2 = V # m>a. As we will discuss in the next section, 1 V>A is clled one ohm ( 1 Æ; we use the Greek letter Æ, or omeg, which is llitertive with ohm ). o the I units for r re Æ # m (ohm-meters). Tle 25.1 lists some representtive vlues of resistivity. A perfect conductor would hve zero resistivity, nd perfect insultor would hve n infinite resistivity. Metls nd lloys hve the smllest resistivities nd re the est conductors. The resistivities of insultors re greter thn those of the metls y n enormous fctor, on the order of The reciprocl of resistivity is conductivity. Its units re 1Æ # m2-1. Good conductors of electricity hve lrger conductivity thn insultors. Conductivity is the direct electricl nlog of therml conductivity. Compring Tle 25.1 with Tle 17.5 (Therml Conductivities), we note tht good electricl conductors, such s metls, re usully lso good conductors of het. Poor electricl conductors, such s cermic nd plstic mterils, re lso poor therml conductors. In metl the free electrons tht crry chrge in electricl conduction lso provide the principl mechnism for het conduction, so we should expect correltion etween electricl nd therml conductivity. Becuse of the enormous difference in conductivity etween electricl conductors nd insultors, it is esy to confine electric currents to well-defined pths or circuits (Fig. 25.5). The vrition in therml conductivity is much less, only fctor of 10 3 or so, nd it is usully impossile to confine het currents to tht extent. emiconductors hve resistivities intermedite etween those of metls nd those of insultors. These mterils re importnt ecuse of the wy their resistivities re ffected y temperture nd y smll mounts of impurities. A mteril tht oeys Ohm s lw resonly well is clled n ohmic conductor or liner conductor. For such mterils, t given temperture, r is constnt tht does not depend on the vlue of E. Mny mterils show sustntil deprtures from Ohm s-lw ehvior; they re nonohmic, or nonliner. In these mterils, J depends on E in more complicted mnner. Anlogies with fluid flow cn e ig help in developing intuition out electric current nd circuits. For exmple, in the mking of wine or mple syrup, the product is sometimes filtered to remove sediments. A pump forces the fluid through the filter under pressure; if the flow rte (nlogous to J) is proportionl to the pressure difference etween the upstrem nd downstrem sides (nlogous to E), the ehvior is nlogous to Ohm s lw The copper wires, or trces, on this circuit ord re printed directly onto the surfce of the drk-colored insulting ord. Even though the trces re very close to ech other (only out millimeter prt), the ord hs such high resistivity (nd low conductivity) tht no current cn flow etween the trces. Conducting pths (trces)

138 824 CHAPTER 25 Current, Resistnce, nd Electromotive Force Appliction Resistivity nd Nerve Conduction This flse-color imge from n electron microscope shows cross section through nerve fier out 1 μm ( 10-6 m) in dimeter. A lyer of n insulting ftty sustnce clled myelin is wrpped round the conductive mteril of the xon. The resistivity of myelin is much greter thn tht of the xon, so n electric signl trveling long the nerve fier remins confined to the xon. This mkes it possile for signl to trvel much more rpidly thn if the myelin were sent Vrition of resistivity r with solute temperture T for () norml metl, () semiconductor, nd (c) superconductor. In () the liner pproximtion to r s function of T is shown s green line; the pproximtion grees exctly t T = T 0, where r = r 0. () () (c) r 0 O O O Myelin Axon r r r Metl: Resistivity increses with incresing temperture. T 0 lope 5r 0 emiconductor: Resistivity decreses with incresing temperture. T T uperconductor: At tempertures elow T c, the resistivity is zero. T c T Resistivity nd Temperture The resistivity of metllic conductor nerly lwys increses with incresing temperture, s shown in Fig As temperture increses, the ions of the conductor virte with greter mplitude, mking it more likely tht moving electron will collide with n ion s in Fig. 25.1; this impedes the drift of electrons through the conductor nd hence reduces the current. Over smll temperture rnge (up to 100 C or so), the resistivity of metl cn e represented pproximtely y the eqution r1t2 = r T - T 0 24 (temperture dependence of resistivity) (25.6) where r 0 is the resistivity t reference temperture T 0 (often tken s 0 C or 20 C2 nd r1t2 is the resistivity t temperture T, which my e higher or lower thn T 0. The fctor is clled the temperture coefficient of resistivity. ome representtive vlues re given in Tle The resistivity of the lloy mngnin is prcticlly independent of temperture. Tle 25.2 Temperture Coefficients of Resistivity (Approximte Vlues Ner Room Temperture) Mteril A 31 C2 1 4 Mteril A 31 C2 1 4 Aluminum Led Brss Mngnin Cron (grphite) Mercury Constntn Nichrome Copper ilver Iron Tungsten The resistivity of grphite ( nonmetl) decreses with incresing temperture, since t higher tempertures, more electrons re shken loose from the toms nd ecome moile; hence the temperture coefficient of resistivity of grphite is negtive. This sme ehvior occurs for semiconductors (Fig. 25.6). Mesuring the resistivity of smll semiconductor crystl is therefore sensitive mesure of temperture; this is the principle of type of thermometer clled thermistor. ome mterils, including severl metllic lloys nd oxides, show phenomenon clled superconductivity. As the temperture decreses, the resistivity t first decreses smoothly, like tht of ny metl. But then t certin criticl temperture T c phse trnsition occurs nd the resistivity suddenly drops to zero, s shown in Fig. 25.6c. Once current hs een estlished in superconducting ring, it continues indefinitely without the presence of ny driving field. uperconductivity ws discovered in 1911 y the Dutch physicist Heike Kmerlingh Onnes ( ). He discovered tht t very low tempertures, elow 4.2 K, the resistivity of mercury suddenly dropped to zero. For the next 75 yers, the highest T c ttined ws out 20 K. This ment tht superconductivity occurred only when the mteril ws cooled using expensive liquid helium, with oiling-point temperture of 4.2 K, or explosive liquid hydrogen, with oiling point of 20.3 K. But in 1986 Krl Müller nd Johnnes Bednorz discovered n oxide of rium, lnthnum, nd copper with T c of nerly 40 K, nd the rce ws on to develop high-temperture superconducting mterils. By 1987 complex oxide of yttrium, copper, nd rium hd een found tht hs vlue of T c well ove the 77 K oiling temperture of liquid nitrogen, refrigernt tht is oth inexpensive nd sfe. The current (2010) record for T c t tmospheric pressure is 138 K, nd mterils tht re superconductors t room temperture my ecome relity. The implictions of these discoveries for power-distriution systems, computer design, nd trnsporttion re enormous. Menwhile, superconducting electromgnets cooled y liquid helium re used in prticle ccelertors nd some experimentl mgnetic-levittion rilrods.

139 25.3 Resistnce 825 uperconductors hve other exotic properties tht require n understnding of mgnetism to explore; we will discuss these further in Chpter 29. Test Your Understnding of ection 25.2 You mintin constnt electric field inside piece of semiconductor while lowering the semiconductor s temperture. Wht hppens to the current density in the semiconductor? (i) It increses; (ii) it decreses; (iii) it remins the sme Resistnce For conductor with resistivity r, the current density t point where the electric field is E is given y Eq. (25.5), which we cn write s E rj (25.7) When Ohm s lw is oeyed, r is constnt nd independent of the mgnitude of the electric field, so E is directly proportionl to J. Often, however, we re more interested in the totl current in conductor thn in J nd more interested in the potentil difference etween the ends of the conductor thn in E. This is so lrgely ecuse current nd potentil difference re much esier to mesure thn re J nd E. uppose our conductor is wire with uniform cross-sectionl re A nd length L, s shown in Fig Let V e the potentil difference etween the higher-potentil nd lower-potentil ends of the conductor, so tht V is positive. The direction of the current is lwys from the higher-potentil end to the lowerpotentil end. Tht s ecuse current in conductor flows in the direction of E, no mtter wht the sign of the moving chrges (Fig. 25.2), nd ecuse E points in the direction of decresing electric potentil (see ection 23.2). As the current flows through the potentil difference, electric potentil energy is lost; this energy is trnsferred to the ions of the conducting mteril during collisions. We cn lso relte the vlue of the current I to the potentil difference etween the ends of the conductor. If the mgnitudes of the current density J nd the electric field E re uniform throughout the conductor, the totl current I is given y I = JA, nd the potentil difference V etween the ends is V = EL. When we solve these equtions for J nd E, respectively, nd sustitute the results in Eq. (25.7), we otin V (25.8) L = ri A or V = rl A I This shows tht when r is constnt, the totl current I is proportionl to the potentil difference V. The rtio of V to I for prticulr conductor is clled its resistnce R: J PhET: Resistnce in Wire 25.7 A conductor with uniform cross section. The current density is uniform over ny cross section, nd the electric field is constnt long the length. Current flows from higher to lower electric potentil. Higher potentil A I L J E I Lower potentil V 5 potentil difference etween ends R = V I (25.9) Compring this definition of R to Eq. (25.8), we see tht the resistnce R of prticulr conductor is relted to the resistivity r of its mteril y R = rl A (reltionship etween resistnce nd resistivity) (25.10) If r is constnt, s is the cse for ohmic mterils, then so is R. The eqution V = IR (reltionship mong voltge, current, nd resistnce) (25.11) is often clled Ohm s lw, ut it is importnt to understnd tht the rel content of Ohm s lw is the direct proportionlity (for some mterils) of V to I or of J to E.

140 826 CHAPTER 25 Current, Resistnce, nd Electromotive Force Eqution (25.9) or (25.11) defines resistnce R for ny conductor, whether or not it oeys Ohm s lw, ut only when R is constnt cn we correctly cll this reltionship Ohm s lw A long fire hose offers sustntil resistnce to wter flow. To mke wter pss through the hose rpidly, the upstrem end of the hose must e t much higher pressure thn the end where the wter emerges. In n nlogous wy, there must e lrge potentil difference etween the ends of long wire in order to cuse sustntil electric current through the wire This resistor hs resistnce of 5.7 kæ with precision (tolernce) of 10%. econd digit First digit Multiplier Tolernce Tle 25.3 Color Codes for Resistors Vlue s Vlue s Color Digit Multiplier Blck 0 1 Brown 1 10 Red Ornge Yellow Green Blue Violet Gry White Interpreting Resistnce Eqution (25.10) shows tht the resistnce of wire or other conductor of uniform cross section is directly proportionl to its length nd inversely proportionl to its cross-sectionl re. It is lso proportionl to the resistivity of the mteril of which the conductor is mde. The flowing-fluid nlogy is gin useful. In nlogy to Eq. (25.10), nrrow wter hose offers more resistnce to flow thn ft one, nd long hose hs more resistnce thn short one (Fig. 25.8). We cn increse the resistnce to flow y stuffing the hose with cotton or snd; this corresponds to incresing the resistivity. The flow rte is pproximtely proportionl to the pressure difference etween the ends. Flow rte is nlogous to current, nd pressure difference is nlogous to potentil difference ( voltge ). Let s not stretch this nlogy too fr, though; the wter flow rte in pipe is usully not proportionl to its crosssectionl re (see ection 14.6). The I unit of resistnce is the ohm, equl to one volt per mpere 11 Æ= 1 V>A2. The kilohm 11 kæ =10 3 Æ2 nd the megohm 11 MÆ =10 6 Æ2 re lso in common use. A 100-m length of 12-guge copper wire, the size usully used in household wiring, hs resistnce t room temperture of out 0.5 Æ. A 100-W, 120-V light ul hs resistnce (t operting temperture) of 140 Æ. If the sme current I flows in oth the copper wire nd the light ul, the potentil difference V = IR is much greter cross the light ul, nd much more potentil energy is lost per chrge in the light ul. This lost energy is converted y the light ul filment into light nd het. You don t wnt your household wiring to glow white-hot, so its resistnce is kept low y using wire of low resistivity nd lrge cross-sectionl re. Becuse the resistivity of mteril vries with temperture, the resistnce of specific conductor lso vries with temperture. For temperture rnges tht re not too gret, this vrition is pproximtely liner reltionship, nlogous to Eq. (25.6): R1T2 = R T - T 0 24 (25.12) In this eqution, R1T2 is the resistnce t temperture T nd R 0 is the resistnce t temperture T 0, often tken to e 0 C or 20 C. The temperture coefficient of resistnce is the sme constnt tht ppers in Eq. (25.6) if the dimensions L nd A in Eq. (25.10) do not chnge pprecily with temperture; this is indeed the cse for most conducting mterils (see Prolem 25.67). Within the limits of vlidity of Eq. (25.12), the chnge in resistnce resulting from temperture chnge T - T 0 is given y R 0 1T - T 0 2. A circuit device mde to hve specific vlue of resistnce etween its ends is clled resistor. Resistors in the rnge 0.01 to 10 7 Æ cn e ought off the shelf. Individul resistors used in electronic circuitry re often cylindricl, few millimeters in dimeter nd length, with wires coming out of the ends. The resistnce my e mrked with stndrd code using three or four color nds ner one end (Fig. 25.9), ccording to the scheme shown in Tle The first two nds (strting with the nd nerest n end) re digits, nd the third is power-of-10 multiplier, s shown in Fig For exmple, greenvioletred mens 57 * 10 2 Æ, or 5.7 kæ. The fourth nd, if present, indictes the precision (tolernce) of the vlue; no nd mens 20%, silver nd 10%, nd gold nd 5%. Another importnt chrcteristic of resistor is the mximum power it cn dissipte without dmge. We ll return to this point in ection 25.5.

141 25.3 Resistnce Currentvoltge reltionships for two devices. Only for resistor tht oeys Ohm s lw s in () is current I proportionl to voltge V. () Ohmic resistor (e.g., typicl metl wire): At given temperture, current is proportionl to voltge. I O lope 5 1 R V () emiconductor diode: nonohmic resistor In the direction of negtive current nd voltge, little current flows. I O In the direction of positive current nd voltge, I increses nonlinerly with V. V For resistor tht oeys Ohm s lw, grph of current s function of potentil difference (voltge) is stright line (Fig ). The slope of the line is 1>R. If the sign of the potentil difference chnges, so does the sign of the current produced; in Fig this corresponds to interchnging the highernd lower-potentil ends of the conductor, so the electric field, current density, nd current ll reverse direction. In devices tht do not oey Ohm s lw, the reltionship of voltge to current my not e direct proportion, nd it my e different for the two directions of current. Figure shows the ehvior of semiconductor diode, device used to convert lternting current to direct current nd to perform wide vriety of logic functions in computer circuitry. For positive potentils V of the node (one of two terminls of the diode) with respect to the cthode (the other terminl), I increses exponentilly with incresing V; for negtive potentils the current is extremely smll. Thus positive V cuses current to flow in the positive direction, ut potentil difference of the other sign cuses little or no current. Hence diode cts like one-wy vlve in circuit. Exmple 25.2 Electric field, potentil difference, nd resistnce in wire The 18-guge copper wire of Exmple 25.1 hs cross-sectionl re of 8.20 * 10-7 m 2. It crries current of 1.67 A. Find () the electric-field mgnitude in the wire; () the potentil difference etween two points in the wire 50.0 m prt; (c) the resistnce of 50.0-m length of this wire. OLUTION IDENTIFY nd ET UP: We re given the cross-sectionl re A nd current I. Our trget vriles re the electric-field mgnitude E, potentil difference V, nd resistnce R. The current density is J = I>A. We find E from Eq. (25.5), E = rj (Tle 25.1 gives the resistivity r for copper). The potentil difference is then the product of E nd the length of the wire. We cn use either Eq. (25.10) or Eq. (25.11) to find R. EXECUTE: () From Tle 25.1, r 1.72 * 10-8 Æ # m. Hence, using Eq. (25.5), E = rj = ri A = V>m * 10-8 Æ # m A2 = 8.20 * 10-7 m 2 () The potentil difference is V = EL = V>m m2 = 1.75 V (c) From Eq. (25.10) the resistnce of 50.0 m of this wire is R = rl A = * 10-8 Æ # m m * 10-7 m 2 = 1.05 Æ Alterntively, we cn find R using Eq. (25.11): R = V I = 1.75 V 1.67 A = 1.05 Æ EVALUATE: We emphsize tht the resistnce of the wire is defined to e the rtio of voltge to current. If the wire is mde of nonohmic mteril, then R is different for different vlues of V ut is lwys given y R = V>I. Resistnce is lso lwys given y R = rl>a; if the mteril is nonohmic, r is not constnt ut depends on E (or, equivlently, on V = EL).

142 828 CHAPTER 25 Current, Resistnce, nd Electromotive Force Exmple 25.3 Temperture dependence of resistnce uppose the resistnce of copper wire is 1.05 Æ t 20 C. Find the resistnce t 0 C nd 100 C. OLUTION IDENTIFY nd ET UP: We re given the resistnce R 0 = 1.05 Æ t reference temperture T 0 = 20 C. We use Eq. (25.12) to find the resistnces t T = 0 C nd T = 100 C (our trget vriles), tking the temperture coefficient of resistivity from Tle EXECUTE: From Tle 25.2, = C 2-1 for copper. Then from Eq. (25.12), R = R T - T 0 24 = Æ C C - 20 C46 = 0.97 Æ t T = 0 C R = Æ C C - 20 C46 = 1.38 Æ t T = 100 C EVALUATE: The resistnce t 100 C is greter thn tht t 0 C y fctor of Æ2>10.97 Æ2 = 1.42: Rising the temperture of copper wire from 0 C to 100 C increses its resistnce y 42%. From Eq. (25.11), V = IR, this mens tht 42% more voltge is required to produce the sme current t 100 C thn t 0 C. Designers of electric circuits tht must operte over wide temperture rnge must tke this sustntil effect into ccount. Test Your Understnding of ection 25.3 uppose you increse the voltge cross the copper wire in Exmples 25.2 nd The incresed voltge cuses more current to flow, which mkes the temperture of the wire increse. (The sme thing hppens to the coils of n electric oven or toster when voltge is pplied to them. We ll explore this issue in more depth in ection 25.5.) If you doule the voltge cross the wire, the current in the wire increses. By wht fctor does it increse? (i) 2; (ii) greter thn 2; (iii) less thn Electromotive Force nd Circuits If n electric field is produced inside conductor tht is not prt of complete circuit, current flows for only very short time. () An electric field E 1 produced inside n isolted conductor cuses current. I I J J E 1 () The current cuses chrge to uild up t the ends. E 1 (c) After very short time E 2 hs the sme mgnitude s E 1 ; then the totl field is E totl 5 0 nd the current stops completely. E 2 E totl I I The chrge uildup produces n opposing field E 2, thus reducing the current. I 0 E 1 E 2 J 0 E totl 0 For conductor to hve stedy current, it must e prt of pth tht forms closed loop or complete circuit. Here s why. If you estlish n electric field E 1 inside n isolted conductor with resistivity r tht is not prt of complete circuit, current egins to flow with current density J E 1>r (Fig ). As result net positive chrge quickly ccumultes t one end of the conductor nd net negtive chrge ccumultes t the other end (Fig ). These chrges themselves produce n electric field E in the direction opposite to E 2 1, cusing the totl electric field nd hence the current to decrese. Within very smll frction of second, enough chrge uilds up on the conductor ends tht the totl electric field E E 1 E 2 0 inside the conductor. Then J 0 s well, nd the current stops ltogether (Fig c). o there cn e no stedy motion of chrge in such n incomplete circuit. To see how to mintin stedy current in complete circuit, we recll sic fct out electric potentil energy: If chrge q goes round complete circuit nd returns to its strting point, the potentil energy must e the sme t the end of the round trip s t the eginning. As descried in ection 25.3, there is lwys decrese in potentil energy when chrges move through n ordinry conducting mteril with resistnce. o there must e some prt of the circuit in which the potentil energy increses. The prolem is nlogous to n ornmentl wter fountin tht recycles its wter. The wter pours out of openings t the top, cscdes down over the terrces nd spouts (moving in the direction of decresing grvittionl potentil energy), nd collects in sin in the ottom. A pump then lifts it ck to the top (incresing the potentil energy) for nother trip. Without the pump, the wter would just fll to the ottom nd sty there. Electromotive Force In n electric circuit there must e device somewhere in the loop tht cts like the wter pump in wter fountin (Fig ). In this device chrge trvels

143 25.4 Electromotive Force nd Circuits 829 uphill, from lower to higher potentil energy, even though the electrosttic force is trying to push it from higher to lower potentil energy. The direction of current in such device is from lower to higher potentil, just the opposite of wht hppens in n ordinry conductor. The influence tht mkes current flow from lower to higher potentil is clled electromotive force (revited emf nd pronounced ee-em-eff ). This is poor term ecuse emf is not force ut n energy-per-unit-chrge quntity, like potentil. The I unit of emf is the sme s tht for potentil, the volt 11 V = 1 J>C2. A typicl flshlight ttery hs n emf of 1.5 V; this mens tht the ttery does 1.5 J of work on every coulom of chrge tht psses through it. We ll use the symol E ( script cpitl E) for emf. Every complete circuit with stedy current must include some device tht provides emf. uch device is clled source of emf. Btteries, electric genertors, solr cells, thermocouples, nd fuel cells re ll exmples of sources of emf. All such devices convert energy of some form (mechnicl, chemicl, therml, nd so on) into electric potentil energy nd trnsfer it into the circuit to which the device is connected. An idel source of emf mintins constnt potentil difference etween its terminls, independent of the current through it. We define electromotive force quntittively s the mgnitude of this potentil difference. As we will see, such n idel source is mythicl est, like the frictionless plne nd the mssless rope. We will discuss lter how rel-life sources of emf differ in their ehvior from this idelized model. Figure is schemtic digrm of n idel source of emf tht mintins potentil difference etween conductors nd, clled the terminls of the device. Terminl, mrked, is mintined t higher potentil thn terminl, mrked -. Associted with this potentil difference is n electric field E in the region round the terminls, oth inside nd outside the source. The electric field inside the device is directed from to, s shown. A chrge q within the source experiences n electric force F e qe. But the source lso provides n dditionl influence, which we represent s nonelectrosttic force F n. This force, operting inside the device, pushes chrge from to in n uphill direction ginst the electric force F e. Thus F mintins the potentil difference etween the terminls. If F n n were not present, chrge would flow etween the terminls until the potentil difference ws zero. The origin of the dditionl influence F n depends on the kind of source. In genertor it results from mgnetic-field forces on moving chrges. In ttery or fuel cell it is ssocited with diffusion processes nd vrying electrolyte concentrtions resulting from chemicl rections. In n electrosttic mchine such s Vn de Grff genertor (see Fig ), n ctul mechnicl force is pplied y moving elt or wheel. If positive chrge q is moved from to inside the source, the nonelectrosttic force F n does positive mount of work W n = qe on the chrge. This displcement is opposite to the electrosttic force F e, so the potentil energy ssocited with the chrge increses y n mount equl to qv, where V = V - V is the (positive) potentil of point with respect to point. For the idel source of emf tht we ve descried, F nd F e n re equl in mgnitude ut opposite in direction, so the totl work done on the chrge q is zero; there is n increse in potentil energy ut no chnge in the kinetic energy of the chrge. It s like lifting ook from the floor to high shelf t constnt speed. The increse in potentil energy is just equl to the nonelectrosttic work W n, so qe = qv, or V = E (idel source of emf) (25.13) Now let s mke complete circuit y connecting wire with resistnce R to the terminls of source (Fig ). The potentil difference etween terminls nd sets up n electric field within the wire; this cuses current to flow round the loop from towrd, from higher to lower potentil. Where the wire ends, equl mounts of positive nd negtive chrge persist on the inside nd outside Just s wter fountin requires pump, n electric circuit requires source of electromotive force to sustin stedy current chemtic digrm of source of emf in n open-circuit sitution. The electric-field force F e qe nd the nonelectrosttic force F n re shown for positive chrge q. V V 5 E V Idel emf source E F n q Terminl t higher potentil F e 5 qe Nonelectrosttic force tending to move chrge to higher potentil Force due to electric field Terminl t lower potentil When the emf source is not prt of closed circuit, F n 5 F e nd there is no net motion of chrge etween the terminls. PhET: Bttery Voltge PhET: ignl Circuit ActivPhysics 12.1: DC eries Circuits (Qulittive)

144 830 CHAPTER 25 Current, Resistnce, nd Electromotive Force chemtic digrm of n idel source of emf in complete circuit. The electric-field force F nd the nonelectrosttic force F e qe n re shown for positive chrge q. The current is in the direction from to in the externl circuit nd from to within the source. Potentil cross terminls cretes electric field in circuit, cusing chrges to move. V V 5 E V Idel emf source E F n F e When rel (s opposed I to idel) emf source is connected to circuit, V nd thus F e fll, so tht F n. F e nd F n does work on the chrges. Appliction Dnger: Electric Ry! Electric rys deliver electric shocks to stun their prey nd to discourge predtors. (In ncient Rome, physicins prcticed primitive form of electroconvulsive therpy y plcing electric rys on their ptients to cure hedches nd gout.) The shocks re produced y specilized flttened cells clled electroplques. uch cell moves ions cross memrnes to produce n emf of out 0.05 V. Thousnds of electroplques re stcked on top of ech other, so their emfs dd to totl of s much s 200 V. These stcks mke up more thn hlf of n electric ry s ody mss. A ry cn use these to deliver n impressive current of up to 30 A for few milliseconds. I E E E I of the end. These chrges exert the forces tht cuse the current to follow the ends in the wire. From Eq. (25.11) the potentil difference etween the ends of the wire in Fig is given y V = IR. Comining with Eq. (25.13), we hve E = V = IR (idel source of emf) (25.14) Tht is, when positive chrge q flows round the circuit, the potentil rise E s it psses through the idel source is numericlly equl to the potentil drop V = IR s it psses through the reminder of the circuit. Once E nd R re known, this reltionship determines the current in the circuit. CAUTION Current is not used up in circuit It s common misconception tht in closed circuit, current is something tht squirts out of the positive terminl of ttery nd is consumed or used up y the time it reches the negtive terminl. In fct the current is the sme t every point in simple loop circuit like tht in Fig , even if the thickness of the wires is different t different points in the circuit. This hppens ecuse chrge is conserved (tht is, it cn e neither creted nor destroyed) nd ecuse chrge cnnot ccumulte in the circuit devices we hve descried. If chrge did ccumulte, the potentil differences would chnge with time. It s like the flow of wter in n ornmentl fountin; wter flows out of the top of the fountin t the sme rte t which it reches the ottom, no mtter wht the dimensions of the fountin. None of the wter is used up long the wy! Internl Resistnce Rel sources of emf in circuit don t ehve in exctly the wy we hve descried; the potentil difference cross rel source in circuit is not equl to the emf s in Eq. (25.14). The reson is tht chrge moving through the mteril of ny rel source encounters resistnce. We cll this the internl resistnce of the source, denoted y r. If this resistnce ehves ccording to Ohm s lw, r is constnt nd independent of the current I. As the current moves through r, it experiences n ssocited drop in potentil equl to Ir. Thus, when current is flowing through source from the negtive terminl to the positive terminl, the potentil difference etween the terminls is V V = E - Ir (terminl voltge, source with internl resistnce) (25.15) The potentil V, clled the terminl voltge, is less thn the emf E ecuse of the term Ir representing the potentil drop cross the internl resistnce r. Expressed nother wy, the increse in potentil energy qv s chrge q moves from to within the source is now less thn the work qe done y the nonelectrosttic force F n, since some potentil energy is lost in trversing the internl resistnce. A 1.5-V ttery hs n emf of 1.5 V, ut the terminl voltge V of the ttery is equl to 1.5 V only if no current is flowing through it so tht I = 0 in Eq. (25.15). If the ttery is prt of complete circuit through which current is flowing, the terminl voltge will e less thn 1.5 V. For rel source of emf, the terminl voltge equls the emf only if no current is flowing through the source (Fig ). Thus we cn descrie the ehvior of source in terms of two properties: n emf E, which supplies constnt potentil difference independent of current, in series with n internl resistnce r. The current in the externl circuit connected to the source terminls nd is still determined y V = IR. Comining this with Eq. (25.15), we find? E - Ir = IR E (current, source with or I = (25.16) R r internl resistnce)

145 25.4 Electromotive Force nd Circuits 831 Tht is, the current equls the source emf divided y the totl circuit resistnce 1R r2. CAUTION A ttery is not current source You might hve thought tht ttery or other source of emf lwys produces the sme current, no mtter wht circuit it s used in. Eqution (25.16) shows tht this isn t so! The greter the resistnce R of the externl circuit, the less current the source will produce. It s nlogous to pushing n oject through thick, viscous liquid such s oil or molsses; if you exert certin stedy push (emf), you cn move smll oject t high speed (smll R, lrge I) or lrge oject t low speed (lrge R, smll I) The emf of this ttery tht is, the terminl voltge when it s not connected to nything is 12 V. But ecuse the ttery hs internl resistnce, the terminl voltge of the ttery is less thn 12 V when it is supplying current to light ul. ymols for Circuit Digrms An importnt prt of nlyzing ny electric circuit is drwing schemtic circuit digrm. Tle 25.4 shows the usul symols used in circuit digrms. We will use these symols extensively in this chpter nd the next. We usully ssume tht the wires tht connect the vrious elements of the circuit hve negligile resistnce; from Eq. (25.11), V = IR, the potentil difference etween the ends of such wire is zero. Tle 25.4 includes two meters tht re used to mesure the properties of circuits. Idelized meters do not distur the circuit in which they re connected. A voltmeter, introduced in ection 23.2, mesures the potentil difference etween its terminls; n idelized voltmeter hs infinitely lrge resistnce nd mesures potentil difference without hving ny current diverted through it. An mmeter mesures the current pssing through it; n idelized mmeter hs zero resistnce nd hs no potentil difference etween its terminls. Becuse meters ct s prt of the circuit in which they re connected, these properties re importnt to rememer. Tle 25.4 ymols for Circuit Digrms R E Conductor with negligile resistnce Resistor ource of emf (longer verticl line lwys represents the positive terminl, usully the terminl with higher potentil) or E E ource of emf with internl resistnce r ( r cn e plced on either side) V Voltmeter (mesures potentil difference etween its terminls) A Ammeter (mesures current through it) Conceptul Exmple 25.4 A source in n open circuit Figure shows source ( ttery) with emf E = 12 V nd internl resistnce r 2 Æ. (For comprison, the internl resistnce of commercil 12-V led storge ttery is only few thousndths of n ohm.) The wires to the left of nd to the right of the mmeter A re not connected to nything. Determine the respective redings V nd I of the idelized voltmeter V nd the idelized mmeter A A source of emf in n open circuit. V V A r 5 2 V, E 5 12 V Continued

146 832 CHAPTER 25 Current, Resistnce, nd Electromotive Force OLUTION There is zero current ecuse there is no complete circuit. (Our idelized voltmeter hs n infinitely lrge resistnce, so no current flows through it.) Hence the mmeter reds I = 0. Becuse there is no current through the ttery, there is no potentil difference cross its internl resistnce. From Eq. (25.15) with I = 0, the potentil difference V cross the ttery terminls is equl to the emf. o the voltmeter reds V = E = 12 V. The terminl voltge of rel, nonidel source equls the emf only if there is no current flowing through the source, s in this exmple. Exmple 25.5 A source in complete circuit We dd 4-Æ resistor to the ttery in Conceptul Exmple 25.4, forming complete circuit (Fig ). Wht re the voltmeter EXECUTE: The idel mmeter hs zero resistnce, so the totl resistnce externl to the source is R = 4 Æ. From Eq. (25.16), the nd mmeter redings nd I now? current through the circuit is then V OLUTION IDENTIFY nd ET UP: Our trget vriles re the current I through the circuit nd the potentil difference V. We first find I using Eq. (25.16). To find V, we cn use either Eq. (25.11) or Eq. (25.15) A source of emf in complete circuit. I V 5 V V r 5 2 V, E 5 12 V R 5 4 V A I E I = R r = 12 V 4 Æ2Æ = 2A Our idelized conducting wires nd the idelized mmeter hve zero resistnce, so there is no potentil difference etween points nd or etween points nd ; tht is, V = V. We find V y considering nd s the terminls of the resistor: From Ohm s lw, Eq. (25.11), we then hve V = IR = 12 A214 Æ2 = 8V Alterntively, we cn consider nd s the terminls of the source. Then, from Eq. (25.15), V = E - Ir = 12 V - 12 A212 Æ2 = 8V Either wy, we see tht the voltmeter reding is 8 V. EVALUATE: With current flowing through the source, the terminl voltge V is less thn the emf E. The smller the internl resistnce r, the less the difference etween nd E. V Conceptul Exmple 25.6 Using voltmeters nd mmeters We move the voltmeter nd mmeter in Exmple 25.5 to different positions in the circuit. Wht re the redings of the idel voltmeter nd mmeter in the situtions shown in () Fig nd () Fig ? OLUTION () The voltmeter now mesures the potentil difference etween points nd. As in Exmple 25.5, V = V, so the voltmeter reds the sme s in Exmple 25.5: V = 8V Different plcements of voltmeter nd n mmeter in complete circuit. () I A r 5 2 V, E 5 12 V R 5 4 V V V I () A r 5 2 V, E 5 12 V R 5 4 V V V CAUTION Current in simple loop As chrges move through resistor, there is decrese in electric potentil energy, ut there is no chnge in the current. The current in simple loop is the sme t every point; it is not used up s it moves through resistor. Hence the mmeter in Fig ( downstrem of the 4-Æ resistor) nd the mmeter in Fig ( upstrem of the resistor) oth red I = 2A. () There is no current through the idel voltmeter ecuse it hs infinitely lrge resistnce. ince the voltmeter is now prt of the circuit, there is no current t ll in the circuit, nd the mmeter reds I = 0. The voltmeter mesures the potentil difference V etween points nd. ince I = 0, the potentil difference cross the resistor is V = IR = 0, nd the potentil difference etween the ends nd of the idelized mmeter is lso zero. o V is equl to V, the terminl voltge of the source. As in Conceptul Exmple 25.4, there is no current, so the terminl voltge equls the emf, nd the voltmeter reding is V = E = 12 V. This exmple shows tht mmeters nd voltmeters re circuit elements, too. Moving the voltmeter from the position in Fig to tht in Fig mkes lrge chnges in the current nd potentil differences in the circuit. If you wnt to mesure the potentil difference etween two points in circuit without disturing the circuit, use voltmeter s in Fig or 25.18, not s in Fig

147 25.4 Electromotive Force nd Circuits 833 Exmple 25.7 A source with short circuit In the circuit of Exmple 25.5 we replce the 4-Æ resistor with EXECUTE: We must hve V = IR = I102 = 0, no mtter wht zero-resistnce conductor. Wht re the meter redings now? the current. We cn therefore find the current I from Eq. (25.15): OLUTION IDENTIFY nd ET UP: Figure shows the new circuit. Our trget vriles re gin I nd V. There is now zero-resistnce pth etween points nd, through the lower loop, so the potentil difference etween these points must e zero Our sketch for this prolem. V = E - Ir = 0 I = E r = 12 V 2 Æ = 6A EVALUATE: The current hs different vlue thn in Exmple 25.5, even though the sme ttery is used; the current depends on oth the internl resistnce r nd the resistnce of the externl circuit. The sitution here is clled short circuit. The externl-circuit resistnce is zero, ecuse terminls of the ttery re connected directly to ech other. The short-circuit current is equl to the emf E divided y the internl resistnce r. Wrning: hort circuits cn e dngerous! An utomoile ttery or household power line hs very smll internl resistnce (much less thn in these exmples), nd the short-circuit current cn e gret enough to melt smll wire or cuse storge ttery to explode. Potentil Chnges Around Circuit The net chnge in potentil energy for chrge q mking round trip round complete circuit must e zero. Hence the net chnge in potentil round the circuit must lso e zero; in other words, the lgeric sum of the potentil differences nd emfs round the loop is zero. We cn see this y rewriting Eq. (25.16) in the form E - Ir - IR = 0 A potentil gin of E is ssocited with the emf, nd potentil drops of Ir nd IR re ssocited with the internl resistnce of the source nd the externl circuit, respectively. Figure is grph showing how the potentil vries s we go round the complete circuit of Fig The horizontl xis doesn t necessrily represent ctul distnces, ut rther vrious points in the loop. If we tke the potentil to e zero t the negtive terminl of the ttery, then we hve rise E nd drop Ir in the ttery nd n dditionl drop IR in the externl resistor, nd s we finish our trip round the loop, the potentil is ck where it strted. In this section we hve considered only situtions in which the resistnces re ohmic. If the circuit includes nonliner device such s diode (see Fig ), Eq. (25.16) is still vlid ut cnnot e solved lgericlly ecuse R is not constnt. In such sitution, the current I cn e found y using numericl techniques. Finlly, we remrk tht Eq. (25.15) is not lwys n dequte representtion of the ehvior of source. The emf my not e constnt, nd wht we hve descried s n internl resistnce my ctully e more complex voltgecurrent reltionship tht doesn t oey Ohm s lw. Nevertheless, the concept of internl resistnce frequently provides n dequte description of tteries, genertors, nd other energy converters. The principl difference etween fresh flshlight ttery nd n old one is not in the emf, which decreses only slightly with use, ut in the internl resistnce, which my increse from less thn n ohm when the ttery is fresh to s much s 1000 Æ or more fter long use. imilrly, cr ttery cn deliver less current to the strter motor on cold morning thn when the ttery is wrm, not ecuse the emf is pprecily less ut ecuse the internl resistnce increses with decresing temperture Potentil rises nd drops in circuit. 12 V 8 V 12 V O V 2 V E 5 12 V 2 A 2 A 2 A 2 A 4 V Ir 5 4 V IR 5 8 V

148 834 CHAPTER 25 Current, Resistnce, nd Electromotive Force Test Your Understnding of ection 25.4 Rnk the following circuits in order from highest to lowest current. (i) 1.4-Æ resistor connected to 1.5-V ttery tht hs n internl resistnce of 0.10 Æ; (ii) 1.8-Æ resistor connected to 4.0-V ttery tht hs terminl voltge of 3.6 V ut n unknown internl resistnce; (iii) n unknown resistor connected to 12.0-V ttery tht hs n internl resistnce of 0.20 Æ nd terminl voltge of 11.0 V Energy nd Power in Electric Circuits The power input to the circuit element etween nd is P = 1V - V 2I = V I. I V Circuit element V PhET: Bttery-Resistor Circuit PhET: Circuit Construction Kit (ACDC) PhET: Circuit Construction Kit (DC Only) PhET: Ohm s Lw I Let s now look t some energy nd power reltionships in electric circuits. The ox in Fig represents circuit element with potentil difference V - V = V etween its terminls nd current I pssing through it in the direction from towrd. This element might e resistor, ttery, or something else; the detils don t mtter. As chrge psses through the circuit element, the electric field does work on the chrge. In source of emf, dditionl work is done y the force F n tht we mentioned in ection As n mount of chrge q psses through the circuit element, there is chnge in potentil energy equl to qv. For exmple, if q 7 0 nd V = V - V is positive, potentil energy decreses s the chrge flls from potentil V to lower potentil V. The moving chrges don t gin kinetic energy, ecuse the current (the rte of chrge flow) out of the circuit element must e the sme s the current into the element. Insted, the quntity qv represents energy trnsferred into the circuit element. This sitution occurs in the coils of toster or electric oven, in which electricl energy is converted to therml energy. If the potentil t is lower thn t, then V is negtive nd there is net trnsfer of energy out of the circuit element. The element then cts s source, delivering electricl energy into the circuit to which it is ttched. This is the usul sitution for ttery, which converts chemicl energy into electricl energy nd delivers it to the externl circuit. Thus qv cn denote either quntity of energy delivered to circuit element or quntity of energy extrcted from tht element. In electric circuits we re most often interested in the rte t which energy is either delivered to or extrcted from circuit element. If the current through the element is I, then in time intervl dt n mount of chrge dq = Idt psses through the element. The potentil energy chnge for this mount of chrge is V dq = V Idt. Dividing this expression y dt, we otin the rte t which energy is trnsferred either into or out of the circuit element. The time rte of energy trnsfer is power, denoted y P, so we write P = V I (rte t which energy is delivered to or extrcted from circuit element) (25.17) The unit of V is one volt, or one joule per coulom, nd the unit of I is one mpere, or one coulom per second. Hence the unit of P = V I is one wtt, s it should e: 11 J>C211 C>s2 = 1 J>s = 1 W Let s consider few specil cses. Power Input to Pure Resistnce If the circuit element in Fig is resistor, the potentil difference is V = IR. From Eq. (25.17) the electricl power delivered to the resistor y the circuit is P = V I = I 2 R = V 2 R (power delivered to resistor) (25.18)

149 25.5 Energy nd Power in Electric Circuits 835 In this cse the potentil t (where the current enters the resistor) is lwys higher thn tht t (where the current exits). Current enters the higher-potentil terminl of the device, nd Eq. (25.18) represents the rte of trnsfer of electric potentil energy into the circuit element. Wht ecomes of this energy? The moving chrges collide with toms in the resistor nd trnsfer some of their energy to these toms, incresing the internl energy of the mteril. Either the temperture of the resistor increses or there is flow of het out of it, or oth. In ny of these cses we sy tht energy is dissipted in the resistor t rte I 2 R. Every resistor hs power rting, the mximum power the device cn dissipte without ecoming overheted nd dmged. ome devices, such s electric heters, re designed to get hot nd trnsfer het to their surroundings. But if the power rting is exceeded, even such device my melt or even explode. Power Output of ource The upper rectngle in Fig represents source with emf E nd internl resistnce r, connected y idel (resistnceless) conductors to n externl circuit represented y the lower ox. This could descrie cr ttery connected to one of the cr s hedlights (Fig ). Point is t higher potentil thn point, so V 7 V nd V is positive. Note tht the current I is leving the source t the higher-potentil terminl (rther thn entering there). Energy is eing delivered to the externl circuit, t rte given y Eq. (25.17): For source tht cn e descried y n emf E nd n internl resistnce r, we my use Eq. (25.15): Multiplying this eqution y I, we find (25.19) Wht do the terms EI nd I 2 r men? In ection 25.4 we defined the emf E s the work per unit chrge performed on the chrges y the nonelectrosttic force s the chrges re pushed uphill from to in the source. In time dt, chrge dq = Idtflows through the source; the work done on it y this nonelectrosttic force is E dq = EIdt. Thus EI is the rte t which work is done on the circulting chrges y whtever gency cuses the nonelectrosttic force in the source. This term represents the rte of conversion of nonelectricl energy to electricl energy within the source. The term I 2 r is the rte t which electricl energy is dissipted in the internl resistnce of the source. The difference EI - I 2 r is the net electricl power output of the source tht is, the rte t which the source delivers electricl energy to the reminder of the circuit. Power Input to ource uppose tht the lower rectngle in Fig is itself source, with n emf lrger thn tht of the upper source nd with its emf opposite to tht of the upper source. Figure shows prcticl exmple, n utomoile ttery (the upper circuit element) eing chrged y the cr s lterntor (the lower element). The current I in the circuit is then opposite to tht shown in Fig ; the lower source is pushing current ckwrd through the upper source. Becuse of this reversl of current, insted of Eq. (25.15) we hve for the upper source nd insted of Eq. (25.19), we hve P = V I V = E - Ir P = V I = EI - I 2 r V = E Ir P = V I = EI I 2 r (25.20) Energy conversion in simple circuit. () Digrmmtic circuit The emf source converts nonelectricl to electricl energy t rte EI. Its internl resistnce dissiptes energy t rte I 2 r. The difference EI 2 I 2 r is its power output. E, r I Hedlight F e F n emf source with internl resistnce r Externl circuit () A rel circuit of the type shown in () I v Bttery When two sources re connected in simple loop circuit, the source with the lrger emf delivers energy to the other source. I q Bttery (smll emf) v r F n F e q Alterntor (lrge emf) I I I

150 836 CHAPTER 25 Current, Resistnce, nd Electromotive Force Work is eing done on, rther thn y, the gent tht cuses the nonelectrosttic force in the upper source. There is conversion of electricl energy into nonelectricl energy in the upper source t rte EI. The term I 2 r in Eq. (25.20) is gin the rte of dissiption of energy in the internl resistnce of the upper source, nd the sum EI I 2 r is the totl electricl power input to the upper source. This is wht hppens when rechrgele ttery ( storge ttery) is connected to chrger. The chrger supplies electricl energy to the ttery; prt of it is converted to chemicl energy, to e reconverted lter, nd the reminder is dissipted (wsted) in the ttery s internl resistnce, wrming the ttery nd cusing het flow out of it. If you hve power tool or lptop computer with rechrgele ttery, you my hve noticed tht it gets wrm while it is chrging. Prolem-olving trtegy 25.1 Power nd Energy in Circuits IDENTIFY the relevnt concepts: The ides of electric power input nd output cn e pplied to ny electric circuit. Mny prolems will sk you to explicitly consider power or energy. ET UP the prolem using the following steps: 1. Mke drwing of the circuit. 2. Identify the circuit elements, including sources of emf nd resistors. We will introduce other circuit elements lter, including cpcitors (Chpter 26) nd inductors (Chpter 30). 3. Identify the trget vriles. Typiclly they will e the power input or output for ech circuit element, or the totl mount of energy put into or tken out of circuit element in given time. EXECUTE the solution s follows: 1. A source of emf E delivers power EI into circuit when current I flows through the source in the direction from - to. (For exmple, energy is converted from chemicl energy in ttery, or from mechnicl energy in genertor.) In this cse there is positive power output to the circuit or, equivlently, negtive power input to the source. 2. A source of emf tkes power EI from circuit when current psses through the source from to -. (This occurs in chrging storge ttery, when electricl energy is converted to chemicl energy.) In this cse there is negtive power output to the circuit or, equivlently, positive power input to the source. 3. There is lwys positive power input to resistor through which current flows, irrespective of the direction of current flow. This process removes energy from the circuit, converting it to het t the rte VI = I 2 R = V 2 >R, where V is the potentil difference cross the resistor. 4. Just s in item 3, there lwys is positive power input to the internl resistnce r of source through which current flows, irrespective of the direction of current flow. This process likewise removes energy from the circuit, converting it into het t the rte I 2 r. 5. If the power into or out of circuit element is constnt, the energy delivered to or extrcted from tht element is the product of power nd elpsed time. (In Chpter 26 we will encounter situtions in which the power is not constnt. In such cses, clculting the totl energy requires n integrl over the relevnt time intervl.) EVALUATE your nswer: Check your results; in prticulr, check tht energy is conserved. This conservtion cn e expressed in either of two forms: net power input = net power output or the lgeric sum of the power inputs to the circuit elements is zero. Exmple 25.8 Power input nd output in complete circuit For the circuit tht we nlyzed in Exmple 25.5, find the rtes of energy conversion (chemicl to electricl) nd energy dissiption in the ttery, the rte of energy dissiption in the 4-Æ resistor, nd the ttery s net power output. OLUTION IDENTIFY nd ET UP: Figure shows the circuit, gives vlues of quntities known from Exmple 25.5, nd indictes how we find the trget vriles. We use Eq. (25.19) to find the ttery s net power output, the rte of chemicl-to-electricl energy conversion, nd the rte of energy dissiption in the ttery s internl resistnce. We use Eq. (25.18) to find the power delivered to (nd dissipted in) the 4-Æ resistor. EXECUTE: From the first term in Eq. (25.19), the rte of energy conversion in the ttery is EI = 112 V212 A2 = 24 W Our sketch for this prolem. From the second term in Eq. (25.19), the rte of dissiption of energy in the ttery is I 2 r = 12 A Æ2 = 8W

151 25.5 Energy nd Power in Electric Circuits 837 The net electricl power output of the ttery is the difference etween these: EI - I 2 r = 16 W. From Eq. (25.18), the electricl power input to, nd the equl rte of dissiption of electricl energy in, the 4-Æ resistor re V I = 18 V212 A2 = 16 W nd I 2 R = 12 A Æ2 = 16 W EVALUATE: The rte V I t which energy is supplied to the 4-Æ resistor equls the rte I 2 R t which energy is dissipted there. This is lso equl to the ttery s net power output: P = V I = 18 V212 A2 = 16 W. In summry, the rte t which the source of emf supplies energy is EI = 24 W, of which I 2 r = 8Wis dissipted in the ttery s internl resistor nd I 2 R = 16 W is dissipted in the externl resistor. Exmple 25.9 Incresing the resistnce uppose we replce the externl 4-Æ resistor in Fig with n The greter resistnce cuses the current to decrese. The potentil 8-Æ resistor. Hiow does this ffect the electricl power dissipted difference cross the resistor is in this resistor? V = IR = 11.2 A218 Æ2 = 9.6 V OLUTION IDENTIFY nd ET UP: Our trget vrile is the power dissipted in the resistor to which the ttery is connected. The sitution is the sme s in Exmple 25.8, ut with higher externl resistnce R. EXECUTE: According to Eq. (25.18), the power dissipted in the resistor is P = I 2 R. You might conclude tht mking the resistnce R twice s gret s in Exmple 25.8 should lso mke the power twice s gret, or 2116 W2 = 32 W. If insted you used the formul P = V 2 >R, you might conclude tht the power should e one-hlf s gret s in the preceding exmple, or 116 W2>2 = 8W. Which nswer is correct? In fct, oth of these nswers re incorrect. The first is wrong ecuse chnging the resistnce R lso chnges the current in the circuit (rememer, source of emf does not generte the sme current in ll situtions). The second nswer is wrong ecuse the potentil difference V cross the resistor chnges when the current chnges. To get the correct nswer, we first find the current just s we did in Exmple 25.5: I = E R r = 12 V 8 Æ2 Æ = 1.2 A which is greter thn tht with the 4-Æ resistor. We cn then find the power dissipted in the resistor in either of two wys: P = I 2 R = 11.2 A Æ2 = 12 W or P = V V2 = R 8 Æ = 12 W EVALUATE: Incresing the resistnce R cuses reduction in the power input to the resistor. In the expression P = I 2 R the decrese in current is more importnt thn the increse in resistnce; in the expression P = V 2 >R the increse in resistnce is more importnt thn the increse in V. This sme principle pplies to ordinry light uls; 50-W light ul hs greter resistnce thn does 100-W light ul. Cn you show tht replcing the 4-Æ resistor with n 8-Æ resistor decreses oth the rte of energy conversion (chemicl to electricl) in the ttery nd the rte of energy dissiption in the ttery? Exmple Power in short circuit For the short-circuit sitution of Exmple 25.7, find the rtes of energy conversion nd energy dissiption in the ttery nd the net power output of the ttery. OLUTION IDENTIFY nd ET UP: Our trget vriles re gin the power inputs nd outputs ssocited with the ttery. Figure shows Our sketch for this prolem. the circuit. This is the sme sitution s in Exmple 25.8, ut now the externl resistnce R is zero. EXECUTE: We found in Exmple 25.7 tht the current in this sitution is I = 6A. From Eq. (25.19), the rte of energy conversion (chemicl to electricl) in the ttery is then EI = 112 V216 A2 = 72 W nd the rte of dissiption of energy in the ttery is I 2 r = 16 A Æ2 = 72 W The net power output of the source is EI - I 2 r = 0. We get this sme result from the expression P = V I, ecuse the terminl voltge V of the source is zero. EVALUATE: With idel wires nd n idel mmeter, so tht R = 0, ll of the converted energy from the source is dissipted within the source. This is why short-circuited ttery is quickly ruined nd my explode.

152 838 CHAPTER 25 Current, Resistnce, nd Electromotive Force Test Your Understnding of ection 25.5 Rnk the following circuits in order from highest to lowest vlues of the net power output of the ttery. (i) 1.4-Æ resistor connected to 1.5-V ttery tht hs n internl resistnce of 0.10 Æ; (ii) 1.8-Æ resistor connected to 4.0-V ttery tht hs terminl voltge of 3.6 V ut n unknown internl resistnce; (iii) n unknown resistor connected to 12.0-V ttery tht hs n internl resistnce of 0.20 Æ nd terminl voltge of 11.0 V Theory of Metllic Conduction Rndom motions of n electron in metllic crystl () with zero electric field nd () with n electric field tht cuses drift. The curvtures of the pths re gretly exggerted. () () Collision with crystl With E field: rndom motion plus drift END TART TART E Without E field: rndom motion E E Net displcement END We cn gin dditionl insight into electricl conduction y looking t the microscopic origin of conductivity. We ll consider very simple model tht trets the electrons s clssicl prticles nd ignores their quntum-mechnicl ehvior in solids. Using this model, we ll derive n expression for the resistivity of metl. Even though this model is not entirely correct, it will still help you to develop n intuitive ide of the microscopic sis of conduction. In the simplest microscopic model of conduction in metl, ech tom in the metllic crystl gives up one or more of its outer electrons. These electrons re then free to move through the crystl, colliding t intervls with the sttionry positive ions. The motion of the electrons is nlogous to the motion of molecules of gs moving through porous ed of snd. If there is no electric field, the electrons move in stright lines etween collisions, the directions of their velocities re rndom, nd on verge they never get nywhere (Fig ). But if n electric field is present, the pths curve slightly ecuse of the ccelertion cused y electric-field forces. Figure shows few pths of n electron in n electric field directed from right to left. As we mentioned in ection 25.1, the verge speed of rndom motion is of the order of 10 6 m>s, while the verge drift speed is much slower, of the order of 10-4 m>s. The verge time etween collisions is clled the men free time, denoted y t. Figure shows mechnicl nlog of this electron motion. We would like to derive from this model n expression for the resistivity r of mteril, defined y Eq. (25.5): r = E J (25.21) where E nd J re the mgnitudes of electric field nd current density, respectively. The current density J is in turn given y Eq. (25.4): J nqvd (25.22) PhET: Conductivity The motion of ll rolling down n inclined plne nd ouncing off pegs in its pth is nlogous to the motion of n electron in metllic conductor with n electric field present. where n is the numer of free electrons per unit volume, q = -e is the chrge of ech, nd v d is their verge drift velocity. We need to relte the drift velocity v to the electric field E d. The vlue of v d is determined y stedy-stte condition in which, on verge, the velocity gins of the chrges due to the force of the E field re just lnced y the velocity losses due to collisions. To clrify this process, let s imgine turning on the two effects one t time. uppose tht efore time t = 0 there is no field. The electron motion is then completely rndom. A typicl electron hs velocity v 0 t time t = 0, nd the vlue of v 0 verged over mny electrons (tht is, the initil velocity of n verge electron) is zero: 1v 02 v 0. Then t time t = 0 we turn on constnt electric field E. The field exerts force F qe on ech chrge, nd this cuses n ccelertion in the direction of the force, given y F m qe m where m is the electron mss. Every electron hs this ccelertion.

153 25.6 Theory of Metllic Conduction 839 We wit for time t, the verge time etween collisions, nd then turn on the collisions. An electron tht hs velocity v 0 t time t = 0 hs velocity t time t = t equl to v v v v 0 t The velocity of n verge electron t this time is the sum of the verges of the two terms on the right. As we hve pointed out, the initil velocity v 0 is zero for n verge electron, so v v t qt (25.23) m E After time t = t, the tendency of the collisions to decrese the velocity of n verge electron (y mens of rndomizing collisions) just lnces the tendency of the E field to increse this velocity. Thus the velocity of n verge electron, given y Eq. (25.23), is mintined over time nd is equl to the drift velocity v d : v d qt m E Now we sustitute this eqution for the drift velocity v into Eq. (25.22): J nqvd nq 2 t m Compring this with Eq. (25.21), which we cn rewrite s J E>r, nd sustituting q = -e for n electron, we see tht the resistivity r is given y E d r = m ne 2 t (25.24) If n nd t re independent of E, then the resistivity is independent of nd the conducting mteril oeys Ohm s lw. Turning the interctions on one t time my seem rtificil. But the derivtion would come out the sme if ech electron hd its own clock nd the t = 0 times were different for different electrons. If t is the verge time etween collisions, then v d is still the verge electron drift velocity, even though the motions of the vrious electrons ren t ctully correlted in the wy we postulted. Wht out the temperture dependence of resistivity? In perfect crystl with no toms out of plce, correct quntum-mechnicl nlysis would let the free electrons move through the crystl with no collisions t ll. But the toms virte out their equilirium positions. As the temperture increses, the mplitudes of these virtions increse, collisions ecome more frequent, nd the men free time t decreses. o this theory predicts tht the resistivity of metl increses with temperture. In superconductor, roughly speking, there re no inelstic collisions, t is infinite, nd the resistivity r is zero. In pure semiconductor such s silicon or germnium, the numer of chrge crriers per unit volume, n, is not constnt ut increses very rpidly with incresing temperture. This increse in n fr outweighs the decrese in the men free time, nd in semiconductor the resistivity lwys decreses rpidly with incresing temperture. At low tempertures, n is very smll, nd the resistivity ecomes so lrge tht the mteril cn e considered n insultor. Electrons gin energy etween collisions through the work done on them y the electric field. During collisions they trnsfer some of this energy to the toms of the mteril of the conductor. This leds to n increse in the mteril s internl energy nd temperture; tht s why wires crrying current get wrm. If the electric field in the mteril is lrge enough, n electron cn gin enough energy etween collisions to knock off electrons tht re normlly ound to toms in the mteril. These cn then knock off more electrons, nd so on, leding to n vlnche of current. This is the sis of dielectric rekdown in insultors (see ection 24.4). E

154 840 CHAPTER 25 Current, Resistnce, nd Electromotive Force Exmple Men free time in copper Clculte the men free time etween collisions in copper t room temperture. OLUTION IDENTIFY nd ET UP: We cn otin n expression for men free time t in terms of n, r, e, nd m y rerrnging Eq. (25.24). From Exmple 25.1 nd Tle 25.1, for copper n = 8.5 * 10 nd r = 1.72 * 10-8 Æ # 28 m -3 m. In ddition, e = 1.60 * C nd m = 9.11 * kg for electrons. EXECUTE: From Eq. (25.24), we get t = m ne 2 r 9.11 * kg = 18.5 * m * C * 10-8 Æ # m2 = 2.4 * s EVALUATE: The men free time is the verge time etween collisions for given electron. Tking the reciprocl of this time, we find tht ech electron verges 1>t = 4.2 * collisions per second! Test Your Understnding of ection 25.6 Which of the following fctors will, if incresed, mke it more difficult to produce certin mount of current in conductor? (There my e more thn one correct nswer.) (i) the mss of the moving chrged prticles in the conductor; (ii) the numer of moving chrged prticles per cuic meter; (iii) the mount of chrge on ech moving prticle; (iv) the verge time etween collisions for typicl moving chrged prticle.

155 CHAPTER 25 UMMARY Current nd current density: Current is the mount of chrge flowing through specified re, per unit time. The I unit of current is the mpere 11 A = 1 C>s2. The current I through n re A depends on the concentrtion n nd chrge q of the chrge crriers, s well s on the mgnitude of their drift velocity v d. The current density is current per unit cross-sectionl re. Current is usully descried in terms of flow of positive chrge, even when the chrges re ctully negtive or of oth signs. (ee Exmple 25.1.) I = dq dt J nqvd = nƒqƒv d A (25.2) (25.4) v d v d I v d v d v d v d E Resistivity: The resistivity r of mteril is the rtio of the mgnitudes of electric field nd current density. Good conductors hve smll resistivity; good insultors hve lrge resistivity. Ohm s lw, oeyed pproximtely y mny mterils, sttes tht r is constnt independent of the vlue of E. Resistivity usully increses with temperture; for smll temperture chnges this vrition is represented pproximtely y Eq. (25.6), where is the temperture coefficient of resistivity. r = E J r1t2 = r T - T 0 24 (25.5) (25.6) r 0 r O T 0 Metl: r increses with incresing T. lope 5 r 0 T Resistors: The potentil difference V cross smple of mteril tht oeys Ohm s lw is proportionl to the current I through the smple. The rtio V>I = R is the resistnce of the smple. The I unit of resistnce is the ohm 11 Æ=1 V>A2. The resistnce of cylindricl conductor is relted to its resistivity r, length L, nd cross-sectionl re A. (ee Exmples 25.2 nd 25.3.) V = IR R = rl A (25.11) (25.10) Higher potentil A I E L J V Lower potentil I Circuits nd emf: A complete circuit hs continuous current-crrying pth. A complete circuit crrying stedy current must contin source of electromotive force (emf) E. The I unit of electromotive force is the volt (1 V). Every rel source of emf hs some internl resistnce r, so its terminl potentil difference V depends on current. (ee Exmples ) V = E - Ir (25.15) (source with internl resistnce) I V 5 V V r 5 2 V, E 5 12 V A R 5 4 V I Energy nd power in circuits: A circuit element with potentil difference V - V = V nd current I puts energy into circuit if the current direction is from P = V I (generl circuit element) (25.17) lower to higher potentil in the device, nd it tkes P = V energy out of the circuit if the current is opposite. The I = I 2 R = V 2 R (25.18) power P equls the product of the potentil difference (power into resistor) nd the current. A resistor lwys tkes electricl energy out of circuit. (ee Exmples ) I V Circuit element V I Conduction in metls: The microscopic sis of conduction in metls is the motion of electrons tht move freely through the metllic crystl, umping into ion cores in the crystl. In crude clssicl model of this motion, the resistivity of the mteril cn e relted to the electron mss, chrge, speed of rndom motion, density, nd men free time etween collisions. (ee Exmple ) E Net displcement 841

156 842 CHAPTER 25 Current, Resistnce, nd Electromotive Force BRIDGING PROBLEM Resistivity, Temperture, nd Power A toster using Nichrome heting element opertes on 120 V. When it is switched on t 20 C, the heting element crries n initil current of 1.35 A. A few seconds lter the current reches the stedy vlue of 1.23 A. () Wht is the finl temperture of the element? The verge vlue of the temperture coefficient of resistivity for Nichrome over the relevnt temperture rnge is 4.5 * C 2-1. () Wht is the power dissipted in the heting element initilly nd when the current reches 1.23 A? OLUTION GUIDE ee MsteringPhysics study re for Video Tutor solution. IDENTIFY nd ET UP 1. A heting element cts s resistor tht converts electricl energy into therml energy. The resistivity ρ of Nichrome depends on temperture, nd hence so does the resistnce R = rl>a of the heting element nd the current I = V>R tht it crries. 2. We re given V = 120 V nd the initil nd finl vlues of I. elect n eqution tht will llow you to find the initil nd finl vlues of resistnce, nd n eqution tht reltes resistnce to temperture [the trget vrile in prt ()]. 3. The power P dissipted in the heting element depends on I nd V. elect n eqution tht will llow you to clculte the initil nd finl vlues of P. EXECUTE 4. Comine your equtions from step 2 to give reltionship etween the initil nd finl vlues of I nd the initil nd finl tempertures (20 C nd ). T finl T finl 5. olve your expression from step 4 for. 6. Use your eqution from step 3 to find the initil nd finl powers. EVALUATE 7. Is the finl temperture greter thn or less thn 20 C? Does this mke sense? 8. Is the finl resistnce greter thn or less thn the initil resistnce? Agin, does this mke sense? 9. Is the finl power greter thn or less thn the initil power? Does this gree with your oservtions in step 8? Prolems For instructor-ssigned homework, go to : Prolems of incresing difficulty. CP: Cumultive prolems incorporting mteril from erlier chpters. CALC: Prolems requiring clculus. BIO: Biosciences prolems. DICUION QUETION Q25.1 The definition of resistivity 1r = E>J) implies tht n electric field exists inside conductor. Yet we sw in Chpter 21 tht there cn e no electric field inside conductor. Is there contrdiction here? Explin. Q25.2 A cylindricl rod hs resistnce R. If we triple its length nd dimeter, wht is its resistnce, in terms of R? Q25.3 A cylindricl rod hs resistivity r. If we triple its length nd dimeter, wht is its resistivity, in terms of r? Q25.4 Two copper wires with different dimeters re joined end to end. If current flows in the wire comintion, wht hppens to electrons when they move from the lrger-dimeter wire into the smller-dimeter wire? Does their drift speed increse, decrese, or sty the sme? If the drift speed chnges, wht is the force tht cuses the chnge? Explin your resoning. Q25.5 When is 1.5-V AAA ttery not ctully 1.5-V ttery? Tht is, when do its terminls provide potentil difference of less thn 1.5 V? Q25.6 Cn the potentil difference etween the terminls of ttery ever e opposite in direction to the emf? If it cn, give n exmple. If it cnnot, explin why not. Q25.7 A rule of thum used to determine the internl resistnce of source is tht it is the open-circuit voltge divided y the shortcircuit current. Is this correct? Why or why not? Q25.8 Btteries re lwys leled with their emf; for instnce, n AA flshlight ttery is leled 1.5 volts. Would it lso e pproprite to put lel on tteries stting how much current they provide? Why or why not? Q25.9 We hve seen tht coulom is n enormous mount of chrge; it is virtully impossile to plce chrge of 1 C on n oject. Yet, current of 10 A, 10 C>s, is quite resonle. Explin this pprent discrepncy. Q25.10 Electrons in n electric circuit pss through resistor. The wire on either side of the resistor hs the sme dimeter. () How does the drift speed of the electrons efore entering the resistor compre to the speed fter leving the resistor? Explin your resoning. () How does the potentil energy for n electron efore entering the resistor compre to the potentil energy fter leving the resistor? Explin your resoning. Q25.11 Current cuses the temperture of rel resistor to increse. Why? Wht effect does this heting hve on the resistnce? Explin. Q25.12 Which of the grphs in Fig. Q25.12 est illustrtes the current I in rel resistor s function of the potentil difference V cross it? Explin. (Hint: ee Discussion Question Q25.11.) Figure Q25.12 () () (c) (d) O I V O I I V O I V O V

157 Exercises 843 Q25.13 Why does n electric light ul nerly lwys urn out just s you turn on the light, lmost never while the light is shining? Q25.14 A light ul glows ecuse it hs resistnce. The rightness of light ul increses with the electricl power dissipted in the ul. () In the circuit shown in Fig. Q25.14, the two uls A nd B re identicl. Compred to ul A, does ul B glow more rightly, just s rightly, or less rightly? Explin your resoning. () Bul B is removed from the circuit nd the circuit is completed s shown in Fig. Q Compred to the rightness of ul A in Fig. Q25.14, does ul A now glow more rightly, just s rightly, or less rightly? Explin your resoning. Figure Q25.14 () E () E Bul A Bul B Bul A Q25.15 (ee Discussion Question Q25.14.) An idel mmeter A is plced in circuit with ttery nd light ul s shown in Fig. Q25.15, nd the mmeter reding is noted. The circuit is then reconnected s in Fig. Q25.15, so tht the positions of the mmeter nd light ul re reversed. () How does the mmeter reding in the sitution shown in Fig. Q25.15 compre to the reding in the sitution shown in Fig. Q25.15? Explin your resoning. () In which sitution does the light ul glow more rightly? Explin your resoning. Figure Q25.15 () Light ul E () E Light ul Q25.16 (ee Discussion Question Q25.14.) Will light ul glow more rightly when it is connected to ttery s shown in Fig. Q25.16, in which n idel mmeter A is plced in the circuit, or when it is connected s shown in Fig , in which n idel voltmeter V is plced in the circuit? Explin your resoning. Figure Q25.16 () Light ul A E () E A Q25.17 The energy tht cn e extrcted from storge ttery is lwys less thn the energy tht goes into it while it is eing chrged. Why? Q25.18 Eight flshlight tteries in series hve n emf of out 12 V, similr to tht of cr ttery. Could they e used to strt cr with ded ttery? Why or why not? Q25.19 mll ircrft often hve 24-V electricl systems rther thn the 12-V systems in utomoiles, even though the electricl A Light ul V power requirements re roughly the sme in oth pplictions. The explntion given y ircrft designers is tht 24-V system weighs less thn 12-V system ecuse thinner wires cn e used. Explin why this is so. Q25.20 Long-distnce, electric-power, trnsmission lines lwys operte t very high voltge, sometimes s much s 750 kv. Wht re the dvntges of such high voltges? Wht re the disdvntges? Q25.21 Ordinry household electric lines in North Americ usully operte t 120 V. Why is this desirle voltge, rther thn vlue considerly lrger or smller? On the other hnd, utomoiles usully hve 12-V electricl systems. Why is this desirle voltge? Q25.22 A fuse is device designed to rek circuit, usully y melting when the current exceeds certin vlue. Wht chrcteristics should the mteril of the fuse hve? Q25.23 High-voltge power supplies re sometimes designed intentionlly to hve rther lrge internl resistnce s sfety precution. Why is such power supply with lrge internl resistnce sfer thn supply with the sme voltge ut lower internl resistnce? Q25.24 The text sttes tht good therml conductors re lso good electricl conductors. If so, why don t the cords used to connect tosters, irons, nd similr het-producing pplinces get hot y conduction of het from the heting element? EXERCIE ection 25.1 Current Lightning trikes. During lightning strikes from cloud to the ground, currents s high s 25,000 A cn occur nd lst for out 40 ms. How much chrge is trnsferred from the cloud to the erth during such strike? A silver wire 2.6 mm in dimeter trnsfers chrge of 420 C in 80 min. ilver contins 5.8 * free electrons per cuic meter. () Wht is the current in the wire? () Wht is the mgnitude of the drift velocity of the electrons in the wire? A 5.00-A current runs through 12-guge copper wire (dimeter 2.05 mm) nd through light ul. Copper hs 8.5 * free electrons per cuic meter. () How mny electrons pss through the light ul ech second? () Wht is the current density in the wire? (c) At wht speed does typicl electron pss y ny given point in the wire? (d) If you were to use wire of twice the dimeter, which of the ove nswers would chnge? Would they increse or decrese? An 18-guge copper wire (dimeter 1.02 mm) crries current with current density of 1.50 * 10 6 A>m 2. The density of free electrons for copper is 8.5 * electrons per cuic meter. Clculte () the current in the wire nd () the drift velocity of electrons in the wire Copper hs 8.5 * free electrons per cuic meter. A 71.0-cm length of 12-guge copper wire tht is 2.05 mm in dimeter crries 4.85 A of current. () How much time does it tke for n electron to trvel the length of the wire? () Repet prt () for 6-guge copper wire (dimeter 4.12 mm) of the sme length tht crries the sme current. (c) Generlly speking, how does chnging the dimeter of wire tht crries given mount of current ffect the drift velocity of the electrons in the wire? Consider the 18-guge wire in Exmple How mny toms re in 1.00 m 3 of copper? With the density of free electrons given in the exmple, how mny free electrons re there per copper tom?

158 844 CHAPTER 25 Current, Resistnce, nd Electromotive Force CALC The current in wire vries with time ccording to the reltionship I = 55 A A>s 2 2t 2. () How mny couloms of chrge pss cross section of the wire in the time intervl etween t = 0 nd t = 8.0 s? () Wht constnt current would trnsport the sme chrge in the sme time intervl? Current psses through solution of sodium chloride. In 1.00 s, 2.68 * N ions rrive t the negtive electrode nd 3.92 * Cl - ions rrive t the positive electrode. () Wht is the current pssing etween the electrodes? () Wht is the direction of the current? BIO Trnsmission of Nerve Impulses. Nerve cells trnsmit electric signls through their long tuulr xons. These signls propgte due to sudden rush of N ions, ech with chrge e, into the xon. Mesurements hve reveled tht typiclly out 5.6 * N ions enter ech meter of the xon during time of 10 ms. Wht is the current during this inflow of chrge in meter of xon? ection 25.2 Resistivity nd ection 25.3 Resistnce () At room temperture wht is the strength of the electric field in 12-guge copper wire (dimeter 2.05 mm) tht is needed to cuse 2.75-A current to flow? () Wht field would e needed if the wire were mde of silver insted? A 1.50-m cylindricl rod of dimeter cm is connected to power supply tht mintins constnt potentil difference of 15.0 V cross its ends, while n mmeter mesures the current through it. You oserve tht t room temperture C2 the mmeter reds 18.5 A, while t 92.0 C it reds 17.2 A. You cn ignore ny therml expnsion of the rod. Find () the resistivity t 20.0 C nd () the temperture coefficient of resistivity t 20 C for the mteril of the rod A copper wire hs squre cross section 2.3 mm on side. The wire is 4.0 m long nd crries current of 3.6 A. The density of free electrons is 8.5 * >m 3. Find the mgnitudes of () the current density in the wire nd () the electric field in the wire. (c) How much time is required for n electron to trvel the length of the wire? A 14-guge copper wire of dimeter mm crries current of 12.5 ma. () Wht is the potentil difference cross 2.00-m length of the wire? () Wht would the potentil difference in prt () e if the wire were silver insted of copper, ut ll else were the sme? A wire 6.50 m long with dimeter of 2.05 mm hs resistnce of Æ. Wht mteril is the wire most likely mde of? A cylindricl tungsten filment 15.0 cm long with dimeter of 1.00 mm is to e used in mchine for which the temperture will rnge from room temperture 120 C2 up to 120 C. It will crry current of 12.5 A t ll tempertures (consult Tles 25.1 nd 25.2). () Wht will e the mximum electric field in this filment, nd () wht will e its resistnce with tht field? (c) Wht will e the mximum potentil drop over the full length of the filment? A ductile metl wire hs resistnce R. Wht will e the resistnce of this wire in terms of R if it is stretched to three times its originl length, ssuming tht the density nd resistivity of the mteril do not chnge when the wire is stretched? (Hint: The mount of metl does not chnge, so stretching out the wire will ffect its cross-sectionl re.) In household wiring, copper wire 2.05 mm in dimeter is often used. Find the resistnce of 24.0-m length of this wire Wht dimeter must copper wire hve if its resistnce is to e the sme s tht of n equl length of luminum wire with dimeter 3.26 mm? You need to produce set of cylindricl copper wires 3.50 m long tht will hve resistnce of Æ ech. Wht will e the mss of ech of these wires? A tightly coiled spring hving 75 coils, ech 3.50 cm in dimeter, is mde of insulted metl wire 3.25 mm in dimeter. An ohmmeter connected cross its opposite ends reds 1.74 Æ. Wht is the resistivity of the metl? An luminum cue hs sides of length 1.80 m. Wht is the resistnce etween two opposite fces of the cue? You pply potentil difference of 4.50 V etween the ends of wire tht is 2.50 m in length nd mm in rdius. The resulting current through the wire is 17.6 A. Wht is the resistivity of the wire? A current-crrying gold wire hs dimeter 0.84 mm. The electric field in the wire is 0.49 V>m. Wht re () the current crried y the wire; () the potentil difference etween two points in the wire 6.4 m prt; (c) the resistnce of 6.4-m length of this wire? A hollow luminum cylinder is 2.50 m long nd hs n inner rdius of 3.20 cm nd n outer rdius of 4.60 cm. Tret ech surfce (inner, outer, nd the two end fces) s n equipotentil surfce. At room temperture, wht will n ohmmeter red if it is connected etween () the opposite fces nd () the inner nd outer surfces? () Wht is the resistnce of Nichrome wire t 0.0 C if its resistnce is Æ t 11.5 C? () Wht is the resistnce of cron rod t 25.8 C if its resistnce is Æ t 0.0 C? A cron resistor is to e used s thermometer. On winter dy when the temperture is 4.0 C, the resistnce of the cron resistor is Æ. Wht is the temperture on spring dy when the resistnce is Æ? (Tke the reference temperture T 0 to e 4.0 C.) A strnd of wire hs resistnce 5.60 mæ. Find the net resistnce of 120 such strnds if they re () plced side y side to form cle of the sme length s single strnd, nd () connected end to end to form wire 120 times s long s single strnd. ection 25.4 Electromotive Force nd Circuits Consider the circuit Figure E25.28 shown in Fig. E The terminl voltge of the 24.0-V ttery is 21.2 V. Wht re () the internl resistnce r of the ttery nd 4.00 A () the resistnce R of the circuit resistor? R A copper trnsmission cle 100 km long nd 10.0 cm in dimeter crries current of 125 A. () Wht is the potentil drop cross the cle? () How much electricl energy is dissipted s therml energy every hour? An idelized mmeter is connected to ttery s shown in Fig. E Find () the reding of the mmeter, () the current through the 4.00-Æ resistor, (c) the terminl voltge of the ttery. r 24.0 V Figure E25.30 A 2.00 V 10.0 V 4.00 V 4.00 A

159 Exercises An idel voltmeter V is connected Figure E25.31 to 2.0-Æ resistor nd ttery 0.5 V 5.0 V with emf 5.0 V nd internl resistnce 0.5 Æ s shown in Fig. E () Wht is the current in the 2.0-Æ resistor? () Wht is the terminl voltge of the V ttery? (c) Wht is the reding on the 2.0 V voltmeter? Explin your nswers The circuit shown in Figure E25.32 Fig. E25.32 contins two tteries, ech with n emf nd n internl resistnce, nd two 1.6 V 16.0 V resistors. Find () the current in the circuit (mgnitude nd 5.0 V 9.0 V direction); () the terminl 1.4 V 8.0 V voltge V of the 16.0-V ttery; c (c) the potentil differ- ence V c of point with respect to point c. (d) Using Fig s model, grph the potentil rises nd drops in this circuit When switch in Fig. E25.33 is open, the voltmeter V of the ttery reds 3.08 V. When the switch is closed, the voltmeter reding drops to 2.97 V, Figure E25.33 V nd the mmeter A reds 1.65 A. Find r E the emf, the internl resistnce of the ttery, nd the circuit resistnce R. Assume tht the two meters re idel, R A so they don t ffect the circuit In the circuit of Fig. E25.32, the 5.0-Æ resistor is removed nd replced y resistor of unknown resistnce R. When this is done, n idel voltmeter connected cross the points nd c reds 1.9 V. Find () the current in the circuit nd () the resistnce R. (c) Grph the potentil rises nd drops in this circuit (see Fig ) In the circuit shown in Fig. E25.32, the 16.0-V ttery is removed nd reinserted with the opposite polrity, so tht its negtive terminl is now next to point. Find () the current in the circuit (mgnitude nd direction); () the terminl voltge V of the 16.0-V ttery; (c) the potentil difference V c of point with respect to point c. (d) Grph the potentil rises nd drops in this circuit (see Fig ) The following mesurements were mde on Thyrite resistor: I 1A2 V 1V () Grph V s function of I. () Does Thyrite oey Ohm s lw? How cn you tell? (c) Grph the resistnce R = V >I s function of I The following mesurements of current nd potentil difference were mde on resistor constructed of Nichrome wire: I 1A2 V 1V2 V () Grph s function of I. () Does Nichrome oey Ohm s lw? How cn you tell? (c) Wht is the resistnce of the resistor in ohms? The circuit shown in Fig. E25.38 contins two tteries, ech with n emf nd n internl resistnce, nd two resistors. Find () the current in the circuit (mgnitude nd direction) nd () the terminl voltge V of the 16.0-V ttery. Figure E V 1.6 V 16.0 V 1.4 V 8.0 V 9.0 V ection 25.5 Energy nd Power in Electric Circuits Light Buls. The power rting of light ul (such s 100-W ul) is the power it dissiptes when connected cross 120-V potentil difference. Wht is the resistnce of () 100-W ul nd () 60-W ul? (c) How much current does ech ul drw in norml use? If 75-W ul (see Prolem 25.39) is connected cross 220-V potentil difference (s is used in Europe), how much power does it dissipte? Europen Light Bul. In Europe the stndrd voltge in homes is 220 V insted of the 120 V used in the United ttes. Therefore 100-W Europen ul would e intended for use with 220-V potentil difference (see Prolem 25.40). () If you ring 100-W Europen ul home to the United ttes, wht should e its U.. power rting? () How much current will the 100-W Europen ul drw in norml use in the United ttes? A ttery-powered glol positioning system (GP) receiver operting on 9.0 V drws current of 0.13 A. How much electricl energy does it consume during 1.5 h? Consider resistor with length L, uniform cross-sectionl re A, nd uniform resistivity r tht is crrying current with uniform current density J. Use Eq. (25.18) to find the electricl power dissipted per unit volume, p. Express your result in terms of () E nd J; () J nd r; (c) E nd r BIO Electric Eels. Electric eels generte electric pulses long their skin tht cn e used to stun n enemy when they come into contct with it. Tests hve shown tht these pulses cn e up to 500 V nd produce currents of 80 ma (or even lrger). A typicl pulse lsts for 10 ms. Wht power nd how much energy re delivered to the unfortunte enemy with single pulse, ssuming stedy current? BIO Tretment of Hert Filure. A hert defirilltor is used to enle the hert to strt eting if it hs stopped. This is done y pssing lrge current of 12 A through the ody t 25 V for very short time, usully out 3.0 ms. () Wht power does the defirilltor deliver to the ody, nd () how much energy is trnsferred? Consider the circuit of Fig. E () Wht is the totl rte t which electricl energy is dissipted in the 5.0-Æ nd 9.0-Æ resistors? () Wht is the power output of the 16.0-V ttery? (c) At wht rte is electricl energy eing converted to other forms in the 8.0-V ttery? (d) how tht the power output of the 16.0-V ttery equls the overll rte of dissiption of electricl energy in the rest of the circuit The cpcity of storge ttery, such s those used in utomoile electricl systems, is rted in mpere-hours 1A # h2. A 50-A # h ttery cn supply current of 50 A for 1.0 h, or 25 A for 2.0 h, nd so on. () Wht totl energy cn e supplied y 12-V, 60-A # h ttery if its internl resistnce is negligile? () Wht

160 846 CHAPTER 25 Current, Resistnce, nd Electromotive Force volume (in liters) of gsoline hs totl het of comustion equl to the energy otined in prt ()? (ee ection 17.6; the density of gsoline is 900 kg>m 3.) (c) If genertor with n verge electricl power output of 0.45 kw is connected to the ttery, how much time will e required for it to chrge the ttery fully? In the circuit nlyzed in Exmple 25.8 the 4.0- Æ resistor is replced y 8.0-Æ resistor, s in Exmple () Clculte the rte of conversion of chemicl energy to electricl energy in the ttery. How does your nswer compre to the result clculted in Exmple 25.8? () Clculte the rte of electricl energy dissiption in the internl resistnce of the ttery. How does your nswer compre to the result clculted in Exmple 25.8? (c) Use the results of prts () nd () to clculte the net power output of the ttery. How does your result compre to the electricl power dissipted in the 8.0-Æ resistor s clculted for this circuit in Exmple 25.9? A Æ ul is connected cross the terminls of 12.0-V ttery hving 3.50 Æ of internl resistnce. Wht percentge of the power of the ttery is dissipted cross the internl resistnce nd hence is not ville to the ul? An idelized voltmeter is connected cross the terminls of 15.0-V ttery, nd 75.0-Æ pplince is lso connected cross its terminls. If the voltmeter reds 11.3 V: () how much power is eing dissipted y the pplince, nd () wht is the internl resistnce of the ttery? In the circuit in Fig. E25.51, find () the rte of conversion of internl (chemicl) energy to electricl energy within the ttery; () the rte of dissiption of electricl energy in the ttery; (c) the rte of dissiption of electricl energy in the externl resistor A typicl smll flshlight Figure E25.51 contins two tteries, ech hving n emf of 1.5 V, connected in series with ul hving resistnce 17 Æ. () If the internl resistnce of the tteries is negligile, wht power is delivered to the ul? () If the tteries lst for 5.0 h, wht is the totl energy delivered to the ul? (c) The resistnce of rel tteries increses s they run down. If the initil internl resistnce is negligile, wht is the comined internl resistnce of oth tteries when the power to the ul hs decresed to hlf its initil vlue? (Assume tht the resistnce of the ul is constnt. Actully, it will chnge somewht when the current through the filment chnges, ecuse this chnges the temperture of the filment nd hence the resistivity of the filment wire.) A 540-W electric heter is designed to operte from 120-V lines. () Wht is its resistnce? () Wht current does it drw? (c) If the line voltge drops to 110 V, wht power does the heter tke? (Assume tht the resistnce is constnt. Actully, it will chnge ecuse of the chnge in temperture.) (d) The heter coils re metllic, so tht the resistnce of the heter decreses with decresing temperture. If the chnge of resistnce with temperture is tken into ccount, will the electricl power consumed y the heter e lrger or smller thn wht you clculted in prt (c)? Explin. ection 25.6 Theory of Metllic Conduction Pure silicon contins pproximtely 1.0 * free electrons per cuic meter. () Referring to Tle 25.1, clculte the men free time t for silicon t room temperture. () Your nswer in prt () is much greter thn the men free time for copper given in Exmple Why, then, does pure silicon hve such high resistivity compred to copper? 1.0 V 12.0 V d 5.0 V c PROBLEM An electricl conductor designed to crry lrge currents hs circulr cross section 2.50 mm in dimeter nd is 14.0 m long. The resistnce etween its ends is Æ. () Wht is the resistivity of the mteril? () If the electric-field mgnitude in the conductor is 1.28 V>m, wht is the totl current? (c) If the mteril hs 8.5 * free electrons per cuic meter, find the verge drift speed under the conditions of prt () A plstic tue 25.0 m long nd 3.00 cm in dimeter is dipped into silver solution, depositing lyer of silver mm thick uniformly over the outer surfce of the tue. If this coted tue is then connected cross 12.0-V ttery, wht will e the current? On your first dy t work s n electricl technicin, you re sked to determine the resistnce per meter of long piece of wire. The compny you work for is poorly equipped. You find ttery, voltmeter, nd n mmeter, ut no meter for directly mesuring resistnce (n ohmmeter). You put the leds from the voltmeter cross the terminls of the ttery, nd the meter reds 12.6 V. You cut off 20.0-m length of wire nd connect it to the ttery, with n mmeter in series with it to mesure the current in the wire. The mmeter reds 7.00 A. You then cut off 40.0-m length of wire nd connect it to the ttery, gin with the mmeter in series to mesure the current. The mmeter reds 4.20 A. Even though the equipment you hve ville to you is limited, your oss ssures you of its high qulity: The mmeter hs very smll resistnce, nd the voltmeter hs very lrge resistnce. Wht is the resistnce of 1 meter of wire? A 2.0-mm length of wire is mde y welding the end of 120-cm-long silver wire to the end of n 80-cm-long copper wire. Ech piece of wire is 0.60 mm in dimeter. The wire is t room temperture, so the resistivities re s given in Tle A potentil difference of 5.0 V is mintined etween the ends of the 2.0-m composite wire. () Wht is the current in the copper section? () Wht is the current in the silver section? (c) Wht is the mgnitude of E in the copper? (d) Wht is the mgnitude of E in the silver? (e) Wht is the potentil difference etween the ends of the silver section of wire? A 3.00-m length of copper wire t 20 C hs 1.20-mlong section with dimeter 1.60 mm nd 1.80-m-long section with dimeter 0.80 mm. There is current of 2.5 ma in the mm-dimeter section. () Wht is the current in the 0.80-mmdimeter section? () Wht is the mgnitude of in the 1.60-mm-dimeter section? (c) Wht is the mgnitude of E in the 0.80-mm-dimeter section? (d) Wht is the potentil difference etween the ends of the 3.00-m length of wire? Criticl Current Density in uperconductors. One prolem with some of the newer high-temperture superconductors is getting lrge enough current density for prcticl use without cusing the resistnce to repper. The mximum current density for which the mteril will remin superconductor is clled the criticl current density of the mteril. In 1987, IBM reserch ls hd produced thin films with criticl current densities of 1.0 * 10 5 A>cm 2. () How much current could n 18-guge wire (see Exmple 25.1 in ection 25.1) of this mteril crry nd still remin superconducting? () Reserchers re trying to develop superconductors with criticl current densities of 1.0 * 10 6 A>cm 2. Wht dimeter cylindricl wire of such mteril would e needed to crry 1000 A without losing its superconductivity? CP A Nichrome heting element tht hs resistnce 28.0 Æ is connected to ttery tht hs emf 96.0 V nd internl E

161 Prolems 847 resistnce 1.2 Æ. An luminum cup with mss kg contins kg of wter. The heting element is plced in the wter nd the electricl energy dissipted in the resistnce of the heting element ll goes into the cup nd wter. The element itself hs very eqution for R in terms of r nd L.) (e) Wht is the temperture coefficient of resistnce for the mercury column, s defined in Eq. (25.12)? How does this vlue compre with the temperture coefficient of resistivity? Is the effect of the chnge in length smll mss. How much time does it tke for the temperture of the importnt? cup nd wter to rise from 21.2 o C to 34.5 o C? (The chnge of the () Wht is the potentil Figure P25.68 difference V d in the circuit resistnce of the Nichrome due to its temperture chnge cn e 0.50 V 4.00 V neglected.) of Fig. P25.68? () Wht is the c A resistor with resistnce R is connected to ttery tht terminl voltge of the 4.00-V 9.00 V hs emf 12.0 V nd internl resistnce r = 0.40 Æ. For wht two ttery? (c) A ttery with emf 6.00 V d vlues of R will the power dissipted in the resistor e 80.0 W? V nd internl resistnce V 8.00 V CP BIO truck y Lightning. Lightning strikes cn 0.50 Æ is inserted in the circuit involve currents s high s 25,000 A tht lst for out 40 ms. If person is struck y olt of lightning with these properties, the t d, with its negtive terminl connected to the negtive 8.00 V current will pss through his ody. We shll ssume tht his mss terminl of the 8.00-V ttery. Wht is the difference of poten- is 75 kg, tht he is wet (fter ll, he is in rinstorm) nd therefore til V c etween the terminls of the 4.00-V ttery now? hs resistnce of 1.0 kæ, nd tht his ody is ll wter (which is The potentil difference cross the terminls of ttery resonle for rough, ut plusile, pproximtion). () By how mny degrees Celsius would this lightning olt increse the temperture of 75 kg of wter? () Given tht the internl ody temperture is out 37 C, would the person s temperture ctully is 8.40 V when there is current of 1.50 A in the ttery from the negtive to the positive terminl. When the current is 3.50 A in the reverse direction, the potentil difference ecomes V. () Wht is the internl resistnce of the ttery? () Wht is the emf of the increse tht much? Why not? Wht would hppen first? ttery? In the Bohr model of the hydrogen tom, the electron BIO A person with ody resistnce etween his hnds of mkes 6.0 * rev>s round the nucleus. Wht is the verge 10 kæ ccidentlly grsps the terminls of 14-kV power supply. current t point on the orit of the electron? () If the internl resistnce of the power supply is 2000 Æ, wht CALC A mteril of resistivity r is Figure is the current through the person s ody? () Wht is the power formed into solid, truncted cone of height h P25.65 dissipted in his ody? (c) If the power supply is to e mde sfe y nd rdii r 1 nd r 2 t either end (Fig. P25.65). incresing its internl resistnce, wht should the internl resistnce r 1 () Clculte the resistnce of the cone etween e for the mximum current in the ove sitution to e 1.00 ma the two flt end fces. (Hint: Imgine slicing the or less? cone into very mny thin disks, nd clculte the BIO The verge ulk resistivity of the humn ody resistnce of one such disk.) () how tht your h (prt from surfce resistnce of the skin) is out 5.0 Æ # m. The result grees with Eq. (25.10) when r 1 = r CALC The region etween two concentric conducting spheres with rdii nd is filled with conducting mteril with resistivity r. () how tht the resistnce etween the spheres is given y R = r 4p 1-1 () Derive n expression for the current density s function of rdius, in terms of the potentil difference V etween the spheres. (c) how tht the result in prt () reduces to Eq. (25.10) when the seprtion L = - etween the spheres is smll The temperture coefficient of resistnce in Eq. (25.12) equls the temperture coefficient of resistivity in Eq. (25.6) only if the coefficient of therml expnsion is smll. A cylindricl column of mercury is in verticl glss tue. At 20 C, the length of the mercury column is 12.0 cm. The dimeter of the mercury column is 1.6 mm nd doesn t chnge with temperture ecuse glss hs smll coefficient of therml expnsion. The coefficient of volume expnsion of the mercury is given in Tle 17.2, its resistivity t 20 C is given in Tle 25.1, nd its temperture coefficient of resistivity is given in Tle () At 20 C, wht is the resistnce etween the ends of the mercury column? () The mercury column is heted to 60 C. Wht is the chnge in its resistivity? (c) Wht is the chnge in its length? Explin why the coefficient of volume expnsion, rther thn the coefficient of liner expnsion, determines the chnge in length. (d) Wht is the chnge in its resistnce? (Hint: ince the percentge chnges in r nd L re smll, you my find it helpful to derive from Eq. (25.10) n r 2 conducting pth etween the hnds cn e represented pproximtely s cylinder 1.6 m long nd 0.10 m in dimeter. The skin resistnce cn e mde negligile y soking the hnds in slt wter. () Wht is the resistnce etween the hnds if the skin resistnce is negligile? () Wht potentil difference etween the hnds is needed for lethl shock current of 100 ma? (Note tht your result shows tht smll potentil differences produce dngerous currents when the skin is dmp.) (c) With the current in prt (), wht power is dissipted in the ody? A typicl cost for electric power is $0.120 per kilowtthour. () ome people leve their porch light on ll the time. Wht is the yerly cost to keep 75-W ul urning dy nd night? () uppose your refrigertor uses 400 W of power when it s running, nd it runs 8 hours dy. Wht is the yerly cost of operting your refrigertor? A 12.6-V cr ttery with negligile internl resistnce is connected to series comintion of 3.2-Æ resistor tht oeys Ohm s lw nd thermistor tht does not oey Ohm s lw ut insted hs currentvoltge reltionship V = I I 2, with = 3.8 Æ nd = 1.3 Æ>A. Wht is the current through the 3.2-Æ resistor? A cylindricl copper cle 1.50 km long is connected cross V potentil difference. () Wht should e its dimeter so tht it produces het t rte of 75.0 W? () Wht is the electric field inside the cle under these conditions? A Nonidel Ammeter. Unlike the idelized mmeter descried in ection 25.4, ny rel mmeter hs nonzero resistnce. () An mmeter with resistnce R A is connected in series with resistor R nd ttery of emf E nd internl resistnce r. The current mesured y the mmeter is I A. Find the current

162 848 CHAPTER 25 Current, Resistnce, nd Electromotive Force through the circuit if the mmeter is removed so tht the ttery nd the resistor form complete circuit. Express your nswer in terms of I A, r, R A, nd R. The more idel the mmeter, the smller the difference etween this current nd the current I A. () If R = 3.80 Æ, E = 7.50 V, nd r = 0.45 Æ, find the mximum vlue of the mmeter resistnce R A so tht I A is within 1.0% of the current in the circuit when the mmeter is sent. (c) Explin why your nswer in prt () represents mximum vlue CALC A 1.50-m cylinder of rdius 1.10 cm is mde of complicted mixture of mterils. Its resistivity depends on the distnce x from the left end nd oeys the formul r1x2 = x 2, where nd re constnts. At the left end, the resistivity is 2.25 * 10 while t the right end it is 10-8 Æ # -8 Æ # m, 8.50 * m. () Wht is the resistnce of this rod? () Wht is the electric field t its midpoint if it crries 1.75-A current? (c) If we cut the rod into two 75.0-cm hlves, wht is the resistnce of ech hlf? According to the U.. Ntionl Electricl Code, copper wire used for interior wiring of houses, hotels, office uildings, nd industril plnts is permitted to crry no more thn specified mximum mount of current. The tle elow shows the mximum current I mx for severl common sizes of wire with vrnished cmric insultion. The wire guge is stndrd used to descrie the dimeter of wires. Note tht the lrger the dimeter of the wire, the smller the wire guge. Wire guge Dimeter (cm) I mx 1A () Wht considertions determine the mximum current-crrying cpcity of household wiring? () A totl of 4200 W of power is to e supplied through the wires of house to the household electricl pplinces. If the potentil difference cross the group of pplinces is 120 V, determine the guge of the thinnest permissile wire tht cn e used. (c) uppose the wire used in this house is of the guge found in prt () nd hs totl length 42.0 m. At wht rte is energy dissipted in the wires? (d) The house is uilt in community where the consumer cost of electric energy is $0.11 per kilowtt-hour. If the house were uilt with wire of the next lrger dimeter thn tht found in prt (), wht would e the svings in electricity costs in one yer? Assume tht the pplinces re kept on for n verge of 12 hours dy Compct Fluorescent Buls. Compct fluorescent uls re much more efficient t producing light thn re ordinry incndescent uls. They initilly cost much more, ut they lst fr longer nd use much less electricity. According to one study of these uls, compct ul tht produces s much light s 100-W incndescent ul uses only 23 W of power. The compct ul lsts 10,000 hours, on the verge, nd costs $11.00, wheres the incndescent ul costs only $0.75, ut lsts just 750 hours. The study ssumed tht electricity costs $0.080 per kilowtt-hour nd tht the uls re on for 4.0 h per dy. () Wht is the totl cost (including the price of the uls) to run ech ul for 3.0 yers? () How much do you sve over 3.0 yers if you use compct fluorescent ul insted of n incndescent ul? (c) Wht is the resistnce of 100-W fluorescent ul? (Rememer, it ctully uses only 23 W of power nd opertes cross 120 V.) In the circuit of Fig. Figure P25.79 P25.79, find () the current E through the 8.0-Æ resistor nd V r V () the totl rte of dissiption of electricl energy in the 8.0-Æ R V resistor nd in the internl resistnce of the tteries. (c) In one of the tteries, chemicl E V r V energy is eing converted into electricl energy. In which one is this hppening, nd t wht rte? (d) In one of the tteries, electricl energy is eing converted into chemicl energy. In which one is this hppening, nd t wht rte? (e) how tht the overll rte of production of electricl energy equls the overll rte of consumption of electricl energy in the circuit A lightning olt strikes one end of steel lightning rod, producing 15,000-A current urst tht lsts for 65 ms. The rod is 2.0 m long nd 1.8 cm in dimeter, nd its other end is connected to the ground y 35 m of 8.0-mm-dimeter copper wire. () Find the potentil difference etween the top of the steel rod nd the lower end of the copper wire during the current urst. () Find the totl energy deposited in the rod nd wire y the current urst A 12.0-V ttery hs n internl resistnce of 0.24 Æ nd cpcity of 50.0 A # h (see Exercise 25.47). The ttery is chrged y pssing 10-A current through it for 5.0 h. () Wht is the terminl voltge during chrging? () Wht totl electricl energy is supplied to the ttery during chrging? (c) Wht electricl energy is dissipted in the internl resistnce during chrging? (d) The ttery is now completely dischrged through resistor, gin with constnt current of 10 A. Wht is the externl circuit resistnce? (e) Wht totl electricl energy is supplied to the externl resistor? (f) Wht totl electricl energy is dissipted in the internl resistnce? (g) Why re the nswers to prts () nd (e) not the sme? Repet Prolem with chrge nd dischrge currents of 30 A. The chrging nd dischrging times will now e 1.7 h rther thn 5.0 h. Wht differences in performnce do you see? CP Consider the circuit shown in Fig. P The emf source hs negligile internl resistnce. The resistors hve resistnces R 1 = 6.00 Æ nd R 2 = 4.00 Æ. The cpcitor hs cpcitnce C = 9.00 mf. When the cpcitor is fully E R 1 C chrged, the mgnitude of the chrge on its pltes is Q = 36.0 mc. Clculte the emf E CP Consider the circuit shown in Fig. P The ttery hs emf 60.0 V nd negligile internl resistnce. R 2 = 2.00 Æ, C 1 = 3.00 mf, nd C 2 = 6.00 mf. After the cpcitors hve ttined their finl chrges, the chrge on C 1 is Q 1 = 18.0 mc. () Wht is the finl chrge on? () Wht is the resistnce? Figure P25.84 C 2 R 1 Figure P25.83 R 2 R 2 C 2 E C 1 R 1

163 Answers 849 CHALLENGE PROBLEM The Tolmn-tewrt experiment in 1916 demonstrted Figure P25.85 tht the free chrges in metl hve negtive chrge nd provided quntittive mesurement L c of their chrge-to-mss rtio, ƒqƒ >m. The experiment consisted of ruptly stopping rpidly rotting spool of wire nd mesuring the potentil difference tht this produced etween the ends of the wire. In simplified model of this experiment, consider metl rod of length L tht is given uniform ccelertion to the right. Initilly the free chrges in the metl lg ehind the rod s motion, thus setting up n electric field E in the rod. In the stedy stte this field exerts force on the free chrges tht mkes them ccelerte long with the rod. () Apply gf m to the free chrges to otin n expression for ƒqƒ >m in terms of the mgnitudes of the induced electric field E nd the ccelertion. () If ll the free chrges in the metl rod hve the sme ccelertion, the electric field E is the sme t ll points in the rod. Use this fct to rewrite the expression for ƒqƒ >m in terms of the potentil V c etween the ends of the rod (Fig. P25.85). (c) If the free chrges hve negtive chrge, which end of the rod, or c, is t higher potentil? (d) If the rod is 0.50 m long nd the free chrges re electrons (chrge q = * C, mss 9.11 * kg2, wht mgnitude of ccelertion is required to produce potentil difference of 1.0 mv etween the ends of the rod? (e) Discuss why the ctul experiment used rotting spool of thin wire rther thn moving r s in our simplified nlysis CALC A source with emf E nd internl resistnce r is connected to n externl circuit. () how tht the power output of the source is mximum when the current in the circuit is one-hlf the short-circuit current of the source. () If the externl circuit consists of resistnce R, show tht the power output is mximum when R = r nd tht the mximum power is E 2 >4r CALC The resistivity of semiconductor cn e modified y dding different mounts of impurities. A rod of semiconducting mteril of length L nd cross-sectionl re A lies long the x-xis etween x = 0 nd x = L. The mteril oeys Ohm s lw, nd its resistivity vries long the rod ccording to r1x2 = r 0 exp1-x>l2. The end of the rod t x = 0 is t potentil V 0 greter thn the end t x = L. () Find the totl resistnce of the rod nd the current in the rod. () Find the electric-field mgnitude E1x2 in the rod s function of x. (c) Find the electric potentil V1x2 in the rod s function of x. (d) Grph the functions r1x2, E1x2, nd V1x2 for vlues of x etween x = 0 nd x = L. Answers Chpter Opening Question? The current out equls the current in. In other words, chrge must enter the ul t the sme rte s it exits the ul. It is not used up or consumed s it flows through the ul. Test Your Understnding Questions 25.1 Answer: (v) Douling the dimeter increses the crosssectionl re A y fctor of 4. Hence the current-density mgnitude J = I>A is reduced to 1 4 of the vlue in Exmple 25.1, nd the mgnitude of the drift velocity v d = J>nƒqƒ is reduced y the sme fctor. The new mgnitude is v d = mm>s2>4 = mm>s. This ehvior is the sme s tht of n incompressile fluid, which slows down when it moves from nrrow pipe to roder one (see ection 14.4) Answer: (ii) Figure 25.6 shows tht the resistivity r of semiconductor increses s the temperture decreses. From Eq. (25.5), the mgnitude of the current density is J = E>r, so the current density decreses s the temperture drops nd the resistivity increses Answer: (iii) olving Eq. (25.11) for the current shows tht I = V>R. If the resistnce R of the wire remined the sme, douling the voltge V would mke the current I doule s well. However, we sw in Exmple 25.3 tht the resistnce is not constnt: As the current increses nd the temperture increses, R increses s well. Thus douling the voltge produces current tht is less thn doule the originl current. An ohmic conductor is one for which R = V>I hs the sme vlue no mtter wht the voltge, so the wire is nonohmic. (In mny prcticl prolems the temperture chnge of the wire is so smll tht it cn e ignored, so we cn sfely regrd the wire s eing ohmic. We do so in lmost ll exmples in this ook.) 25.4 Answer: (iii), (ii), (i) For circuit (i), we find the current from Eq. (25.16): I = E>1R r2 = 11.5 V2>11.4 Æ0.10 Æ2 = 1.0 A. For circuit (ii), we note tht the terminl voltge v = 3.6 V equls the voltge IR cross the 1.8-Æ resistor: V = IR, so I = V >R = 13.6 V2>11.8 Æ2 = 2.0 A. For circuit (iii), we use Eq. (25.15) for the terminl voltge: V = E - Ir, so I = 1E - V 2>r = V V2>10.20 Æ2 = 5.0 A Answer: (iii), (ii), (i) These re the sme circuits tht we nlyzed in Test Your Understnding of ection In ech cse the net power output of the ttery is P = V I, where V is the ttery terminl voltge. For circuit (i), we found tht I = 1.0 A, so V = E - Ir = 1.5 V A Æ2 = 1.4 V, so P = 11.4 V211.0 A2 = 1.4 W. For circuit (ii), we hve V = 3.6 V nd found tht I = 2.0 A, so P = 13.6 V212.0 A2 = 7.2 W. For circuit (iii), we hve V = 11.0 V nd found tht I = 5.0 A, so P = V215.0 A2 = 55 A Answer: (i) The difficulty of producing certin mount of current increses s the resistivity r increses. From Eq. (25.24), r = m>ne 2 t, so incresing the mss m will increse the resistivity. Tht s ecuse more mssive chrged prticle will respond more sluggishly to n pplied electric field nd hence drift more slowly. To produce the sme current, greter electric field would e needed. (Incresing n, e, or t would decrese the resistivity nd mke it esier to produce given current.) Bridging Prolem Answers: () 237 C () 162 W initilly, 148 W t 1.23 A

164 26 DIRECT-CURRENT CIRCUIT LEARNING GOAL By studying this chpter, you will lern: How to nlyze circuits with multiple resistors in series or prllel. Rules tht you cn pply to ny circuit with more thn one loop. How to use n mmeter, voltmeter, ohmmeter, or potentiometer in circuit. How to nlyze circuits tht include oth resistor nd cpcitor. How electric power is distriuted in the home.? In complex circuit like the one on this circuit ord, is it possile to connect severl resistors with different resistnces so tht they ll hve the sme potentil difference? If so, will the current e the sme through ll of the resistors? If you look inside your TV, your computer, or under the hood of cr, you will find circuits of much greter complexity thn the simple circuits we studied in Chpter 25. Whether connected y wires or integrted in semiconductor chip, these circuits often include severl sources, resistors, nd other circuit elements interconnected in network. In this chpter we study generl methods for nlyzing such networks, including how to find voltges nd currents of circuit elements. We ll lern how to determine the equivlent resistnce for severl resistors connected in series or in prllel. For more generl networks we need two rules clled Kirchhoff s rules. One is sed on the principle of conservtion of chrge pplied to junction; the other is derived from energy conservtion for chrge moving round closed loop. We ll discuss instruments for mesuring vrious electricl quntities. We ll lso look t circuit contining resistnce nd cpcitnce, in which the current vries with time. Our principl concern in this chpter is with direct-current (dc) circuits, in which the direction of the current does not chnge with time. Flshlights nd utomoile wiring systems re exmples of direct-current circuits. Household electricl power is supplied in the form of lternting current (c), in which the current oscilltes ck nd forth. The sme principles for nlyzing networks pply to oth kinds of circuits, nd we conclude this chpter with look t household wiring systems. We ll discuss lternting-current circuits in detil in Chpter 31. ActivPhysics 12.1: DC eries Circuits (Qulittive) Resistors in eries nd Prllel Resistors turn up in ll kinds of circuits, rnging from hir dryers nd spce heters to circuits tht limit or divide current or reduce or divide voltge. uch circuits often contin severl resistors, so it s pproprite to consider comintions of resistors. A simple exmple is string of light uls used for holidy decortions;

165 26.1 Resistors in eries nd Prllel 851 ech ul cts s resistor, nd from circuit-nlysis perspective the string of uls is simply comintion of resistors. uppose we hve three resistors with resistnces R 1, R 2, nd R 3. Figure 26.1 shows four different wys in which they might e connected etween points nd. When severl circuit elements such s resistors, tteries, nd motors re connected in sequence s in Fig. 26.1, with only single current pth etween the points, we sy tht they re connected in series. We studied cpcitors in series in ection 24.2; we found tht, ecuse of conservtion of chrge, cpcitors in series ll hve the sme chrge if they re initilly unchrged. In circuits we re often more interested in the current, which is chrge flow per unit time. The resistors in Fig re sid to e connected in prllel etween points nd. Ech resistor provides n lterntive pth etween the points. For circuit elements tht re connected in prllel, the potentil difference is the sme cross ech element. We studied cpcitors in prllel in ection In Fig. 26.1c, resistors R 2 nd R 3 re in prllel, nd this comintion is in series with R 1. In Fig. 26.1d, R 2 nd R 3 re in series, nd this comintion is in prllel with R 1. For ny comintion of resistors we cn lwys find single resistor tht could replce the comintion nd result in the sme totl current nd potentil difference. For exmple, string of holidy light uls could e replced y single, ppropritely chosen light ul tht would drw the sme current nd hve the sme potentil difference etween its terminls s the originl string of uls. The resistnce of this single resistor is clled the equivlent resistnce of the comintion. If ny one of the networks in Fig were replced y its equivlent resistnce R eq, we could write V V = IR eq or R eq = V I where is the potentil difference etween terminls nd of the network nd I is the current t point or. To compute n equivlent resistnce, we ssume potentil difference V cross the ctul network, compute the corresponding current I, nd tke the rtio V >I Four different wys of connecting three resistors. () R 1, R 2, nd R 3 in series R1 R2 R3 x y () R 1, R 2, nd R 3 in prllel R1 R2 R3 (c) R 1 in series with prllel comintion of R 2 nd R 3 R2 I I I R1 R3 (d) R 1 in prllel with series comintion of R 2 nd R 3 R2 R3 I R1 I I I I Resistors in eries We cn derive generl equtions for the equivlent resistnce of series or prllel comintion of resistors. If the resistors re in series, s in Fig. 26.1, the current I must e the sme in ll of them. (As we discussed in ection 25.4, current is not used up s it psses through circuit.) Applying V = IR to ech resistor, we hve The potentil differences cross ech resistor need not e the sme (except for the specil cse in which ll three resistnces re equl). The potentil difference V cross the entire comintion is the sum of these individul potentil differences: nd so V x = IR 1 V xy = IR 2 V y = IR 3 V = V x V xy V y = I1R 1 R 2 R 3 2 V I = R 1 R 2 R 3 The rtio V >I is, y definition, the equivlent resistnce Therefore R eq = R 1 R 2 R 3 It is esy to generlize this to ny numer of resistors: R eq. R eq = R 1 R 2 R 3 Á (resistors in series) (26.1)

166 852 CHAPTER 26 Direct-Current Circuits The equivlent resistnce of ny numer of resistors in series equls the sum of their individul resistnces. The equivlent resistnce is greter thn ny individul resistnce. Let s compre this result with Eq. (24.5) for cpcitors in series. Resistors in series dd directly ecuse the voltge cross ech is directly proportionl to its resistnce nd to the common current. Cpcitors in series dd reciproclly ecuse the voltge cross ech is directly proportionl to the common chrge ut inversely proportionl to the individul cpcitnce A cr s hedlights nd tillights re connected in prllel. Hence ech light is exposed to the full potentil difference supplied y the cr s electricl system, giving mximum rightness. Another dvntge is tht if one hedlight or tillight urns out, the other one keeps shining (see Exmple 26.2). Resistors in Prllel If the resistors re in prllel, s in Fig. 26.1, the current through ech resistor need not e the sme. But the potentil difference etween the termi-nls of ech resistor must e the sme nd equl to V (Fig. 26.2). (Rememer tht the potentil difference etween ny two points does not depend on the pth tken etween the points.) Let s cll the currents in the three resistors I 1, I 2, nd I 3. Then from I = V>R, I 1 = V R 1 I 2 = V R 2 I 3 = V R 3 In generl, the current is different through ech resistor. Becuse chrge is not ccumulting or drining out of point, the totl current I must equl the sum of the three currents in the resistors: I = I 1 I 2 I 3 = V 1 R 1 1 R 2 1 R 3 or I V = 1 R 1 1 R 2 1 R 3 But y the definition of the equivlent resistnce R eq, I>V = 1>R eq, so 1 R eq = 1 R 1 1 R 2 1 R 3 Agin it is esy to generlize to ny numer of resistors in prllel: ActivPhysics 12.2: DC Prllel Circuits 1 R eq = 1 R 1 1 R 2 1 R 3 Á (resistors in prllel) (26.2) For ny numer of resistors in prllel, the reciprocl of the equivlent resistnce equls the sum of the reciprocls of their individul resistnces. The equivlent resistnce is lwys less thn ny individul resistnce. Compre this with Eq. (24.7) for cpcitors in prllel. Resistors in prllel dd reciproclly ecuse the current in ech is proportionl to the common voltge cross them nd inversely proportionl to the resistnce of ech. Cpcitors in prllel dd directly ecuse the chrge on ech is proportionl to the common voltge cross them nd directly proportionl to the cpcitnce of ech. For the specil cse of two resistors in prllel, 1 R eq = 1 R 1 1 R 2 = R 1 R 2 R 1 R 2 nd R eq = R 1 R 2 R 1 R 2 (two resistors in prllel) (26.3)

167 26.1 Resistors in eries nd Prllel 853 Becuse V = I 1 R 1 = I 2 R 2, it follows tht I 1 I 2 = R 2 R 1 (two resistors in prllel) (26.4) This shows tht the currents crried y two resistors in prllel re inversely proportionl to their resistnces. More current goes through the pth of lest resistnce. Prolem-olving trtegy 26.1 Resistors in eries nd Prllel IDENTIFY the relevnt concepts: As in Fig. 26.1, mny resistor networks re mde up of resistors in series, in prllel, or comintion thereof. uch networks cn e replced y single equivlent resistor. The logic is similr to tht of Prolem-olving trtegy 24.1 for networks of cpcitors. ET UP the prolem using the following steps: 1. Mke drwing of the resistor network. 2. Identify groups of resistors connected in series or prllel. 3. Identify the trget vriles. They could include the equivlent resistnce of the network, the potentil difference cross ech resistor, or the current through ech resistor. EXECUTE the solution s follows: 1. Use Eq. (26.1) or (26.2), respectively, to find the equivlent resistnce for series or prllel comintions. 2. If the network is more complex, try reducing it to series nd prllel comintions. For exmple, in Fig. 26.1c we first replce the prllel comintion of nd with its equivlent resistnce; R 2 R 3 this then forms series comintion with R 1. In Fig. 26.1d, the comintion of R 2 nd R 3 in series forms prllel comintion with R Keep in mind tht the totl potentil difference cross resistors connected in series is the sum of the individul potentil differences. The potentil difference cross resistors connected in prllel is the sme for every resistor nd equls the potentil difference cross the comintion. 4. The current through resistors connected in series is the sme through every resistor nd equls the current through the comintion. The totl current through resistors connected in prllel is the sum of the currents through the individul resistors. EVALUATE your nswer: Check whether your results re consistent. The equivlent resistnce of resistors connected in series should e greter thn tht of ny individul resistor; tht of resistors in prllel should e less thn tht of ny individul resistor. Exmple 26.1 Equivlent resistnce Find the equivlent resistnce of the network in Fig nd the current in ech resistor. The source of emf hs negligile internl resistnce. OLUTION IDENTIFY nd ET UP: This network of three resistors is comintion of series nd prllel resistnces, s in Fig. 26.1c. We determine 26.3 teps in reducing comintion of resistors to single equivlent resistor nd finding the current in ech resistor. () E 18 V, r 0 6 V 4 V c 3 V () (c) (d) (e) (f) Continued

168 854 CHAPTER 26 Direct-Current Circuits the equivlent resistnce of the prllel 6-Æ nd 3-Æ resistors, nd then tht of their series comintion with the 4-Æ resistor: This is the equivlent resistnce R eq of the network s whole. We then find the current in the emf, which is the sme s tht in the 4-Æ resistor. The potentil difference is the sme cross ech of the prllel 6-Æ nd 3-Æ resistors; we use this to determine how the current is divided etween these. EXECUTE: Figures 26.3 nd 26.3c show successive steps in reducing the network to single equivlent resistnce R eq. From Eq. (26.2), the 6-Æ nd 3-Æ resistors in prllel in Fig re equivlent to the single 2-Æ resistor in Fig. 26.3: 1 = 1 R 6 Æ3 Æ 6 Æ 1 3 Æ = 1 2 Æ [Eqution (26.3) gives the sme result.] From Eq. (26.1) the series comintion of this 2-Æ resistor with the 4-Æ resistor is equivlent to the single 6-Æ resistor in Fig. 26.3c. We reverse these steps to find the current in ech resistor of the originl network. In the circuit shown in Fig. 26.3d (identicl to Fig. 26.3c), the current is I = V >R = 118 V2>16 Æ2 = 3 A. o the current in the 4-Æ nd 2-Æ resistors in Fig. 26.3e (identicl to Fig. 26.3) is lso 3 A. The potentil difference V c cross the 2-Æ resistor is therefore V c = IR = 13 A212 Æ2 = 6 V. This potentil difference must lso e 6 V in Fig. 26.3f (identicl to Fig. 26.3). From I = V c >R, the currents in the 6-Æ nd 3-Æ resistors in Fig. 26.3f re respectively 16 V2>16 Æ2 = 1 A nd 16 V2>13 Æ2 = 2 A. EVALUATE: Note tht for the two resistors in prllel etween points c nd in Fig. 26.3f, there is twice s much current through the 3-Æ resistor s through the 6-Æ resistor; more current goes through the pth of lest resistnce, in ccordnce with Eq. (26.4). Note lso tht the totl current through these two resistors is 3 A, the sme s it is through the 4-Æ resistor etween points nd c. Exmple 26.2 eries versus prllel comintions Two identicl light uls, ech with resistnce R = 2 Æ, re connected to source with E = 8V nd negligile internl resistnce. Find the current through ech ul, the potentil difference cross ech ul, nd the power delivered to ech ul nd to the entire network if the uls re connected () in series nd () in prllel. (c) uppose one of the uls urns out; tht is, its filment reks nd current cn no longer flow through it. Wht hppens to the other ul in the series cse? In the prllel cse? OLUTION IDENTIFY nd ET UP: The light uls re just resistors in simple series nd prllel connections (Figs nd 26.4). Once we find the current I through ech ul, we cn find the power delivered to ech ul using Eq. (25.18), P = I 2 R = V 2 >R. EXECUTE: () From Eq. (26.1) the equivlent resistnce of the two uls etween points nd c in Fig is R eq = 2R = 212 Æ2 = 4 Æ. In series, the current is the sme through ech ul: I = V c R eq = 8V 4 Æ = 2A I = V de R nd the power delivered to ech ul is P = I 2 R = 14 A Æ2 = 32 W or P = V de V2 = R 2 Æ = 32 W Both the potentil difference cross ech ul nd the current through ech ul re twice s gret s in the series cse. Hence the power delivered to ech ul is four times greter, nd ech ul is righter. The totl power delivered to the prllel network is P totl = 2P = 64 W, four times greter thn in the series cse. The 26.4 Our sketches for this prolem. () Light uls in series = 8V 2 Æ = 4A ince the uls hve the sme resistnce, the potentil difference is the sme cross ech ul: V = V c = IR = 12 A212 Æ2 = 4V From Eq. (25.18), the power delivered to ech ul is P = I 2 R = 12 A Æ2 = 8W P = V 2 R = V c 2 R 14 V2 = 2 Æ = 8W The totl power delivered to oth uls is P tot = 2P = 16 W. () If the uls re in prllel, s in Fig. 26.4, the potentil difference V de cross ech ul is the sme nd equl to 8 V, the terminl voltge of the source. Hence the current through ech light ul is 2 or () Light uls in prllel

169 26.2 Kirchhoff s Rules When connected to the sme source, two light uls in series (shown t top) drw less power nd glow less rightly thn when they re in prllel (shown t ottom). incresed power compred to the series cse isn t otined for free ; energy is extrcted from the source four times more rpidly in the prllel cse thn in the series cse. If the source is ttery, it will e used up four times s fst. (c) In the series cse the sme current flows through oth uls. If one ul urns out, there will e no current in the circuit, nd neither ul will glow. In the prllel cse the potentil difference cross either ul is unchnged if ul urns out. The current through the functionl ul nd the power delivered to it re unchnged. EVALUATE: Our clcultion isn t completely ccurte, ecuse the resistnce R = V>I of rel light uls depends on the potentil difference V cross the ul. Tht s ecuse the filment resistnce increses with incresing operting temperture nd therefore with incresing V. But uls connected in series cross source do in fct glow less rightly thn when connected in prllel cross the sme source (Fig. 26.5). Test Your Understnding of ection 26.1 uppose ll three of the resistors shown in Fig hve the sme resistnce, so R 1 = R 2 = R 3 = R. Rnk the four rrngements shown in prts ()(d) of Fig in order of their equivlent resistnce, from highest to lowest Kirchhoff s Rules Mny prcticl resistor networks cnnot e reduced to simple series-prllel comintions. Figure 26.6 shows dc power supply with emf E 1 chrging ttery with smller emf E 2 nd feeding current to light ul with resistnce R. Figure 26.6 is ridge circuit, used in mny different types of mesurement nd control systems. (Prolem descries one importnt ppliction of ridge circuit.) To compute the currents in these networks, we ll use the techniques developed y the Germn physicist Gustv Roert Kirchhoff ( ). First, here re two terms tht we will use often. A junction in circuit is point where three or more conductors meet. A loop is ny closed conducting pth. In Fig points nd re junctions, ut points c nd d re not; in Fig the points,, c, nd d re junctions, ut points e nd f re not. The lue lines in Figs nd 26.6 show some possile loops in these circuits. Kirchhoff s rules re the following two sttements: Kirchhoff s junction rule: The lgeric sum of the currents into ny junction is zero. Tht is, I = 0 (junction rule, vlid t ny junction) (26.5) Kirchhoff s loop rule: The lgeric sum of the potentil differences in ny loop, including those ssocited with emfs nd those of resistive elements, must equl zero. Tht is, V = 0 (loop rule, vlid for ny closed loop) (26.6) 26.6 Two networks tht cnnot e reduced to simple series-prllel comintions of resistors. () r 1 E 1 Not junction () r E f e c Loop 2 (2) Loop 1 r 2 E 2 Junction (1) Junction Loop 3 R 1 (3) R 2 R m R Not junction R 3 (4) R 4 d d c

170 856 CHAPTER 26 Direct-Current Circuits 26.7 Kirchhoff s junction rule sttes tht s much current flows into junction s flows out of it. () Kirchhoff s junction rule Junction I 1 I 1 I 2 I 2 The current leving junction equls the current entering it. () Wter-pipe nlogy The flow rte of wter leving the pipe equls the flow rte entering it. The junction rule is sed on conservtion of electric chrge. No chrge cn ccumulte t junction, so the totl chrge entering the junction per unit time must equl the totl chrge leving per unit time (Fig. 26.7). Chrge per unit time is current, so if we consider the currents entering junction to e positive nd those leving to e negtive, the lgeric sum of currents into junction must e zero. It s like T rnch in wter pipe (Fig. 26.7); if you hve totl of 1 liter per minute coming in the two pipes, you cn t hve 3 liters per minute going out the third pipe. We my s well confess tht we used the junction rule (without sying so) in ection 26.1 in the derivtion of Eq. (26.2) for resistors in prllel. The loop rule is sttement tht the electrosttic force is conservtive. uppose we go round loop, mesuring potentil differences cross successive circuit elements s we go. When we return to the strting point, we must find tht the lgeric sum of these differences is zero; otherwise, we could not sy tht the potentil t this point hs definite vlue. ign Conventions for the Loop Rule In pplying the loop rule, we need some sign conventions. Prolem-olving trtegy 26.2 descries in detil how to use these, ut here s quick overview. We first ssume direction for the current in ech rnch of the circuit nd mrk it on digrm of the circuit. Then, strting t ny point in the circuit, we imgine trveling round loop, dding emfs nd IR terms s we come to them. When we trvel through source in the direction from - to, the emf is considered to e positive; when we trvel from to -, the emf is considered to e negtive (Fig. 26.8). When we trvel through resistor in the sme direction s the ssumed current, the IR term is negtive ecuse the current goes in the direction of decresing potentil. When we trvel through resistor in the direction opposite to the ssumed current, the IR term is positive ecuse this represents rise of potentil (Fig. 26.8). Kirchhoff s two rules re ll we need to solve wide vriety of network prolems. Usully, some of the emfs, currents, nd resistnces re known, nd others re unknown. We must lwys otin from Kirchhoff s rules numer of independent equtions equl to the numer of unknowns so tht we cn solve the equtions simultneously. Often the hrdest prt of the solution is not understnding the sic principles ut keeping trck of lgeric signs! 26.8 Use these sign conventions when you pply Kirchhoff s loop rule. In ech prt of the figure Trvel is the direction tht we imgine going round the loop, which is not necessrily the direction of the current. () ign conventions for emfs 1E: Trvel direction from to : () ign conventions for resistors Trvel Trvel Trvel Trvel I I E 2E: Trvel direction from to : E 1IR: Trvel opposite to current direction: R 2IR: Trvel in current direction: R

171 26.2 Kirchhoff s Rules 857 Prolem-olving trtegy 26.2 Kirchhoff s Rules IDENTIFY the relevnt concepts: Kirchhoff s rules re useful for nlyzing ny electric circuit. ET UP the prolem using the following steps: 1. Drw circuit digrm, leving room to lel ll quntities, known nd unknown. Indicte n ssumed direction for ech unknown current nd emf. (Kirchhoff s rules will yield the mgnitudes nd directions of unknown currents nd emfs. If the ctul direction of quntity is opposite to your ssumption, the resulting quntity will hve negtive sign.) 2. As you lel currents, it helpful to use Kirchhoff s junction rule, s in Fig. 26.9, so s to express the currents in terms of s few quntities s possile. 3. Identify the trget vriles. EXECUTE the solution s follows: 1. Choose ny loop in the network nd choose direction (clockwise or counterclockwise) to trvel round the loop s you pply Kirchhoff s loop rule. The direction need not e the sme s ny ssumed current direction. 2. Trvel round the loop in the chosen direction, dding potentil differences lgericlly s you cross them. Use the sign conventions of Fig Equte the sum otined in step 2 to zero in ccordnce with the loop rule. 4. If you need more independent equtions, choose nother loop nd repet steps 13; continue until you hve s mny independent equtions s unknowns or until every circuit element hs een included in t lest one loop. 5. olve the equtions simultneously to determine the unknowns. 6. You cn use the loop-rule ookkeeping system to find the potentil V of ny point with respect to ny other point. trt t nd dd the potentil chnges you encounter in going from to, using the sme sign rules s in step 2. The lgeric sum of these chnges is V = V - V. EVALUATE your nswer: Check ll the steps in your lger. Apply steps 1 nd 2 to loop you hve not yet considered; if the sum of potentil drops isn t zero, you ve mde n error somewhere Applying the junction rule to point reduces the numer of unknown currents from three to two. () Three unknown currents: I 1, I 2, I 3 () Applying the junction rule to point elimintes I 3. r 1 E 1 r 2 E 2 r 1 E 1 r 2 E 2 I 1 I 2 I 3 R 3 I 1 I 2 R 1 R 2 I 1 I 2 I 1 1 I 2 R 3 I 1 I 2 R 1 R 2 Exmple 26.3 A single-loop circuit The circuit shown in Fig contins two tteries, ech with n emf nd n internl resistnce, nd two resistors. Find () the current in the circuit, () the potentil difference V, nd (c) the power output of the emf of ech ttery. OLUTION IDENTIFY nd ET UP: There re no junctions in this single-loop circuit, so we don t need Kirchhoff s junction rule. To pply Kirchhoff s loop rule, we first ssume direction for the current; let s ssume counterclockwise direction s shown in Fig EXECUTE: () trting t nd trveling counterclockwise with the current, we dd potentil increses nd decreses nd equte the sum to zero s in Eq. (26.6): -I14 Æ2-4V- I17 Æ2 12 V - I12 Æ2 - I13 Æ2 = 0 Collecting like terms nd solving for I, we find 8 V = I116 Æ2 nd I = 0.5 A The positive result for I shows tht our ssumed current direction is correct. () To find V, the potentil t with respect to, we strt t nd dd potentil chnges s we go towrd. There re two pths from to ; tking the lower one, we find V = 10.5 A217 Æ2 4V 10.5 A214 Æ2 = 9.5 V Point is t 9.5 V higher potentil thn. All the terms in this sum, including the IR terms, re positive ecuse ech represents n increse in potentil s we go from to. Tking the upper pth, we find V = 12 V A212 Æ A213 Æ2 = 9.5 V Here the IR terms re negtive ecuse our pth goes in the direction of the current, with potentil decreses through the resistors. The results for V re the sme for oth pths, s they must e in order for the totl potentil chnge round the loop to e zero. Continued

172 858 CHAPTER 26 Direct-Current Circuits (c) The power outputs of the emf of the 12-V nd 4-V tteries re P 12V = EI = 112 V210.5 A2 = 6W P 4V = EI = 1-4 V210.5 A2 = -2W The negtive sign in E for the 4-V ttery ppers ecuse the current ctully runs from the higher-potentil side of the ttery to the lower-potentil side. The negtive vlue of P mens tht we re storing energy in tht ttery; the 12-V ttery is rechrging it (if it is in fct rechrgele; otherwise, we re destroying it). EVALUATE: By pplying the expression P = I 2 R to ech of the four resistors in Fig , you cn show tht the totl power dissipted in ll four resistors is 4 W. Of the 6 W provided y the emf of the 12-V ttery, 2 W goes into storing energy in the 4-V ttery nd 4 W is dissipted in the resistnces. The circuit shown in Fig is much like tht used when fully chrged 12-V storge ttery (in cr with its engine running) is used to jump-strt cr with run-down ttery (Fig ). The run-down ttery is slightly rechrged in the process. The 3-Æ nd 7-Æ resistors in Fig represent the resistnces of the jumper cles nd of the conducting pth through the utomoile with the run-down ttery. (The vlues of the resistnces in ctul utomoiles nd jumper cles re considerly lower.) () In this exmple we trvel round the loop in the sme direction s the ssumed current, so ll the IR terms re negtive. The potentil decreses s we trvel from to - through the ottom emf ut increses s we trvel from - to through the top emf. () A rel-life exmple of circuit of this kind. () 2 V 12 V () Ded ttery Live ttery I Trvel 3 V I I 7 V I 4 V 4 V Exmple 26.4 Chrging ttery In the circuit shown in Fig , 12-V power supply with unknown internl resistnce r is connected to run-down rechrgele ttery with unknown emf E nd internl resistnce 1 Æ nd to n indictor light ul of resistnce 3 Æ crrying current of 2 A. The current through the run-down ttery is 1 A in the direction shown. Find r, E, nd the current I through the power supply. OLUTION IDENTIFY nd ET UP: This circuit hs more thn one loop, so we must pply oth the junction nd loop rules. We ssume the direction of the current through the 12-V power supply, nd the polrity of the run-down ttery, to e s shown in Fig There re three trget vriles, so we need three equtions In this circuit power supply chrges run-down ttery nd lights ul. An ssumption hs een mde out the polrity of the emf E of the run-down ttery. Is this ssumption correct? 2 A 3 V (2) (3) E 1 A 1 V I (1) 12 V r EXECUTE: We pply the junction rule, Eq. (26.5), to point : -I 1A 2A= 0 so I = 3A To determine r, we pply the loop rule, Eq. (26.6), to the lrge, outer loop (1): 12 V - 13 A2r - 12 A213 Æ2 = 0 so r = 2 Æ To determine E, we pply the loop rule to the left-hnd loop (2): -E 11 A211 Æ2-12 A213 Æ2 = 0 so E = -5V The negtive vlue for E shows tht the ctul polrity of this emf is opposite to tht shown in Fig As in Exmple 26.3, the ttery is eing rechrged. EVALUATE: Try pplying the junction rule t point insted of point, nd try pplying the loop rule y trveling counterclockwise rther thn clockwise round loop (1). You ll get the sme results for I nd r. We cn check our result for E y using the righthnd loop (3): 12 V - 13 A212 Æ2-11 A211 Æ2 E = 0 which gin gives us E = -5 V. As n dditionl check, we note tht V = V - V equls the voltge cross the 3-Æ resistnce, which is 12 A213 Æ2 = 6V. Going from to y the top rnch, we encounter potentil differences 12 V - 13 A212 Æ2 = 6V, nd going y the middle rnch, we find 1-5 V2 11 A211 Æ2 = 6V. The three wys of getting V ā give the sme results.

173 26.2 Kirchhoff s Rules 859 Exmple 26.5 Power in ttery-chrging circuit In the circuit of Exmple 26.4 (shown in Fig ), find the power delivered y the 12-V power supply nd y the ttery eing rechrged, nd find the power dissipted in ech resistor. OLUTION IDENTIFY nd ET UP: We use the results of ection 25.5, in which we found tht the power delivered from n emf to circuit is EI nd the power delivered to resistor from circuit is V I = I 2 R. We know the vlues of ll relevnt quntities from Exmple EXECUTE: The power output from the emf of the power supply is P s P supply = E supply I supply = 112 V213 A2 = 36 W The power dissipted in the power supply s internl resistnce r is P r-supply = I supply 2 r supply = 13 A Æ2 = 18 W so the power supply s net power output is P net = 36 W - 18 W = 18 W. Alterntively, from Exmple 26.4 the terminl voltge of the ttery is V = 6V, so the net power output is P net = V I supply = 16 V213 A2 = 18 W The power output of the emf E of the ttery eing chrged is P emf = EI ttery = 1-5 V211 A2 = -5W This is negtive ecuse the 1-A current runs through the ttery from the higher-potentil side to the lower-potentil side. (As we mentioned in Exmple 26.4, the polrity ssumed for this ttery in Fig ws wrong.) We re storing energy in the ttery s we chrge it. Additionl power is dissipted in the ttery s internl resistnce; this power is P r-ttery = I ttery 2 r ttery = 11 A Æ2 = 1W The totl power input to the ttery is thus 1W ƒ -5 Wƒ = 6W. Of this, 5 W represents useful energy stored in the ttery; the reminder is wsted in its internl resistnce. The power dissipted in the light ul is P ul = I ul 2 R ul = 12 A Æ2 = 12 W EVALUATE: As check, note tht ll of the power from the supply is ccounted for. Of the 18 W of net power from the power supply, 5 W goes to rechrge the ttery, 1 W is dissipted in the ttery s internl resistnce, nd 12 W is dissipted in the light ul. Exmple 26.6 A complex network Figure shows ridge circuit of the type descried t the eginning of this section (see Fig. 26.6). Find the current in ech resistor nd the equivlent resistnce of the network of five resistors. OLUTION IDENTIFY nd ET UP: This network is neither series comintion nor prllel comintion. Hence we must use Kirchhoff s rules to find the vlues of the trget vriles. There re five unknown currents, ut y pplying the junction rule to junctions nd, we cn represent them in terms of three unknown currents I 1, I 2, nd, s shown in Fig I A network circuit with severl resistors. (1) 13 V I 1 I 2 (2) I 1 (3) I 2 1 V 1 V I 1 I 3 I 3 c 1 V d 1 V 2 V I 2 I 3 EXECUTE: We pply the loop rule to the three loops shown: 13 V - I 1 11 Æ2-1I 1 - I Æ2 = 0 (1) -I 2 11 Æ2-1I 2 I Æ2 13 V = 0 (2) -I 1 11 Æ2 - I 3 11 Æ2 I 2 11 Æ2 = 0 (3) One wy to solve these simultneous equtions is to solve Eq. (3) for I 2, otining I 2 = I 1 I 3, nd then sustitute this expression into Eq. (2) to eliminte I 2. We then hve 13 V = I 1 12 Æ2 - I 3 11 Æ2 (1 ) 13 V = I 1 13 Æ2 I 3 15 Æ2 (2 ) Now we cn eliminte I 3 y multiplying Eq. (1 ) y 5 nd dding the two equtions. We otin 78 V = I Æ2 I 1 = 6A We sustitute this result into Eq. (1 ) to otin I 3 = -1A, nd from Eq. (3) we find I 2 = 5A. The negtive vlue of I 3 tells us tht its direction is opposite to the direction we ssumed. The totl current through the network is I 1 I 2 = 11 A, nd the potentil drop cross it is equl to the ttery emf, 13 V. The equivlent resistnce of the network is therefore R eq = 13 V 11 A = 1.2 Æ EVALUATE: You cn check our results for I 1, I 2, nd I 3 y sustituting them ck into Eqs. (1)(3). Wht do you find?

174 860 CHAPTER 26 Direct-Current Circuits Exmple 26.7 A potentil difference in complex network In the circuit of Exmple 26.6 (Fig ), find the potentil difference V. OLUTION IDENTIFY nd ET UP: Our trget vrile V = V - V is the potentil t point with respect to point. To find it, we strt t point nd follow pth to point, dding potentil rises nd drops s we go. We cn follow ny of severl pths from to ; the result must e the sme for ll such pths, which gives us wy to check our result. EXECUTE: The simplest pth is through the center 1-Æ resistor. In Exmple 26.6 we found I 3 = -1A, showing tht the ctul current direction through this resistor is from right to left. Thus, s we go from to, there is drop of potentil with mgnitude ƒ I 3 ƒ R = 11 A211 Æ2 = 1V. Hence V = -1V, nd the potentil t is 1 V less thn t point. EVALUATE: To check our result, let s try pth from to tht goes through the lower two resistors. The currents through these re nd so I 2 I 3 = 5A 1-1 A2 = 4A nd I 1 - I 3 = 6A- 1-1 A2 = 7A V = -14 A212 Æ2 17 A211 Æ2 = -1V You cn confirm this result using some other pths from to This mmeter (top) nd voltmeter (ottom) re oth d Arsonvl glvnometers. The difference hs to do with their internl connections (see Fig ). Test Your Understnding of ection 26.2 utrct Eq. (1) from Eq. (2) in Exmple To which loop in Fig does this eqution correspond? Would this eqution hve simplified the solution of Exmple 26.6? 26.3 Electricl Mesuring Instruments A d Arsonvl glvnometer, showing pivoted coil with ttched pointer, permnent mgnet supplying mgnetic field tht is uniform in mgnitude, nd spring to provide restoring torque, which opposes mgnetic-field torque. Mgnetic-field torque tends to push pointer wy from zero. 0 5 pring torque tends to push pointer towrd zero. 10 We ve een tlking out potentil difference, current, nd resistnce for two chpters, so it s out time we sid something out how to mesure these quntities. Mny common devices, including cr instrument pnels, ttery chrgers, nd inexpensive electricl instruments, mesure potentil difference (voltge), current, or resistnce using d Arsonvl glvnometer (Fig ). In the following discussion we ll often cll it just meter. A pivoted coil of fine wire is plced in the mgnetic field of permnent mgnet (Fig ). Attched to the coil is spring, similr to the hirspring on the lnce wheel of wtch. In the equilirium position, with no current in the coil, the pointer is t zero. When there is current in the coil, the mgnetic field exerts torque on the coil tht is proportionl to the current. (We ll discuss this mgnetic interction in detil in Chpter 27.) As the coil turns, the spring exerts restoring torque tht is proportionl to the ngulr displcement. Thus the ngulr deflection of the coil nd pointer is directly proportionl to the coil current, nd the device cn e clirted to mesure current. The mximum deflection, typiclly 90 or so, is clled full-scle deflection. The essentil electricl chrcteristics of the meter re the current I fs required for full-scle deflection (typiclly on the order of 10 ma to 10 ma) nd the resistnce R c of the coil (typiclly on the order of 10 to 1000 Æ). The meter deflection is proportionl to the current in the coil. If the coil oeys Ohm s lw, the current is proportionl to the potentil difference etween the terminls of the coil, nd the deflection is lso proportionl to this potentil difference. For exmple, consider meter whose coil hs resistnce R c = 20.0 Æ nd tht deflects full scle when the current in its coil is I fs = 1.00 ma. The corresponding potentil difference for full-scle deflection is V = I fs R c = * 10-3 A Æ2 = V pring Permnent mgnet oft-iron core Mgnetic field Pivoted coil Ammeters A current-mesuring instrument is usully clled n mmeter (or millimmeter, micrommeter, nd so forth, depending on the rnge). An mmeter lwys mesures the current pssing through it. An idel mmeter, discussed in ection 25.4, would hve zero resistnce, so including it in rnch of circuit would not

175 26.3 Electricl Mesuring Instruments 861 ffect the current in tht rnch. Rel mmeters lwys hve some finite resistnce, ut it is lwys desirle for n mmeter to hve s little resistnce s possile. We cn dpt ny meter to mesure currents tht re lrger thn its full-scle reding y connecting resistor in prllel with it (Fig ) so tht some of the current ypsses the meter coil. The prllel resistor is clled shunt resistor or simply shunt, denoted s R sh. uppose we wnt to mke meter with full-scle current I fs nd coil resistnce R c into n mmeter with full-scle reding I. To determine the shunt resistnce R sh needed, note tht t full-scle deflection the totl current through the prllel comintion is I, the current through the coil of the meter is I fs, nd the current through the shunt is the difference I - I fs. The potentil difference V is the sme for oth pths, so I fs R c = 1I - I fs 2R sh (for n mmeter) (26.7) Using the sme meter to mesure () current nd () voltge. () I Moving-coil mmeter R c R sh I () V I Moving-coil voltmeter R c R s Circuit element V I Exmple 26.8 Designing n mmeter Wht shunt resistnce is required to mke the 1.00-mA, 20.0-Æ EVALUATE: It s useful to consider the equivlent resistnce R eq of meter descried ove into n mmeter with rnge of 0 to the mmeter s whole. From Eq. (26.2), 50.0 ma? R eq = = R c R sh 20.0 Æ Æ OLUTION = Æ IDENTIFY nd ET UP: ince the meter is eing used s n mmeter, its internl connections re s shown in Fig Our trget The shunt resistnce is so smll in comprison to the coil resistnce tht the equivlent resistnce is very nerly equl to the shunt vrile is the shunt resistnce R sh, which we will find using Eq. (26.7). The mmeter must hndle mximum current resistnce. The result is n mmeter with low equivlent resistnce nd the desired mA rnge. At full-scle deflection, I = 50.0 * 10-3 A. The coil resistnce is R c = 20.0 Æ, nd the meter shows full-scle deflection when the current through the coil I = I = 50.0 ma, the current through the glvnometer is 1.00 is I fs = 1.00 * 10-3 A. ma, the current through the shunt resistor is 49.0 ma, nd V If the current I is less thn 50.0 ma, the coil current nd the deflection re proportionlly less. EXECUTE: olving Eq. (26.7) for R sh, we find = V. I fs R c * 10-3 A Æ2 R sh = = I - I fs 50.0 * 10-3 A * 10-3 A = Æ Voltmeters This sme sic meter my lso e used to mesure potentil difference or voltge. A voltge-mesuring device is clled voltmeter. A voltmeter lwys mesures the potentil difference etween two points, nd its terminls must e connected to these points. (Exmple 25.6 in ection 25.4 descried wht cn hppen if voltmeter is connected incorrectly.) As we discussed in ection 25.4, n idel voltmeter would hve infinite resistnce, so connecting it etween two points in circuit would not lter ny of the currents. Rel voltmeters lwys hve finite resistnce, ut voltmeter should hve lrge enough resistnce tht connecting it in circuit does not chnge the other currents pprecily. For the meter descried in Exmple 26.8 the voltge cross the meter coil t full-scle deflection is only I fs R c = * 10-3 A Æ2 = V. We cn extend this rnge y connecting resistor R s in series with the coil (Fig ). Then only frction of the totl potentil difference ppers cross the coil itself, nd the reminder ppers cross R s. For voltmeter with full-scle reding V, we need series resistor in Fig such tht R s V V = I fs 1R c R s 2 (for voltmeter) (26.8) Appliction Electromyogrphy A fine needle contining two electrodes is eing inserted into muscle in this ptient s hnd. By using sensitive voltmeter to mesure the potentil difference etween these electrodes, physicin cn proe the muscle s electricl ctivity. This is n importnt technique for dignosing neurologicl nd neuromusculr diseses.

176 862 CHAPTER 26 Direct-Current Circuits Exmple 26.9 Designing voltmeter Wht series resistnce is required to mke the 1.00-mA, 20.0-Æ EVALUATE: At full-scle deflection, V = 10.0 V, the voltge meter descried ove into voltmeter with rnge of 0 to 10.0 V? cross the meter is V, the voltge cross R s is 9.98 V, nd the current through the voltmeter is A. Most of the voltge OLUTION ppers cross the series resistor. The meter s equivlent resistnce is desirly high R eq 20.0 Æ9980 Æ 10,000 Æ. uch IDENTIFY nd ET UP: ince this meter is eing used s voltmeter, its internl connections re s shown in Fig Our meter is clled 1000 ohms-per-volt meter, referring to the rtio of resistnce to full-scle deflection. In norml opertion the current through the circuit element eing mesured (I in Fig ) trget vrile is the series resistnce R s. The mximum llowle voltge cross the voltmeter is V = 10.0 V. We wnt this to occur when the current through the coil is I fs = 1.00 * 10-3 is much greter thn A, nd the resistnce etween points A. nd in the circuit is much less thn 10,000 Æ. The voltmeter Our trget vrile is the series resistnce R s, which we find using drws off only smll frction of the current nd thus disturs the Eq. (26.8). circuit eing mesured only slightly. EXECUTE: From Eq. (26.8), R s = V V I fs - R c = 10.0 V Æ=9980 Æ A ActivPhysics 12.4: Using Ammeters nd Voltmeters Ammeters nd Voltmeters in Comintion A voltmeter nd n mmeter cn e used together to mesure resistnce nd power. The resistnce R of resistor equls the potentil difference V etween its terminls divided y the current I; tht is, R = V >I. The power input P to ny circuit element is the product of the potentil difference cross it nd the current through it: P = V I. In principle, the most strightforwrd wy to mesure R or P is to mesure V nd I simultneously. With prcticl mmeters nd voltmeters this isn t quite s simple s it seems. In Fig , mmeter A reds the current I in the resistor R. Voltmeter V, however, reds the sum of the potentil difference V cross the resistor nd the potentil difference V c cross the mmeter. If we trnsfer the voltmeter terminl from c to, s in Fig , then the voltmeter reds the potentil difference V correctly, ut the mmeter now reds the sum of the current I in the resistor nd the current I V in the voltmeter. Either wy, we hve to correct the reding of one instrument or the other unless the corrections re smll enough to e negligile Ammetervoltmeter method for mesuring resistnce. () () R R A c R A A c I I I V V V R V R V Exmple Mesuring resistnce I The voltmeter in the circuit of Fig reds 12.0 V nd the mmeter reds A. The meter resistnces re R V = 10,000 Æ (for the voltmeter) nd R A = 2.00 Æ (for the mmeter). Wht re the resistnce R nd the power dissipted in the resistor? OLUTION IDENTIFY nd ET UP: The mmeter reds the current I = A through the resistor, nd the voltmeter reds the potentil difference etween nd c. If the mmeter were idel (tht is, if R A = 0), there would e zero potentil difference etween nd c, the voltmeter reding V = 12.0 V would e equl to the potentil difference V cross the resistor, nd the resistnce would simply e equl to R = V>I = V2> A2 = 120 Æ. The mmeter is not idel, however (its resistnce is R A = 2.00 Æ), so the voltmeter reding V is ctully the sum of the potentil differences V c (cross the mmeter) nd V (cross the resistor). We use Ohm s lw to find the voltge from the known current nd V c

177 26.3 Electricl Mesuring Instruments 863 mmeter resistnce. Then we solve for nd the resistnce R. Given these, we re le to clculte the power P into the resistor. EXECUTE: From Ohm s lw, V c = IR A = A Æ2 = V nd V = IR. The sum of these is V = 12.0 V, so the potentil difference cross the resistor is V = V - V c = V V2 = 11.8 V. Hence the resistnce is V R = V I = 11.8 V A = 118 Æ The power dissipted in this resistor is P = V I = V A2 = 1.18 W EVALUATE: You cn confirm this result for the power y using the lterntive formul P = I 2 R. Do you get the sme nswer? Exmple Mesuring resistnce II uppose the meters of Exmple re connected to different resistor s shown in Fig , nd the redings otined on the meters re the sme s in Exmple Wht is the vlue of this new resistnce R, nd wht is the power dissipted in the resistor? OLUTION IDENTIFY nd ET UP: In Exmple the mmeter red the ctul current through the resistor, ut the voltmeter reding ws not the sme s the potentil difference cross the resistor. Now the sitution is reversed: The voltmeter reding V = 12.0 V shows the ctul potentil difference V cross the resistor, ut the mmeter reding I A = A is not equl to the current I through the resistor. Applying the junction rule t in Fig shows tht I A = I I V, where I V is the current through the voltmeter. We find I V from the given vlues of V nd the voltmeter resistnce R V, nd we use this vlue to find the resistor current I. We then determine the resistnce R from I nd the voltmeter reding, nd clculte the power s in Exmple EXECUTE: We hve I V = V>R V = V2>110,000 Æ2 = 1.20 ma. The ctul current I in the resistor is I = I A - I V = A A = A, nd the resistnce is R = V I The power dissipted in the resistor is = 12.0 V A = 121 Æ P = V I = V A2 = 1.19 W EVALUATE: Hd the meters een idel, our results would hve een R = 12.0 V>0.100 A = 120 Æ nd P = VI = V2 * A2 = 1.2 W oth here nd in Exmple The ctul (correct) results re not too different in either cse. Tht s ecuse the mmeter nd voltmeter re nerly idel: Compred with the resistnce R under test, the mmeter resistnce R A is very smll nd the voltmeter resistnce R V is very lrge. Under these conditions, treting the meters s idel yields pretty good results; ccurte work requires clcultions s in these two exmples. Ohmmeters An lterntive method for mesuring resistnce is to use d Arsonvl meter in n rrngement clled n ohmmeter. It consists of meter, resistor, nd source (often flshlight ttery) connected in series (Fig ). The resistnce R to e mesured is connected etween terminls x nd y. The series resistnce R s is vrile; it is djusted so tht when terminls x nd y re short-circuited (tht is, when R = 0), the meter deflects full scle. When nothing is connected to terminls x nd y, so tht the circuit etween x nd y is open (tht is, when R q), there is no current nd hence no deflection. For ny intermedite vlue of R the meter deflection depends on the vlue of R, nd the meter scle cn e clirted to red the resistnce R directly. Lrger currents correspond to smller resistnces, so this scle reds ckwrd compred to the scle showing the current. In situtions in which high precision is required, instruments contining d Arsonvl meters hve een supplnted y electronic instruments with direct digitl redouts. Digitl voltmeters cn e mde with extremely high internl resistnce, of the order of 100 MÆ. Figure shows digitl multimeter, n instrument tht cn mesure voltge, current, or resistnce over wide rnge Ohmmeter circuit. The resistor hs vrile resistnce, s is indicted y the rrow through the resistor symol. To use the ohmmeter, first connect x directly to y nd djust R s until the meter reds zero. Then connect x nd y cross the resistor R nd red the scle. x R s ` 0 R E y R s The Potentiometer The potentiometer is n instrument tht cn e used to mesure the emf of source without drwing ny current from the source; it lso hs numer of other useful pplictions. Essentilly, it lnces n unknown potentil difference ginst n djustle, mesurle potentil difference.

178 864 CHAPTER 26 Direct-Current Circuits This digitl multimeter cn e used s voltmeter (red rc), mmeter (yellow rc), or ohmmeter (green rc) A potentiometer. () Potentiometer circuit E 1 The principle of the potentiometer is shown schemticlly in Fig A resistnce wire of totl resistnce R is permnently connected to the terminls of source of known emf E 1. A sliding contct c is connected through the glvnometer G to second source whose emf E 2 is to e mesured. As contct c is moved long the resistnce wire, the resistnce R c etween points c nd vries; if the resistnce wire is uniform, R c is proportionl to the length of wire etween c nd. To determine the vlue of E 2, contct c is moved until position is found t which the glvnometer shows no deflection; this corresponds to zero current pssing through E 2. With I 2 = 0, Kirchhoff s loop rule gives E 2 = IR c With I 2 = 0, the current I produced y the emf E 1 hs the sme vlue no mtter wht the vlue of the emf E 2. We clirte the device y replcing E 2 y source of known emf; then ny unknown emf E 2 cn e found y mesuring the length of wire c for which I 2 = 0. Note tht for this to work, V must e greter thn E 2. The term potentiometer is lso used for ny vrile resistor, usully hving circulr resistnce element nd sliding contct controlled y rotting shft nd kno. The circuit symol for potentiometer is shown in Fig I c I I I 2 0 G r G () Circuit symol for potentiometer (vrile resistor) E 2, r I Test Your Understnding of ection 26.3 You wnt to mesure the current through nd the potentil difference cross the 2-Æ resistor shown in Fig (Exmple 26.6 in ection 26.2). () How should you connect n mmeter nd voltmeter to do this? (i) mmeter nd voltmeter oth in series with the 2-Æ resistor; (ii) mmeter in series with the 2-Æ resistor nd voltmeter connected etween points nd d; (iii) mmeter connected etween points nd d nd voltmeter in series with the 2-Æ resistor; (iv) mmeter nd voltmeter oth connected etween points nd d. () Wht resistnces should these meters hve? (i) Ammeter nd voltmeter resistnces should oth e much greter thn 2 Æ; (ii) mmeter resistnce should e much greter thn 2 Æ nd voltmeter resistnce should e much less thn 2 Æ; (iii) mmeter resistnce should e much less thn 2 Æ nd voltmeter resistnce should e much greter thn 2 Æ; (iv) mmeter nd voltmeter resistnces should oth e much less thn 2 Æ R-C Circuits In the circuits we hve nlyzed up to this point, we hve ssumed tht ll the emfs nd resistnces re constnt (time independent) so tht ll the potentils, currents, nd powers re lso independent of time. But in the simple ct of chrging or dischrging cpcitor we find sitution in which the currents, voltges, nd powers do chnge with time. Mny devices incorporte circuits in which cpcitor is lterntely chrged nd dischrged. These include flshing trffic lights, utomoile turn signls, nd electronic flsh units. Understnding wht hppens in such circuits is thus of gret prcticl importnce. Chrging Cpcitor Figure shows simple circuit for chrging cpcitor. A circuit such s this tht hs resistor nd cpcitor in series is clled n R-C circuit. We idelize the ttery (or power supply) to hve constnt emf E nd zero internl resistnce 1r = 02, nd we neglect the resistnce of ll the connecting conductors. We egin with the cpcitor initilly unchrged (Fig ); then t some initil time t = 0 we close the switch, completing the circuit nd permitting current round the loop to egin chrging the cpcitor (Fig ). For ll prcticl purposes, the current egins t the sme instnt in every conducting prt of the circuit, nd t ech instnt the current is the sme in every prt.

179 26.4 R-C Circuits 865 CAUTION Lowercse mens time-vrying Up to this point we hve een working with constnt potentil differences (voltges), currents, nd chrges, nd we hve used cpitl letters V, I, nd Q, respectively, to denote these quntities. To distinguish etween quntities tht vry with time nd those tht re constnt, we will use lowercse letters v, i, nd q for time-vrying voltges, currents, nd chrges, respectively. We suggest tht you follow this sme convention in your own work. Becuse the cpcitor in Fig is initilly unchrged, the potentil difference v c cross it is zero t t = 0. At this time, from Kirchhoff s loop lw, the voltge v cross the resistor R is equl to the ttery emf E. The initil 1t = 02 current through the resistor, which we will cll I 0, is given y Ohm s lw: I 0 = v >R = E>R. As the cpcitor chrges, its voltge v c increses nd the potentil difference v cross the resistor decreses, corresponding to decrese in current. The sum of these two voltges is constnt nd equl to E. After long time the cpcitor ecomes fully chrged, the current decreses to zero, nd the potentil difference v cross the resistor ecomes zero. Then the entire ttery emf E ppers cross the cpcitor nd v c = E. Let q represent the chrge on the cpcitor nd i the current in the circuit t some time t fter the switch hs een closed. We choose the positive direction for the current to correspond to positive chrge flowing onto the left-hnd cpcitor plte, s in Fig The instntneous potentil differences v nd re v c Using these in Kirchhoff s loop rule, we find (26.9) The potentil drops y n mount ir s we trvel from to nd y q>c s we trvel from to c. olving Eq. (26.9) for i, we find (26.10) At time t = 0, when the switch is first closed, the cpcitor is unchrged, nd so q = 0. ustituting q = 0 into Eq. (26.10), we find tht the initil current I 0 is given y I 0 = E>R, s we hve lredy noted. If the cpcitor were not in the circuit, the lst term in Eq. (26.10) would not e present; then the current would e constnt nd equl to E>R. As the chrge q increses, the term q>rc ecomes lrger nd the cpcitor chrge pproches its finl vlue, which we will cll Q f. The current decreses nd eventully ecomes zero. When i = 0, Eq. (26.10) gives Q f v = ir v c = q C E - ir - q C = 0 i = E R - q RC E R = Q f RC Q f = CE (26.11) Note tht the finl chrge does not depend on R. Figure shows the current nd cpcitor chrge s functions of time. At the instnt the switch is closed 1t = 02, the current jumps from zero to its initil vlue I 0 = E>R; fter tht, it grdully pproches zero. The cpcitor chrge strts t zero nd grdully pproches the finl vlue given y Eq. (26.11), Q f = CE. We cn derive generl expressions for the chrge q nd current i s functions of time. With our choice of the positive direction for current (Fig ), i equls the rte t which positive chrge rrives t the left-hnd (positive) Chrging cpcitor. () Just efore the switch is closed, the chrge q is zero. () When the switch closes (t t = 0), the current jumps from zero to E>R. As time psses, q pproches Q f nd the current i pproches zero. () Cpcitor initilly unchrged witch E open i 5 0 R E q 5 0 C () Chrging the cpcitor O i R witch closed 1q i C 2q c c When the switch is closed, the chrge on the cpcitor increses over time while the current decreses Current i nd cpcitor chrge q s functions of time for the circuit of Fig The initil current is I 0 nd the initil cpcitor chrge is zero. The current symptoticlly pproches zero, nd the cpcitor chrge symptoticlly pproches finl vlue of Q f. () Grph of current versus time for chrging cpcitor i I 0 I 0/2 I 0/e Q f Q f/2 O The current decreses exponentilly with time s the cpcitor chrges. RC () Grph of cpcitor chrge versus time for chrging cpcitor q Q f/e RC The chrge on the cpcitor increses exponentilly with time towrd the finl vlue Q f. t t

180 866 CHAPTER 26 Direct-Current Circuits plte of the cpcitor, so i = dq>dt. Mking this sustitution in Eq. (26.10), we hve We cn rerrnge this to dq dt = E R - q RC = - 1 1q - CE2 RC nd then integrte oth sides. We chnge the integrtion vriles to q nd t so tht we cn use q nd t for the upper limits. The lower limits re q =0 nd t =0: dq L0 q -CE = - L0 When we crry out the integrtion, we get q dq q - CE = - dt RC ln q - CE -CE = - t RC Exponentiting oth sides (tht is, tking the inverse logrithm) nd solving for q, we find q - CE = e -t/rc -CE t dt RC PhET: Circuit Construction Kit (ACDC) PhET: Circuit Construction Kit (DC Only) ActivPhysics 12.6: Cpcitnce ActivPhysics 12.7: eries nd Prllel Cpcitors ActivPhysics 12.8: Circuit Time Constnts Appliction Pcemkers nd Cpcitors This x-ry imge shows pcemker implnted in ptient with mlfunctioning sinotril node, the prt of the hert tht genertes the electricl signl to trigger hertets. The pcemker circuit contins ttery, cpcitor, nd computer-controlled switch. To mintin regulr eting, once per second the switch dischrges the cpcitor nd sends n electricl pulse long the led to the hert. The switch then flips to llow the cpcitor to rechrge for the next pulse. Pcemker Lung Electricl led Hert Lung q = CE11 - e -t/rc 2 = Q f 11 - e -t/rc 2 (R-C circuit, chrging cpcitor) The instntneous current i is just the time derivtive of Eq. (26.12): i = dq dt = E R e-t/rc = I 0 e -t/rc (R-C circuit, chrging cpcitor) (26.12) (26.13) The chrge nd current re oth exponentil functions of time. Figure is grph of Eq. (26.13) nd Fig is grph of Eq. (26.12). Time Constnt After time equl to RC, the current in the R-C circuit hs decresed to 1>e (out 0.368) of its initil vlue. At this time, the cpcitor chrge hs reched 11-1>e2 = of its finl vlue Q f = CE. The product RC is therefore mesure of how quickly the cpcitor chrges. We cll RC the time constnt, or the relxtion time, of the circuit, denoted y t: t = RC (time constnt for R-C circuit) (26.14) When t is smll, the cpcitor chrges quickly; when it is lrger, the chrging tkes more time. If the resistnce is smll, it s esier for current to flow, nd the cpcitor chrges more quickly. If R is in ohms nd C in frds, t is in seconds. In Fig the horizontl xis is n symptote for the curve. trictly speking, i never ecomes exctly zero. But the longer we wit, the closer it gets. After time equl to 10RC, the current hs decresed to of its initil vlue. imilrly, the curve in Fig pproches the horizontl dshed line leled Q f s n symptote. The chrge q never ttins exctly this vlue, ut fter time equl to 10RC, the difference etween q nd Q f is only of Q f. We invite you to verify tht the product RC hs units of time.

181 26.4 R-C Circuits 867 Dischrging Cpcitor Now suppose tht fter the cpcitor in Fig hs cquired chrge Q 0, we remove the ttery from our R-C circuit nd connect points nd c to n open switch (Fig ). We then close the switch nd t the sme instnt reset our stopwtch to t = 0; t tht time, q = Q 0. The cpcitor then dischrges through the resistor, nd its chrge eventully decreses to zero. Agin let i nd q represent the time-vrying current nd chrge t some instnt fter the connection is mde. In Fig we mke the sme choice of the positive direction for current s in Fig Then Kirchhoff s loop rule gives Eq. (26.10) ut with E = 0; tht is, (26.15) The current i is now negtive; this is ecuse positive chrge q is leving the lefthnd cpcitor plte in Fig , so the current is in the direction opposite to tht shown in the figure. At time t = 0, when q = Q 0, the initil current is I 0 = -Q 0 >RC. To find q s function of time, we rerrnge Eq. (26.15), gin chnge the nmes of the vriles to q nd t, nd integrte. This time the limits for q re Q 0 to q. We get The instntneous current i is the derivtive of this with respect to time: i = dq dt i = dq dt dq = - 1 LQ 0 q RC L0 ln q = - t Q 0 RC q = Q 0 e -t/rc (R-C circuit, dischrging cpcitor) = - Q 0 RC e-t/rc = I 0 e -t/rc q = - q RC t dt (R-C circuit, dischrging cpcitor) (26.16) (26.17) We grph the current nd the chrge in Fig ; oth quntities pproch zero exponentilly with time. Compring these results with Eqs. (26.12) nd (26.13), we note tht the expressions for the current re identicl, prt from the sign of I 0. The cpcitor chrge pproches zero symptoticlly in Eq. (26.16), while the difference etween q nd Q pproches zero symptoticlly in Eq. (26.12). Energy considertions give us dditionl insight into the ehvior of n R-C circuit. While the cpcitor is chrging, the instntneous rte t which the ttery delivers energy to the circuit is P = Ei. The instntneous rte t which electricl energy is dissipted in the resistor is i 2 R, nd the rte t which energy is stored in the cpcitor is iv c = iq>c. Multiplying Eq. (26.9) y i, we find Ei = i 2 R iq C (26.18) This mens tht of the power Ei supplied y the ttery, prt 1i 2 R2 is dissipted in the resistor nd prt 1iq>C2 is stored in the cpcitor. The totl energy supplied y the ttery during chrging of the cpcitor equls the ttery emf E multiplied y the totl chrge Q f, or EQ f. The totl energy stored in the cpcitor, from Eq. (24.9), is Q f E>2. Thus, of the energy supplied y the ttery, exctly hlf is stored in the cpcitor, nd the other hlf is dissipted in the resistor. This hlf-nd-hlf division of energy doesn t depend on C, R, or E. You cn verify this result y tking the integrl over time of ech of the power quntities in Eq. (26.18) (see Prolem 26.88) Dischrging cpcitor. () Before the switch is closed t time t = 0, the cpcitor chrge is Q 0 nd the current is zero. () At time t fter the switch is closed, the cpcitor chrge is q nd the current is i. The ctul current direction is opposite to the direction shown; i is negtive. After long time, q nd i oth pproch zero. () Cpcitor initilly chrged i 0 R witch open Q 0 Q 0 C () Dischrging the cpcitor witch closed I 0 Q 0 Q 0/2 Q 0/e O i q q R C i c c When the switch is closed, the chrge on the cpcitor nd the current oth decrese over time Current i nd cpcitor chrge q s functions of time for the circuit of Fig The initil current is I 0 nd the initil cpcitor chrge is Q 0. Both i nd q symptoticlly pproch zero. () Grph of current versus time for dischrging cpcitor i RC t O I 0/e I 0/2 The current decreses exponentilly s the cpcitor dischrges. (The current is negtive ecuse its direction is opposite to tht in Fig ) () Grph of cpcitor chrge versus time for dischrging cpcitor q The chrge on the cpcitor decreses exponentilly s the cpcitor dischrges. RC t

182 868 CHAPTER 26 Direct-Current Circuits Exmple Chrging cpcitor A 10-MÆ resistor is connected in series with 1.0-mF cpcitor nd ttery with emf 12.0 V. Before the switch is closed t time t = 0, the cpcitor is unchrged. () Wht is the time constnt? EXECUTE: () From Eq. (26.14), t = RC = 110 * 10 6 Æ211.0 * 10-6 F2 = 10 s () Wht frction of the finl chrge Q f is on the cpcitor t () From Eq. (26.12), t = 46 s? (c) Wht frction of the initil current I 0 is still flowing q t t = 46 s? = 1 - e -t>rc = 1 - e -146 s2>110 s2 = 0.99 Q f OLUTION IDENTIFY nd ET UP: This is the sme sitution s shown in Fig , with R = 10 MÆ, C = 1.0 mf, nd E = 12.0 V. The chrge q nd current i vry with time s shown in Fig Our trget vriles re () the time constnt t, () the rtio q>q f t t = 46 s, nd (c) the rtio i>i 0 t t = 46 s. Eqution (26.14) gives t. For cpcitor eing chrged, Eq. (26.12) gives q nd Eq. (26.13) gives i. (c) From Eq. (26.13), i I 0 = e -t>rc = e -146 s2>110 s2 = EVALUATE: After 4.6 time constnts the cpcitor is 99% chrged nd the chrging current hs decresed to 1.0% of its initil vlue. The circuit would chrge more rpidly if we reduced the time constnt y using smller resistnce. Exmple Dischrging cpcitor The resistor nd cpcitor of Exmple re reconnected s shown in Fig The cpcitor hs n initil chrge of 5.0 mc nd is dischrged y closing the switch t t = 0. () At wht time will the chrge e equl to 0.50 mc? () Wht is the current t this time? OLUTION IDENTIFY nd ET UP: Now the cpcitor is eing dischrged, so q nd i vry with time s in Fig , with Q 0 = 5.0 * 10-6 C. Agin we hve RC t = 10 s. Our trget vriles re () the vlue of t t which q = 0.50 mc nd () the vlue of i t this time. We first solve Eq. (26.16) for t, nd then solve Eq. (26.17) for i. EXECUTE: () olving Eq. (26.16) for the time t gives t = -RC ln q 0.50 mc = -110 s2ln = 23 s = 2.3t Q mc () From Eq. (26.17), with Q 0 = 5.0 mc = 5.0 * 10-6 C, i = - Q 0 RC e-t>rc = * 10-6 C e -2.3 = -5.0 * 10-8 A 10 s EVALUATE: The current in prt () is negtive ecuse i hs the opposite sign when the cpcitor is dischrging thn when it is chrging. Note tht we could hve voided evluting e -t>rc y noticing tht t the time in question, q = 0.10Q 0 ; from Eq. (26.16) this mens tht e -t>rc = Test Your Understnding of ection 26.4 The energy stored in cpcitor is equl to q 2 /2C. When cpcitor is dischrged, wht frction of the initil energy remins fter n elpsed time of one time constnt? (i) 1/e; (ii) 1/e 2 ; (iii) 1-1/e; (iv) 11-1/e2 2 ; (v) nswer depends on how much energy ws stored initilly Power Distriution ystems We conclude this chpter with rief discussion of prcticl household nd utomotive electric-power distriution systems. Automoiles use direct-current (dc) systems, while nerly ll household, commercil, nd industril systems use lternting current (c) ecuse of the ese of stepping voltge up nd down with trnsformers. Most of the sme sic wiring concepts pply to oth. We ll tlk out lternting-current circuits in greter detil in Chpter 31. The vrious lmps, motors, nd other pplinces to e operted re lwys connected in prllel to the power source (the wires from the power compny for houses, or from the ttery nd lterntor for cr). If pplinces were connected in series, shutting one pplince off would shut them ll off (see Exmple 26.2 in ection 26.1). Figure shows the sic ide of house wiring. One side of the line, s the pir of conductors is clled, is clled the neutrl side; it is lwys connected to

183 26.5 Power Distriution ystems chemtic digrm of prt of house wiring system. Only two rnch circuits re shown; n ctul system might hve four to thirty rnch circuits. Lmps nd pplinces my e plugged into the outlets. The grounding wires, which normlly crry no current, re not shown. From power compny Meter Min fuse Ground Fuse Outlets Fuse Outlets witch witch Light Light Hot line Neutrl line Hot line Neutrl line ground t the entrnce pnel. For houses, ground is n ctul electrode driven into the erth (which is usully good conductor) or sometimes connected to the household wter pipes. Electricins spek of the hot side nd the neutrl side of the line. Most modern house wiring systems hve two hot lines with opposite polrity with respect to the neutrl. We ll return to this detil lter. Household voltge is nominlly 120 V in the United ttes nd Cnd, nd often 240 V in Europe. (For lternting current, which vries sinusoidlly with time, these numers represent the root-men-squre voltge, which is 1> 12 times the pek voltge. We ll discuss this further in ection 31.1.) The mount of current I drwn y given device is determined y its power input P, given y Eq. (25.17): P = VI. Hence I = P>V. For exmple, the current in 100-W light ul is The power input to this ul is ctully determined y its resistnce R. Using Eq. (25.18), which sttes tht P = VI = I 2 R = V 2 >R for resistor, the resistnce of this ul t operting temperture is R = V I = 120 V 0.83 A I = P V = 100 W 120 V = 0.83 A = 144 Æ or R = V 2 imilrly, 1500-W wffle iron drws current of W2>1120 V2 = 12.5 A nd hs resistnce, t operting temperture, of 9.6 Æ. Becuse of the temperture dependence of resistivity, the resistnces of these devices re considerly less when they re cold. If you mesure the resistnce of 100-W light ul with n ohmmeter (whose smll current cuses very little temperture rise), you will proly get vlue of out 10 Æ. When light ul is turned on, this low resistnce cuses n initil surge of current until the filment hets up. Tht s why light ul tht s redy to urn out nerly lwys does so just when you turn it on. Circuit Overlods nd hort Circuits The mximum current ville from n individul circuit is limited y the resistnce of the wires. As we discussed in ection 25.5, the I 2 R power loss in the wires cuses them to ecome hot, nd in extreme cses this cn cuse fire or melt the wires. Ordinry lighting nd outlet wiring in houses usully uses 12-guge wire. This hs dimeter of 2.05 mm nd cn crry mximum current of 20 A sfely (without overheting). Lrger-dimeter wires of the sme length hve lower resistnce [see Eq. (25.10)]. Hence 8-guge (3.26 mm) or 6-guge (4.11 mm) re used for high-current pplinces such s clothes dryers, nd 2-guge (6.54 mm) or lrger is used for the min power lines entering house. P = 1120 V W = 144 Æ

184 870 CHAPTER 26 Direct-Current Circuits () Excess current will melt the thin wire of ledtin lloy tht runs long the length of fuse, inside the trnsprent housing. () The switch on this circuit reker will flip if the mximum llowle current is exceeded. () () Protection ginst overloding nd overheting of circuits is provided y fuses or circuit rekers. A fuse contins link of ledtin lloy with very low melting temperture; the link melts nd reks the circuit when its rted current is exceeded (Fig ). A circuit reker is n electromechnicl device tht performs the sme function, using n electromgnet or imetllic strip to trip the reker nd interrupt the circuit when the current exceeds specified vlue (Fig ). Circuit rekers hve the dvntge tht they cn e reset fter they re tripped, while lown fuse must e replced. If your system hs fuses nd you plug too mny high-current pplinces into the sme outlet, the fuse lows. Do not replce the fuse with one of lrger rting; if you do, you risk overheting the wires nd strting fire. The only sfe solution is to distriute the pplinces mong severl circuits. Modern kitchens often hve three or four seprte 20-A circuits. Contct etween the hot nd neutrl sides of the line cuses short circuit. uch sitution, which cn e cused y fulty insultion or y ny of vriety of mechnicl mlfunctions, provides very low-resistnce current pth, permitting very lrge current tht would quickly melt the wires nd ignite their insultion if the current were not interrupted y fuse or circuit reker (see Exmple in ection 25.5). An eqully dngerous sitution is roken wire tht interrupts the current pth, creting n open circuit. This is hzrdous ecuse of the sprking tht cn occur t the point of intermittent contct. In pproved wiring prctice, fuse or reker is plced only in the hot side of the line, never in the neutrl side. Otherwise, if short circuit should develop ecuse of fulty insultion or other mlfunction, the ground-side fuse could low. The hot side would still e live nd would pose shock hzrd if you touched the live conductor nd grounded oject such s wter pipe. For similr resons the wll switch for light fixture is lwys in the hot side of the line, never the neutrl side. Further protection ginst shock hzrd is provided y third conductor clled the grounding wire, included in ll present-dy wiring. This conductor corresponds to the long round or U-shped prong of the three-prong connector plug on n pplince or power tool. It is connected to the neutrl side of the line t the entrnce pnel. The grounding wire normlly crries no current, ut it connects the metl cse or frme of the device to ground. If conductor on the hot side of the line ccidentlly contcts the frme or cse, the grounding conductor provides current pth, nd the fuse lows. Without the ground wire, the frme could ecome live tht is, t potentil 120 V ove ground. Then if you touched it nd wter pipe (or even dmp sement floor) t the sme time, you could get dngerous shock (Fig ). In some situtions, especilly outlets locted outdoors or ner sink or other wter pipes, specil kind of circuit reker clled ground-fult interrupter (GFI or GFCI) is used. This device senses the difference in current etween the hot nd neutrl conductors (which is normlly zero) nd trips when this difference exceeds some very smll vlue, typiclly 5 ma. Household nd Automotive Wiring Most modern household wiring systems ctully use slight elortion of the system descried ove. The power compny provides three conductors. One is neutrl; the other two re oth t 120 V with respect to the neutrl ut with opposite polrity, giving voltge etween them of 240 V. The power compny clls this three-wire line, in contrst to the 120-V two-wire (plus ground wire) line descried ove. With three-wire line, 120-V lmps nd pplinces cn e connected etween neutrl nd either hot conductor, nd high-power devices requiring 240 V, such s electric rnges nd clothes dryers, re connected etween the two hot lines. All of the ove discussion cn e pplied directly to utomoile wiring. The voltge is out 13 V (direct current); the power is supplied y the ttery nd y