Chapter 23. Electric Fields

Transcription

1 Chpter 23 Electric Fields CHAPTER OUTLINE 23.1 Properties of Electric Chrges 23.2 Chrging Objects B Induction 23.3 Coulomb s Lw 23.4 The Electric Field 23.5 Electric Field of Continuous Chrge Distribution 23.6 Electric Field Lines 23.7 Motion of Chrged Prticles in Uniform Electric Field Mother nd dughter re both enjoing the effects of electricll chrging their bodies. Ech individul hir on their heds becomes chrged nd eerts repulsive force on the other hirs, resulting in the stnd-up hirdos tht ou see here. (Courtes of Resonnce Reserch Corportion) 706

2 The electromgnetic force between chrged prticles is one of the fundmentl forces of nture. We begin this chpter b describing some of the bsic properties of one mnifesttion of the electromgnetic force, the electric force. We then discuss Coulomb s lw, which is the fundmentl lw governing the electric force between n two chrged prticles. Net, we introduce the concept of n electric field ssocited with chrge distribution nd describe its effect on other chrged prticles. We then show how to use Coulomb s lw to clculte the electric field for given chrge distribution. We conclude the chpter with discussion of the motion of chrged prticle in uniform electric field Properties of Electric Chrges A number of simple eperiments demonstrte the eistence of electric forces nd chrges. For emple, fter running comb through our hir on dr d, ou will find tht the comb ttrcts bits of pper. The ttrctive force is often strong enough to suspend the pper. The sme effect occurs when certin mterils re rubbed together, such s glss rubbed with silk or rubber with fur. Another simple eperiment is to rub n inflted blloon with wool. The blloon then dheres to wll, often for hours. When mterils behve in this w, the re sid to be electrified, or to hve become electricll chrged. You cn esil electrif our bod b vigorousl rubbing our shoes on wool rug. Evidence of the electric chrge on our bod cn be detected b lightl touching (nd strtling) friend. Under the right conditions, ou will see sprk when ou touch, nd both of ou will feel slight tingle. (Eperiments such s these work best on dr d becuse n ecessive mount of moisture in the ir cn cuse n chrge ou build up to lek from our bod to the Erth.) In series of simple eperiments, it ws found tht there re two kinds of electric chrges, which were given the nmes positive nd negtive b Benjmin Frnklin ( ). We identif negtive chrge s tht tpe possessed b electrons nd positive chrge s tht possessed b protons. To verif tht there re two tpes of chrge, suppose hrd rubber rod tht hs been rubbed with fur is suspended b sewing thred, s shown in Figure When glss rod tht hs been rubbed with silk is brought ner the rubber rod, the two ttrct ech other (Fig. 23.1). On the other hnd, if two chrged rubber rods (or two chrged glss rods) re brought ner ech other, s shown in Figure 23.1b, the two repel ech other. This observtion shows tht the rubber nd glss hve two different tpes of chrge on them. On the bsis of these observtions, we conclude tht chrges of the sme sign repel one nother nd chrges with opposite signs ttrct one nother. Using the convention suggested b Frnklin, the electric chrge on the glss rod is clled positive nd tht on the rubber rod is clled negtive. Therefore, n chrged object ttrcted to chrged rubber rod (or repelled b chrged glss rod) must 707

3 708 CHAPTER 23 Electric Fields Rubber Rubber F F Glss () F F Rubber Figure 23.1 () A negtivel chrged rubber rod suspended b thred is ttrcted to positivel chrged glss rod. (b) A negtivel chrged rubber rod is repelled b nother negtivel chrged rubber rod. (b) Electric chrge is conserved Figure 23.2 When glss rod is rubbed with silk, electrons re trnsferred from the glss to the silk. Becuse of conservtion of chrge, ech electron dds negtive chrge to the silk, nd n eul positive chrge is left behind on the rod. Also, becuse the chrges re trnsferred in discrete bundles, the chrges on the two objects re e, or 2e, or 3e, nd so on. hve positive chrge, nd n chrged object repelled b chrged rubber rod (or ttrcted to chrged glss rod) must hve negtive chrge. Attrctive electric forces re responsible for the behvior of wide vriet of commercil products. For emple, the plstic in mn contct lenses, etfilcon, is mde up of molecules tht electricll ttrct the protein molecules in humn ters. These protein molecules re bsorbed nd held b the plstic so tht the lens ends up being primril composed of the werer s ters. Becuse of this, the lens does not behve s foreign object to the werer s ee, nd it cn be worn comfortbl. Mn cosmetics lso tke dvntge of electric forces b incorporting mterils tht re electricll ttrcted to skin or hir, cusing the pigments or other chemicls to st put once the re pplied. Another importnt spect of electricit tht rises from eperimentl observtions is tht electric chrge is lws conserved in n isolted sstem. Tht is, when one object is rubbed ginst nother, chrge is not creted in the process. The electrified stte is due to trnsfer of chrge from one object to the other. One object gins some mount of negtive chrge while the other gins n eul mount of positive chrge. For emple, when glss rod is rubbed with silk, s in Figure 23.2, the silk obtins negtive chrge tht is eul in mgnitude to the positive chrge on the glss rod. We now know from our understnding of tomic structure tht electrons re trnsferred from the glss to the silk in the rubbing process. Similrl, when rubber is rubbed with fur, electrons re trnsferred from the fur to the rubber, giving the rubber net negtive chrge nd the fur net positive chrge. This process is consistent with the fct tht neutrl, unchrged mtter contins s mn positive chrges (protons within tomic nuclei) s negtive chrges (electrons). In 1909, Robert Millikn ( ) discovered tht electric chrge lws occurs s some integrl multiple of fundmentl mount of chrge e (see Section 25.7). In modern terms, the electric chrge is sid to be untized, where is the stndrd smbol used for chrge s vrible. Tht is, electric chrge eists s discrete pckets, nd we cn write Ne, where N is some integer. Other eperiments in the sme period showed tht the electron hs chrge e nd the proton hs chrge of eul mgnitude but opposite sign e. Some prticles, such s the neutron, hve no chrge. From our discussion thus fr, we conclude tht electric chrge hs the following importnt properties:

4 SECTION 23.2 Chrging Objects B Induction 709 There re two kinds of chrges in nture; chrges of opposite sign ttrct one nother nd chrges of the sme sign repel one nother. Totl chrge in n isolted sstem is conserved. Chrge is untized. Properties of electric chrge Quick Quiz 23.1 If ou rub n inflted blloon ginst our hir, the two mterils ttrct ech other, s shown in Figure Is the mount of chrge present in the sstem of the blloon nd our hir fter rubbing () less thn, (b) the sme s, or (c) more thn the mount of chrge present before rubbing? Quick Quiz 23.2 Three objects re brought close to ech other, two t time. When objects A nd B re brought together, the repel. When objects B nd C re brought together, the lso repel. Which of the following re true? () Objects A nd C possess chrges of the sme sign. (b) Objects A nd C possess chrges of opposite sign. (c) All three of the objects possess chrges of the sme sign. (d) One of the objects is neutrl. (e) We would need to perform dditionl eperiments to determine the signs of the chrges. Chrles D. Winters Figure 23.3 (Quick Quiz 23.1) Rubbing blloon ginst our hir on dr d cuses the blloon nd our hir to become chrged Chrging Objects B Induction It is convenient to clssif mterils in terms of the bilit of electrons to move through the mteril: Electricl conductors re mterils in which some of the electrons re free electrons 1 tht re not bound to toms nd cn move reltivel freel through the mteril; electricl insultors re mterils in which ll electrons re bound to toms nd cnnot move freel through the mteril. Mterils such s glss, rubber, nd wood fll into the ctegor of electricl insultors. When such mterils re chrged b rubbing, onl the re rubbed becomes chrged, nd the chrged prticles re unble to move to other regions of the mteril. In contrst, mterils such s copper, luminum, nd silver re good electricl conductors. When such mterils re chrged in some smll region, the chrge redil distributes itself over the entire surfce of the mteril. If ou hold copper rod in our hnd nd rub it with wool or fur, it will not ttrct smll piece of pper. This might suggest tht metl cnnot be chrged. However, if ou ttch wooden hndle to the rod nd then hold it b tht hndle s ou rub the rod, the rod will remin chrged nd ttrct the piece of pper. The eplntion for this is s follows: without the insulting wood, the electric chrges produced b rubbing redil move from the copper through our bod, which is lso conductor, nd into the Erth. The insulting wooden hndle prevents the flow of chrge into our hnd. Semiconductors re third clss of mterils, nd their electricl properties re somewhere between those of insultors nd those of conductors. Silicon nd 1 A metl tom contins one or more outer electrons, which re wekl bound to the nucleus. When mn toms combine to form metl, the so-clled free electrons re these outer electrons, which re not bound to n one tom. These electrons move bout the metl in mnner similr to tht of gs molecules moving in continer.

5 710 CHAPTER 23 Electric Fields () (b) (c) (d) (e) Figure 23.4 Chrging metllic object b induction (tht is, the two objects never touch ech other). () A neutrl metllic sphere, with eul numbers of positive nd negtive chrges. (b) The electrons on the neutrl sphere re redistributed when chrged rubber rod is plced ner the sphere. (c) When the sphere is grounded, some of its electrons leve through the ground wire. (d) When the ground connection is removed, the sphere hs ecess positive chrge tht is nonuniforml distributed. (e) When the rod is removed, the remining electrons redistribute uniforml nd there is net uniform distribution of positive chrge on the sphere. germnium re well-known emples of semiconductors commonl used in the fbriction of vriet of electronic chips used in computers, cellulr telephones, nd stereo sstems. The electricl properties of semiconductors cn be chnged over mn orders of mgnitude b the ddition of controlled mounts of certin toms to the mterils. To understnd how to chrge conductor b process known s induction, consider neutrl (unchrged) conducting sphere insulted from the ground, s shown in Figure There re n eul number of electrons nd protons in the sphere if the chrge on the sphere is ectl zero. When negtivel chrged rubber rod is brought ner the sphere, electrons in the region nerest the rod eperience repulsive force nd migrte to the opposite side of the sphere. This leves the side of the sphere ner the rod with n effective positive chrge becuse of the diminished number of electrons, s in Figure 23.4b. (The left side of the sphere in Figure 23.4b is positivel chrged s if positive chrges moved into this region, but remember tht it is onl electrons tht re free to move.) This occurs even if the rod never ctull touches the sphere. If the sme eperiment is performed with conducting wire connected from the sphere to the Erth (Fig. 23.4c), some of the electrons in the conductor re so strongl repelled b the presence of the negtive chrge in the rod tht the move out of the sphere through the wire nd into the Erth. The smbol t the end of the wire in Figure 23.4c indictes tht the wire is connected to ground, which mens reservoir, such s the Erth, tht cn ccept or provide electrons freel with negligible effect on its electricl chrcteristics. If the wire to ground is then removed (Fig. 23.4d), the conducting sphere contins n ecess of induced positive chrge becuse it hs fewer electrons thn it needs to cncel out the positive chrge of the protons. When the rubber rod is removed from the vicinit of the sphere (Fig. 23.4e), this induced positive chrge remins on the ungrounded sphere. Note tht the rubber rod loses none of its negtive chrge during this process. Chrging n object b induction reuires no contct with the object inducing the chrge. This is in contrst to chrging n object b rubbing (tht is, b conduction), which does reuire contct between the two objects. A process similr to induction in conductors tkes plce in insultors. In most neutrl molecules, the center of positive chrge coincides with the center of negtive chrge. However, in the presence of chrged object, these centers inside ech molecule in n insultor m shift slightl, resulting in more positive chrge on one side of the molecule thn on the other. This relignment of chrge within individul molecules produces ler of chrge on the surfce of the insultor, s shown in Figure Knowing bout induction in insultors, ou should be ble to eplin wh comb tht hs been rubbed through hir ttrcts bits of electricll neutrl pper nd wh blloon tht hs been rubbed ginst our clothing is ble to stick to n electricll neutrl wll. Chrged object () Insultor Induced chrges 1968 Fundmentl Photogrphs Figure 23.5 () The chrged object on the left induces chrge distribution on the surfce of n insultor due to relignment of chrges in the molecules. (b) A chrged comb ttrcts bits of pper becuse chrges in molecules in the pper re religned. (b)

6 SECTION 23.3 Coulomb s Lw 711 Quick Quiz 23.3 Three objects re brought close to ech other, two t time. When objects A nd B re brought together, the ttrct. When objects B nd C re brought together, the repel. From this, we conclude tht () objects A nd C possess chrges of the sme sign. (b) objects A nd C possess chrges of opposite sign. (c) ll three of the objects possess chrges of the sme sign. (d) one of the objects is neutrl. (e) we need to perform dditionl eperiments to determine informtion bout the chrges on the objects. Suspension hed Fiber 23.3 Coulomb s Lw Chrles Coulomb ( ) mesured the mgnitudes of the electric forces between chrged objects using the torsion blnce, which he invented (Fig. 23.6). Coulomb confirmed tht the electric force between two smll chrged spheres is proportionl to the inverse sure of their seprtion distnce r tht is, F e 1/r 2. The operting principle of the torsion blnce is the sme s tht of the pprtus used b Cvendish to mesure the grvittionl constnt (see Section 13.2), with the electricll neutrl spheres replced b chrged ones. The electric force between chrged spheres A nd B in Figure 23.6 cuses the spheres to either ttrct or repel ech other, nd the resulting motion cuses the suspended fiber to twist. Becuse the restoring torue of the twisted fiber is proportionl to the ngle through which the fiber rottes, mesurement of this ngle provides untittive mesure of the electric force of ttrction or repulsion. Once the spheres re chrged b rubbing, the electric force between them is ver lrge compred with the grvittionl ttrction, nd so the grvittionl force cn be neglected. From Coulomb s eperiments, we cn generlize the following properties of the electric force between two sttionr chrged prticles. The electric force B A Figure 23.6 Coulomb s torsion blnce, used to estblish the inverse-sure lw for the electric force between two chrges. is inversel proportionl to the sure of the seprtion r between the prticles nd directed long the line joining them; is proportionl to the product of the chrges 1 nd 2 on the two prticles; is ttrctive if the chrges re of opposite sign nd repulsive if the chrges hve the sme sign; is conservtive force. We will use the term point chrge to men prticle of zero size tht crries n electric chrge. The electricl behvior of electrons nd protons is ver well described b modeling them s point chrges. From eperimentl observtions on the electric force, we cn epress Coulomb s lw s n eution giving the mgnitude of the electric force (sometimes clled the Coulomb force) between two point chrges: F e k e 1 2 r 2 (23.1) Coulomb s lw where k e is constnt clled the Coulomb constnt. In his eperiments, Coulomb ws ble to show tht the vlue of the eponent of r ws 2 to within n uncertint of few percent. Modern eperiments hve shown tht the eponent is 2 to within n uncertint of few prts in The vlue of the Coulomb constnt depends on the choice of units. The SI unit of chrge is the coulomb (C). The Coulomb constnt k e in SI units hs the vlue This constnt is lso written in the form k e N m 2 /C 2 (23.2) k e (23.3) Coulomb constnt

7 712 CHAPTER 23 Electric Fields Tble 23.1 Chrge nd Mss of the Electron, Proton, nd Neutron Prticle Chrge (C) Mss (kg) Electron (e) Proton (p) Neutron (n) where the constnt 0 (lowercse Greek epsilon) is known s the permittivit of free spce nd hs the vlue C 2 /N m 2 (23.4) Chrles Coulomb French phsicist ( ) Coulomb s mjor contributions to science were in the res of electrosttics nd mgnetism. During his lifetime, he lso investigted the strengths of mterils nd determined the forces tht ffect objects on bems, thereb contributing to the field of structurl mechnics. In the field of ergonomics, his reserch provided fundmentl understnding of the ws in which people nd nimls cn best do work. (Photo courtes of AIP Niels Bohr Librr/E. Scott Brr Collection) The smllest unit of chrge e known in nture 2 is the chrge on n electron ( e) or proton ( e) nd hs mgnitude e C (23.5) Therefore, 1 C of chrge is pproimtel eul to the chrge of electrons or protons. This number is ver smll when compred with the number of free electrons in 1 cm 3 of copper, which is on the order of Still, 1 C is substntil mount of chrge. In tpicl eperiments in which rubber or glss rod is chrged b friction, net chrge on the order of 10 6 C is obtined. In other words, onl ver smll frction of the totl vilble chrge is trnsferred between the rod nd the rubbing mteril. The chrges nd msses of the electron, proton, nd neutron re given in Tble Quick Quiz 23.4 Object A hs chrge of 2 C, nd object B hs chrge of 6 C. Which sttement is true bout the electric forces on the objects? () F AB 3F BA (b) F AB F BA (c) 3F AB F BA (d) F AB 3F BA (e) F AB F BA (f) 3F AB F BA Emple 23.1 The Hdrogen Atom The electron nd proton of hdrogen tom re seprted (on the verge) b distnce of pproimtel m. Find the mgnitudes of the electric force nd the grvittionl force between the two prticles. Solution From Coulomb s lw, we find tht the mgnitude of the electric force is F e k e e e r 2 ( N m 2 /C 2 ) ( C) 2 ( m) N Using Newton s lw of universl grvittion nd Tble 23.1 for the prticle msses, we find tht the mgnitude of the grvittionl force is F g G m em p r 2 ( N m 2 /kg 2 ) ( kg)( kg) ( m) N The rtio F e /F g Thus, the grvittionl force between chrged tomic prticles is negligible when compred with the electric force. Note the similrit of form of Newton s lw of universl grvittion nd Coulomb s lw of electric forces. Other thn mgnitude, wht is fundmentl difference between the two forces? 2 No unit of chrge smller thn e hs been detected on free prticle; however, current theories propose the eistence of prticles clled urks hving chrges e/3 nd 2e/3. Although there is considerble eperimentl evidence for such prticles inside nucler mtter, free urks hve never been detected. We discuss other properties of urks in Chpter 46 of the etended version of this tet.

8 SECTION 23.3 Coulomb s Lw 713 When deling with Coulomb s lw, ou must remember tht force is vector untit nd must be treted ccordingl. The lw epressed in vector form for the electric force eerted b chrge 1 on second chrge 2, written F 12, is F 12 k e 1 2 r 2 ˆr (23.6) Vector form of Coulomb s lw where rˆ is unit vector directed from 1 towrd 2, s shown in Figure Becuse the electric force obes Newton s third lw, the electric force eerted b 2 on 1 is eul in mgnitude to the force eerted b 1 on 2 nd in the opposite direction; tht is, F 21 F 12. Finll, from Eution 23.6, we see tht if 1 nd 2 hve the sme sign, s in Figure 23.7, the product 1 2 is positive. If 1 nd 2 re of opposite sign, s shown in Figure 23.7b, the product 1 2 is negtive. These signs describe the reltive direction of the force but not the bsolute direction. A negtive product indictes n ttrctive force, so tht the chrges ech eperience force towrd the other thus, the force on one chrge is in direction reltive to the other. A positive product indictes repulsive force such tht ech chrge eperiences force w from the other. The bsolute direction of the force in spce is not determined solel b the sign of 1 2 whether the force on n individul chrge is in the positive or negtive direction on coordinte is depends on the loction of the other chrge. For emple, if n is lies long the two chrges in Figure 23.7, the product 1 2 is positive, but F 12 points in the direction nd F 21 points in the direction. Quick Quiz 23.5 Object A hs chrge of 2 C, nd object B hs chrge of 6 C. Which sttement is true bout the electric forces on the objects? () F AB 3F BA (b) F AB F BA (c) 3F AB F BA (d) F AB 3F BA (e) F AB F BA (f) 3F AB F BA When more thn two chrges re present, the force between n pir of them is given b Eution Therefore, the resultnt force on n one of them euls the vector sum of the forces eerted b the vrious individul chrges. For emple, if four chrges re present, then the resultnt force eerted b prticles 2, 3, nd 4 on prticle 1 is F 1 F 21 F 31 F 41 F r rˆ () F 21 (b) 2 F 12 F 12 2 Active Figure 23.7 Two point chrges seprted b distnce r eert force on ech other tht is given b Coulomb s lw. The force F 21 eerted b 2 on 1 is eul in mgnitude nd opposite in direction to the force F 12 eerted b 1 on 2. () When the chrges re of the sme sign, the force is repulsive. (b) When the chrges re of opposite signs, the force is ttrctive. At the Active Figures link t ou cn move the chrges to n position in two-dimensionl spce nd observe the electric forces on them. Emple 23.2 Find the Resultnt Force Consider three point chrges locted t the corners of right tringle s shown in Figure 23.8, where C, C, nd 0.10 m. Find the resultnt force eerted on 3. 2 F 23 3 F 13 Solution First, note the direction of the individul forces eerted b 1 nd 2 on 3. The force F 23 eerted b 2 on 3 is ttrctive becuse 2 nd 3 hve opposite signs. The force F 13 eerted b 1 on 3 is repulsive becuse both chrges re positive. The mgnitude of F 23 is 1 2 Figure 23.8 (Emple 23.2) The force eerted b 1 on 3 is F 13. The force eerted b 2 on 3 is F 23. The resultnt force F 3 eerted on 3 is the vector sum F 13 F 23. F 23 k e ( N m 2 /C 2 ) ( C)( C) (0.10 m) N In the coordinte sstem shown in Figure 23.8, the ttrctive force F 23 is to the left (in the negtive direction).

9 714 CHAPTER 23 Electric Fields The mgnitude of the force F 13 eerted b 1 on 3 is F 13 k e 1 3 ( 2) 2 ( N m 2 /C 2 ) ( C)( C) 2(0.10 m) 2 11 N The repulsive force F 13 mkes n ngle of 45 with the is. Therefore, the nd components of F 13 re eul, with mgnitude given b F 13 cos N. Combining F 13 with F 23 b the rules of vector ddition, we rrive t the nd components of the resultnt force cting on 3 : F 3 F 13 F N ( 9.0 N) 1.1 N F 3 F 13 F N N We cn lso epress the resultnt force cting on 3 in unitvector form s F 3 ( 1.1î 7.9 ĵ) N Wht If? Wht if the signs of ll three chrges were chnged to the opposite signs? How would this ffect the result for F 3? Answer The chrge 3 would still be ttrcted towrd 2 nd repelled from 1 with forces of the sme mgnitude. Thus, the finl result for F 3 would be ectl the sme. Emple 23.3 Where Is the Resultnt Force Zero? Interctive Three point chrges lie long the is s shown in Figure The positive chrge C is t 2.00 m, the positive chrge C is t the origin, nd the resultnt force cting on 3 is zero. Wht is the coordinte of 3? Solution Becuse 3 is negtive nd 1 nd 2 re positive, the forces F 13 nd F 23 re both ttrctive, s indicted in Figure From Coulomb s lw, F 13 nd F 23 hve mgnitudes F 13 k e 1 3 (2.00 ) 2 F 23 k 2 3 e 2 For the resultnt force on 3 to be zero, F 23 must be eul in mgnitude nd opposite in direction to F 13. Setting the mgnitudes of the two forces eul, we hve k e k e 1 3 (2.00 ) 2 of the forces on 3 re eul, but both forces re in the sme direction t this loction. Wht If? Suppose chrge 3 is constrined to move onl long the is. From its initil position t m, it is pulled ver smll distnce long the is. When relesed, will it return to euilibrium or be pulled further from euilibrium? Tht is, is the euilibrium stble or unstble? Answer If the chrge is moved to the right, F 13 becomes lrger nd F 23 becomes smller. This results in net force to the right, in the sme direction s the displcement. Thus, the euilibrium is unstble. Note tht if the chrge is constrined to st t fied coordinte but llowed to move up nd down in Figure 23.9, the euilibrium is stble. In this cse, if the chrge is pulled upwrd (or downwrd) nd relesed, it will move bck towrd the euilibrium position nd undergo oscilltion. Noting tht k e nd 3 re common to both sides nd so cn be dropped, we solve for nd find tht (2.00 ) ( )( C) 2 ( C) 2.00 m 2.00 This cn be reduced to the following udrtic eution: = 0 Solving this udrtic eution for, we find tht the positive root is m. There is lso second root, 3.44 m. This is nother loction t which the mgnitudes 2 F 23 3 F 13 1 Figure 23.9 (Emple 23.3) Three point chrges re plced long the is. If the resultnt force cting on 3 is zero, then the force F 13 eerted b 1 on 3 must be eul in mgnitude nd opposite in direction to the force F 23 eerted b 2 on 3. At the Interctive Worked Emple link t ou cn predict where on the is the electric force is zero for rndom vlues of 1 nd 2. Emple 23.4 Find the Chrge on the Spheres Two identicl smll chrged spheres, ech hving mss of kg, hng in euilibrium s shown in Figure The length of ech string is 0.15 m, nd the ngle is 5.0. Find the mgnitude of the chrge on ech sphere. Solution Figure helps us conceptulize this problem the two spheres eert repulsive forces on ech other. If the re held close to ech other nd relesed, the will move outwrd from the center nd settle into the configurtion in Figure fter the dmped oscilltions

10 SECTION 23.4 The Electric Field 715 The seprtion of the spheres is m. From Coulomb s lw (E. 23.1), the mgnitude of the electric force is L L F e T cos T T sin F e k e 2 r 2 where r m nd is the mgnitude of the chrge on ech sphere. (Note tht the term 2 rises here becuse the chrge is the sme on both spheres.) This eution cn be solved for 2 to give L = 0.15 m = 5.0 () Figure (Emple 23.4) () Two identicl spheres, ech crring the sme chrge, suspended in euilibrium. (b) The free-bod digrm for the sphere on the left. due to ir resistnce hve vnished. The ke phrse in euilibrium helps us ctegorize this s n euilibrium problem, which we pproch s we did euilibrium problems in Chpter 5 with the dded feture tht one of the forces on sphere is n electric force. We nlze this problem b drwing the free-bod digrm for the left-hnd sphere in Figure 23.10b. The sphere is in euilibrium under the ppliction of the forces T from the string, the electric force F e from the other sphere, nd the grvittionl force mg. Becuse the sphere is in euilibrium, the forces in the horizontl nd verticl directions must seprtel dd up to zero: (1) F T sin F e 0 (2) F T cos mg 0 From Eution (2), we see tht T mg/cos ; thus, T cn be eliminted from Eution (1) if we mke this substitution. This gives vlue for the mgnitude of the electric force F e : F e mg tn ( kg)(9.80 m/s 2 ) tn(5.0 ) N Considering the geometr of the right tringle in Figure 23.10, we see tht sin /L. Therefore, (b) mg L sin (0.15 m) sin(5.0 ) m 2 F e r 2 ( N)(0.026 m) 2 k e N m 2 /C C C To finlize the problem, note tht we found onl the mgnitude of the chrge on the spheres. There is no w we could find the sign of the chrge from the informtion given. In fct, the sign of the chrge is not importnt. The sitution will be ectl the sme whether both spheres re positivel chrged or negtivel chrged. Wht If? Suppose our roommte proposes solving this problem without the ssumption tht the chrges re of eul mgnitude. She clims tht the smmetr of the problem is destroed if the chrges re not eul, so tht the strings would mke two different ngles with the verticl, nd the problem would be much more complicted. How would ou respond? Answer You should rgue tht the smmetr is not destroed nd the ngles remin the sme. Newton s third lw reuires tht the electric forces on the two chrges be the sme, regrdless of the eulit or noneulit of the chrges. The solution to the emple remins the sme through the clcultion of 2. In this sitution, the vlue of C 2 corresponds to the product 1 2, where 1 nd 2 re the vlues of the chrges on the two spheres. The smmetr of the problem would be destroed if the msses of the spheres were not the sme. In this cse, the strings would mke different ngles with the verticl nd the problem would be more complicted The Electric Field Two field forces hve been introduced into our discussions so fr the grvittionl force in Chpter 13 nd the electric force here. As pointed out erlier, field forces cn ct through spce, producing n effect even when no phsicl contct occurs between intercting objects. The grvittionl field g t point in spce ws defined in Section 13.5 to be eul to the grvittionl force F g cting on test prticle of mss m divided b tht mss: g F g /m. The concept of field ws developed b Michel Frd ( ) in the contet of electric forces nd is of such prcticl vlue tht we shll devote much ttention to it in the net severl chpters. In this pproch, n electric field is sid to eist in the region of spce round chrged object the source chrge. When nother chrged object the test chrge enters this electric field, n

11 716 CHAPTER 23 Electric Fields Q 0 Figure A smll positive test chrge 0 plced ner n object crring much lrger positive chrge Q eperiences n electric field E directed s shown. E Johnn Auter This drmtic photogrph cptures lightning bolt striking tree ner some rurl homes. Lightning is ssocited with ver strong electric fields in the tmosphere. electric force cts on it. As n emple, consider Figure 23.11, which shows smll positive test chrge 0 plced ner second object crring much greter positive chrge Q. We define the electric field due to the source chrge t the loction of the test chrge to be the electric force on the test chrge per unit chrge, or to be more specific Definition of electric field the electric field vector E t point in spce is defined s the electric force F e cting on positive test chrge 0 plced t tht point divided b the test chrge: E F e 0 (23.7) PITFALL PREVENTION 23.1 Prticles Onl Eution 23.8 is onl vlid for chrged prticle n object of zero size. For chrged object of finite size in n electric field, the field m vr in mgnitude nd direction over the size of the object, so the corresponding force eution m be more complicted. Note tht E is the field produced b some chrge or chrge distribution seprte from the test chrge it is not the field produced b the test chrge itself. Also, note tht the eistence of n electric field is propert of its source the presence of the test chrge is not necessr for the field to eist. The test chrge serves s detector of the electric field. Eution 23.7 cn be rerrnged s F e E (23.8) where we hve used the generl smbol for chrge. This eution gives us the force on chrged prticle plced in n electric field. If is positive, the force is in the sme

12 SECTION 23.4 The Electric Field 717 Tble 23.2 Tpicl Electric Field Vlues Source direction s the field. If is negtive, the force nd the field re in opposite directions. Notice the similrit between Eution 23.8 nd the corresponding eution for prticle with mss plced in grvittionl field, F g mg (E. 5.6). The vector E hs the SI units of newtons per coulomb (N/C). The direction of E, s shown in Figure 23.11, is the direction of the force positive test chrge eperiences when plced in the field. We s tht n electric field eists t point if test chrge t tht point eperiences n electric force. Once the mgnitude nd direction of the electric field re known t some point, the electric force eerted on n chrged prticle plced t tht point cn be clculted from Eution The electric field mgnitudes for vrious field sources re given in Tble When using Eution 23.7, we must ssume tht the test chrge 0 is smll enough tht it does not disturb the chrge distribution responsible for the electric field. If vnishingl smll test chrge 0 is plced ner uniforml chrged metllic sphere, s in Figure 23.12, the chrge on the metllic sphere, which produces the electric field, remins uniforml distributed. If the test chrge is gret enough ( 0 0 ), s in Figure 23.12b, the chrge on the metllic sphere is redistributed nd the rtio of the force to the test chrge is different: (F e / 0 F e / 0 ). Tht is, becuse of this redistribution of chrge on the metllic sphere, the electric field it sets up is different from the field it sets up in the presence of the much smller test chrge 0. To determine the direction of n electric field, consider point chrge s source chrge. This chrge cretes n electric field t ll points in spce surrounding it. A test chrge 0 is plced t point P, distnce r from the source chrge, s in Figure We imgine using the test chrge to determine the direction of the electric force nd therefore tht of the electric field. However, the electric field does not depend on the eistence of the test chrge it is estblished solel b the source chrge. According to Coulomb s lw, the force eerted b on the test chrge is F e k e E (N/C) Fluorescent lighting tube 10 Atmosphere (fir wether) 100 Blloon rubbed on hir Atmosphere (under thundercloud) Photocopier Sprk in ir Ner electron in hdrogen tom r 2 rˆ where rˆ is unit vector directed from towrd 0. This force in Figure is directed w from the source chrge. Becuse the electric field t P, the position of the test chrge, is defined b E F e / 0, we find tht t P, the electric field creted b is E k e (23.9) r 2 rˆ If the source chrge is positive, Figure 23.13b shows the sitution with the test chrge removed the source chrge sets up n electric field t point P, directed w from. If is negtive, s in Figure 23.13c, the force on the test chrge is towrd the source chrge, so the electric field t P is directed towrd the source chrge, s in Figure 23.13d. 0 0 >> 0 () (b) Figure () For smll enough test chrge 0, the chrge distribution on the sphere is undisturbed. (b) When the test chrge 0 is greter, the chrge distribution on the sphere is disturbed s the result of the proimit of 0. rˆ rˆ rˆ r () r (b) F (c) E r rˆ (d) 0 Active Figure A test chrge 0 t point P is distnce r from point chrge. () If is positive, then the force on the test chrge is directed w from. (b) For the positive source chrge, the electric field t P points rdill outwrd from. (c) If is negtive, then the force on the test chrge is directed towrd. (d) For the negtive source chrge, the electric field t P points rdill inwrd towrd. At the Active Figures link t ou cn move point P to n position in two-dimensionl spce nd observe the electric field due to. P P 0 P P F E

13 718 CHAPTER 23 Electric Fields To clculte the electric field t point P due to group of point chrges, we first clculte the electric field vectors t P individull using Eution 23.9 nd then dd them vectorill. In other words, t n point P, the totl electric field due to group of source chrges euls the vector sum of the electric fields of ll the chrges. This superposition principle pplied to fields follows directl from the superposition propert of electric forces, which, in turn, follows from the fct tht we know tht forces dd s vectors from Chpter 5. Thus, the electric field t point P due to group of source chrges cn be epressed s the vector sum Electric field due to finite number of point chrges E k e i i r i 2 rˆi (23.10) where r i is the distnce from the i th source chrge i to the point P nd rˆi is unit vector directed from i towrd P. Quick Quiz 23.6 A test chrge of 3 C is t point P where n eternl electric field is directed to the right nd hs mgnitude of N/C. If the test chrge is replced with nother test chrge of 3 C, the eternl electric field t P () is unffected (b) reverses direction (c) chnges in w tht cnnot be determined Emple 23.5 Electric Field Due to Two Chrges A chrge C is locted t the origin, nd second chrge C is locted on the is, 0.30 m from the origin (Fig ). Find the electric field t the point P, which hs coordintes (0, 0.40) m. Solution First, let us find the mgnitude of the electric field t P due to ech chrge. The fields E 1 due to the 7.0- C chrge nd E 2 due to the 5.0- C chrge re shown in Figure Their mgnitudes re 1 E 1 k e r 2 ( N m 2 /C 2 ) ( C) 1 (0.40 m) N/C E 1 E 2 E 2 k e r 2 ( N m 2 /C 2 ) ( C) 2 (0.50 m) N/C P φ E 2 The vector E 1 hs onl component. The vector E 2 hs n 3 component given b E 2 cos E 2 nd negtive component given b E 2 sin E 2. Hence, we cn epress the vectors s m 0.50 m E ĵ N/C E 2 ( î ĵ) N/C m Figure (Emple 23.5) The totl electric field E t P euls the vector sum E 1 E 2, where E 1 is the field due to the positive chrge 1 nd E 2 is the field due to the negtive chrge 2. 2 The resultnt field E t P is the superposition of E 1 nd E 2 : E E 1 E 2 ( î ĵ) N/C From this result, we find tht E mkes n ngle of 66 with the positive is nd hs mgnitude of N/C.

14 SECTION 23.5 Electric Field of Continuous Chrge Distribution 719 Emple 23.6 Electric Field of Dipole An electric dipole is defined s positive chrge nd negtive chrge seprted b distnce 2. For the dipole shown in Figure 23.15, find the electric field E t P due to the dipole, where P is distnce from the origin. Solution At P, the fields E 1 nd E 2 due to the two chrges re eul in mgnitude becuse P is euidistnt from the chrges. The totl field is E E 1 E 2, where E 1 E 2 k e r P r 2 k e E 1 E E Figure (Emple 23.6) The totl electric field E t P due to two chrges of eul mgnitude nd opposite sign (n electric dipole) euls the vector sum E 1 E 2. The field E 1 is due to the positive chrge, nd E 2 is the field due to the negtive chrge. The components of E 1 nd E 2 cncel ech other, nd the components re both in the positive direction nd hve the sme mgnitude. Therefore, E is prllel to the is nd hs mgnitude eul to 2E 1 cos. From Figure we see tht cos /r /( 2 2 ) 1/2. Therefore, E 2E 1 cos 2k e ( 2 2 ) k e 2 ( 2 2 ) 3/2 Becuse, we cn neglect 2 compred to 2 nd write E k e 2 3 ( 2 2 ) 1/2 Thus, we see tht, t distnces fr from dipole but long the perpendiculr bisector of the line joining the two chrges, the mgnitude of the electric field creted b the dipole vries s 1/r 3, wheres the more slowl vring field of point chrge vries s 1/r 2 (see E. 23.9). This is becuse t distnt points, the fields of the two chrges of eul mgnitude nd opposite sign lmost cncel ech other. The 1/r 3 vrition in E for the dipole lso is obtined for distnt point long the is (see Problem 22) nd for n generl distnt point. The electric dipole is good model of mn molecules, such s hdrochloric cid (HCl). Neutrl toms nd molecules behve s dipoles when plced in n eternl electric field. Furthermore, mn molecules, such s HCl, re permnent dipoles. The effect of such dipoles on the behvior of mterils subjected to electric fields is discussed in Chpter Electric Field of Continuous Chrge Distribution Ver often the distnces between chrges in group of chrges re much smller thn the distnce from the group to some point of interest (for emple, point where the electric field is to be clculted). In such situtions, the sstem of chrges cn be modeled s continuous. Tht is, the sstem of closel spced chrges is euivlent to totl chrge tht is continuousl distributed long some line, over some surfce, or throughout some volume. To evlute the electric field creted b continuous chrge distribution, we use the following procedure: first, we divide the chrge distribution into smll elements, ech of which contins smll chrge, s shown in Figure Net, we use Eution 23.9 to clculte the electric field due to one of these elements t point P. Finll, we evlute the totl electric field t P due to the chrge distribution b summing the contributions of ll the chrge elements (tht is, b ppling the superposition principle). The electric field t P due to one chrge element crring chrge is E k e r 2 rˆ E r P rˆ Figure The electric field t P due to continuous chrge distribution is the vector sum of the fields E due to ll the elements of the chrge distribution.

15 720 CHAPTER 23 Electric Fields where r is the distnce from the chrge element to point P nd rˆ is unit vector directed from the element towrd P. The totl electric field t P due to ll elements in the chrge distribution is pproimtel E k e i i r i 2 rˆi where the inde i refers to the i th element in the distribution. Becuse the chrge distribution is modeled s continuous, the totl field t P in the limit i : 0 is Electric field due to continuous chrge distribution E k e lim i : 0 i i r 2 rˆi k e i d r 2 rˆ (23.11) where the integrtion is over the entire chrge distribution. This is vector opertion nd must be treted ppropritel. We illustrte this tpe of clcultion with severl emples, in which we ssume the chrge is uniforml distributed on line, on surfce, or throughout volume. When performing such clcultions, it is convenient to use the concept of chrge densit long with the following nottions: If chrge Q is uniforml distributed throughout volume V, the volume chrge densit is defined b Volume chrge densit Q V where hs units of coulombs per cubic meter (C/m 3 ). If chrge Q is uniforml distributed on surfce of re A, the surfce chrge densit (lowercse Greek sigm) is defined b Surfce chrge densit Q A where hs units of coulombs per sure meter (C/m 2 ). If chrge Q is uniforml distributed long line of length, the liner chrge densit is defined b Liner chrge densit Q where hs units of coulombs per meter (C/m). If the chrge is nonuniforml distributed over volume, surfce, or line, the mounts of chrge d in smll volume, surfce, or length element re d dv d da d d PROBLEM-SOLVING HINTS Finding the Electric Field Units: in clcultions using the Coulomb constnt k e ( 1/4 0 ), chrges must be epressed in coulombs nd distnces in meters. Clculting the electric field of point chrges: to find the totl electric field t given point, first clculte the electric field t the point due to ech individul chrge. The resultnt field t the point is the vector sum of the fields due to the individul chrges. Continuous chrge distributions: when ou re confronted with problems tht involve continuous distribution of chrge, the vector sums for evluting the

16 SECTION 23.5 Electric Field of Continuous Chrge Distribution 721 totl electric field t some point must be replced b vector integrls. Divide the chrge distribution into infinitesiml pieces, nd clculte the vector sum b integrting over the entire chrge distribution. Emples 23.7 through 23.9 demonstrte this techniue. Smmetr: with both distributions of point chrges nd continuous chrge distributions, tke dvntge of n smmetr in the sstem to simplif our clcultions. Emple 23.7 The Electric Field Due to Chrged Rod A rod of length hs uniform positive chrge per unit length nd totl chrge Q. Clculte the electric field t point P tht is locted long the long is of the rod nd distnce from one end (Fig ). Solution Let us ssume tht the rod is ling long the is, tht d is the length of one smll segment, nd tht d is the chrge on tht segment. Becuse the rod hs chrge per unit length, the chrge d on the smll segment is d d. The field d E t P due to this segment is in the negtive direction (becuse the source of the field crries positive chrge), nd its mgnitude is Becuse ever other element lso produces field in the negtive direction, the problem of summing their contributions E P de k e d 2 k e d d 2 d = d Figure (Emple 23.7) The electric field t P due to uniforml chrged rod ling long the is. The mgnitude of the field t P due to the segment of chrge d is k e d/ 2. The totl field t P is the vector sum over ll segments of the rod. is prticulrl simple in this cse. The totl field t P due to ll segments of the rod, which re t different distnces from P, is given b Eution 23.11, which in this cse becomes 3 where the limits on the integrl etend from one end of the rod ( ) to the other ( ). The constnts k e nd cn be removed from the integrl to ield E k e E d 2 k e 1 k e 1 1 k e d 2 k e Q ( ) where we hve used the fct tht the totl chrge Q. Wht If? Suppose we move to point P ver fr w from the rod. Wht is the nture of the electric field t such point? Answer If P is fr from the rod ( ), then in the denomintor of the finl epression for E cn be neglected, nd E k e Q/ 2. This is just the form ou would epect for point chrge. Therefore, t lrge vlues of /, the chrge distribution ppers to be point chrge of mgnitude Q we re so fr w from the rod tht we cnnot distinguish tht it hs size. The use of the limiting techniue ( / : ) often is good method for checking mthemticl epression. Emple 23.8 The Electric Field of Uniform Ring of Chrge A ring of rdius crries uniforml distributed positive totl chrge Q. Clculte the electric field due to the ring t point P ling distnce from its center long the centrl is perpendiculr to the plne of the ring (Fig ). Solution The mgnitude of the electric field t P due to the segment of chrge d is This field hs n component de de cos long the is nd component de perpendiculr to the is. As we see in Figure 23.18b, however, the resultnt field t P must lie long the is becuse the perpendiculr comd E k e d r 2 3 It is importnt tht ou understnd how to crr out integrtions such s this. First, epress the chrge element d in terms of the other vribles in the integrl. (In this emple, there is one vrible,, nd so we mde the chnge d d.) The integrl must be over sclr untities; therefore, ou must epress the electric field in terms of components, if necessr. (In this emple the field hs onl n component, so we do not bother with this detil.) Then, reduce our epression to n integrl over single vrible (or to multiple integrls, ech over single vrible). In emples tht hve sphericl or clindricl smmetr, the single vrible will be rdil coordinte.

17 722 CHAPTER 23 Electric Fields d () r P de de de Figure (Emple 23.8) A uniforml chrged ring of rdius. () The field t P on the is due to n element of chrge d. (b) The totl electric field t P is long the is. The perpendiculr component of the field t P due to segment 1 is cnceled b the perpendiculr component due to segment (b) de 1 de 2 ponents of ll the vrious chrge segments sum to zero. Tht is, the perpendiculr component of the field creted b n chrge element is cnceled b the perpendiculr component creted b n element on the opposite side of the ring. Becuse r ( 2 2 ) 1/2 nd cos /r, we find tht d E d E cos k e d r 2 r All segments of the ring mke the sme contribution to the field t P becuse the re ll euidistnt from this point. Thus, we cn integrte to obtin the totl field t P: E k e ( 2 2 ) 3/2 d k e ( 2 2 ) 3/2 d k e ( 2 2 ) 3/2 Q k e ( 2 2 ) 3/2 d This result shows tht the field is zero t = 0. Does this finding surprise ou? Wht If? Suppose negtive chrge is plced t the center of the ring in Figure nd displced slightl b distnce long the is. When relesed, wht tpe of motion does it ehibit? Answer In the epression for the field due to ring of chrge, we let, which results in E k e Q 3 Thus, from Eution 23.8, the force on chrge plced ner the center of the ring is F k e Q 3 Becuse this force hs the form of Hooke s lw (E. 15.1), the motion will be simple hrmonic! Emple 23.9 The Electric Field of Uniforml Chrged Disk A disk of rdius R hs uniform surfce chrge densit. Clculte the electric field t point P tht lies long the centrl perpendiculr is of the disk nd distnce from the center of the disk (Fig ). Solution If we consider the disk s set of concentric rings, we cn use our result from Emple 23.8 which gives the field creted b ring of rdius nd sum the contributions of ll rings mking up the disk. B smmetr, the field t n il point must be long the centrl is. The ring of rdius r nd width dr shown in Figure hs surfce re eul to 2 r dr. The chrge d on this ring is eul to the re of the ring multiplied b the surfce chrge densit: d 2 r dr.using this result in the eution given for E in Emple 23.8 (with replced b r), we hve for the field due to the ring r R dr Figure (Emple 23.9) A uniforml chrged disk of rdius R. The electric field t n il point P is directed long the centrl is, perpendiculr to the plne of the disk. d P d E k e ( 2 r 2 ) 3/2 (2 r dr)to obtin the totl field t P, we integrte this epression over the limits r 0 to r R, noting tht is constnt.

18 SECTION 23.6 Electric Field Lines 723 This gives E k e R k e R ( 2 r 2 ) 3/2 d(r 2 ) k e ( 2 r 2 ) 1/2 R 1/ r dr ( 2 r 2 ) 3/2 2 k e 1 ( 2 R 2 ) 1/2 This result is vlid for ll vlues of 0. We cn clculte the field close to the disk long the is b ssuming tht R ; thus, the epression in prentheses reduces to unit to give us the ner-field pproimtion: E 2 k e where 0 is the permittivit of free spce. In the net chpter we shll obtin the sme result for the field creted b uniforml chrged infinite sheet Electric Field Lines We hve defined the electric field mthemticll through Eution We now eplore mens of representing the electric field pictorill. A convenient w of visulizing electric field ptterns is to drw curved lines tht re prllel to the electric field vector t n point in spce. These lines, clled electric field lines nd first introduced b Frd, re relted to the electric field in region of spce in the following mnner: The electric field vector E is tngent to the electric field line t ech point. The line hs direction, indicted b n rrowhed, tht is the sme s tht of the electric field vector. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of the electric field in tht region. Thus, the field lines re close together where the electric field is strong nd fr prt where the field is wek. These properties re illustrted in Figure The densit of lines through surfce A is greter thn the densit of lines through surfce B. Therefore, the mgnitude of the electric field is lrger on surfce A thn on surfce B. Furthermore, the fct tht the lines t different loctions point in different directions indictes tht the field is nonuniform. Is this reltionship between strength of the electric field nd the densit of field lines consistent with Eution 23.9, the epression we obtined for E using Coulomb s lw? To nswer this uestion, consider n imginr sphericl surfce of rdius r concentric with point chrge. From smmetr, we see tht the mgnitude of the electric field is the sme everwhere on the surfce of the sphere. The number of lines N tht emerge from the chrge is eul to the number tht penetrte the sphericl surfce. Hence, the number of lines per unit re on the sphere is N/4 r 2 (where the surfce re of the sphere is 4 r 2 ). Becuse E is proportionl to the number of lines per unit re, we see tht E vries s 1/r 2 ; this finding is consistent with Eution Representtive electric field lines for the field due to single positive point chrge re shown in Figure This two-dimensionl drwing shows onl the field lines tht lie in the plne contining the point chrge. The lines re ctull directed rdill outwrd from the chrge in ll directions; thus, insted of the flt wheel of lines shown, ou should picture n entire sphericl distribution of lines. Becuse positive test chrge plced in this field would be repelled b the positive source chrge, the lines re directed rdill w from the source chrge. The electric field lines representing the field due to single negtive point chrge re directed towrd the chrge (Fig b). In either cse, the lines re long the rdil direction nd etend ll the w to infinit. Note tht the lines become closer together s the pproch the chrge; this indictes tht the strength of the field increses s we move towrd the source chrge. B A Figure Electric field lines penetrting two surfces. The mgnitude of the field is greter on surfce A thn on surfce B. PITFALL PREVENTION 23.2 Electric Field Lines re not Pths of Prticles! Electric field lines represent the field t vrious loctions. Ecept in ver specil cses, the do not represent the pth of chrged prticle moving in n electric field.

19 724 CHAPTER 23 Electric Fields () (b) Figure The electric field lines for point chrge. () For positive point chrge, the lines re directed rdill outwrd. (b) For negtive point chrge, the lines re directed rdill inwrd. Note tht the figures show onl those field lines tht lie in the plne of the pge. (c) The drk res re smll pieces of thred suspended in oil, which lign with the electric field produced b smll chrged conductor t the center. Courtes of Hrold M. Wge, Princeton Universit (c) The rules for drwing electric field lines re s follows: The lines must begin on positive chrge nd terminte on negtive chrge. In the cse of n ecess of one tpe of chrge, some lines will begin or end infinitel fr w. The number of lines drwn leving positive chrge or pproching negtive chrge is proportionl to the mgnitude of the chrge. No two field lines cn cross. PITFALL PREVENTION 23.3 Electric Field Lines re not Rel Electric field lines re not mteril objects. The re used onl s pictoril representtion to provide ulittive description of the electric field. Onl finite number of lines from ech chrge cn be drwn, which mkes it pper s if the field were untized nd eists onl in certin prts of spce. The field, in fct, is continuous eisting t ever point. You should void obtining the wrong impression from two-dimensionl drwing of field lines used to describe three-dimensionl sitution. We choose the number of field lines strting from n positivel chrged object to be C nd the number of lines ending on n negtivel chrged object to be C, where C is n rbitrr proportionlit constnt. Once C is chosen, the number of lines is fied. For emple, if object 1 hs chrge Q 1 nd object 2 hs chrge Q 2, then the rtio of number of lines is N 2 /N 1 Q 2 /Q 1. The electric field lines for two point chrges of eul mgnitude but opposite signs (n electric dipole) re shown in Figure Becuse the chrges re of eul mgnitude, the number of lines tht begin t the positive chrge must eul the number tht terminte t the negtive chrge. At points ver ner the chrges, the lines re nerl rdil. The high densit of lines between the chrges indictes region of strong electric field. Courtes of Hrold M. Wge, Princeton Universit () Figure () The electric field lines for two point chrges of eul mgnitude nd opposite sign (n electric dipole). The number of lines leving the positive chrge euls the number terminting t the negtive chrge. (b) The drk lines re smll pieces of thred suspended in oil, which lign with the electric field of dipole. (b)

20 SECTION 23.7 Motion of Chrged Prticles in Uniform Electric Field 725 A C B Courtes of Hrold M. Wge, Princeton Universit () Figure () The electric field lines for two positive point chrges. (The loctions A, B, nd C re discussed in Quick Quiz 23.7.) (b) Pieces of thred suspended in oil, which lign with the electric field creted b two eul-mgnitude positive chrges. (b) Figure shows the electric field lines in the vicinit of two eul positive point chrges. Agin, the lines re nerl rdil t points close to either chrge, nd the sme number of lines emerge from ech chrge becuse the chrges re eul in mgnitude. At gret distnces from the chrges, the field is pproimtel eul to tht of single point chrge of mgnitude 2. Finll, in Figure we sketch the electric field lines ssocited with positive chrge 2 nd negtive chrge. In this cse, the number of lines leving 2 is twice the number terminting t. Hence, onl hlf of the lines tht leve the positive chrge rech the negtive chrge. The remining hlf terminte on negtive chrge we ssume to be t infinit. At distnces tht re much greter thn the chrge seprtion, the electric field lines re euivlent to those of single chrge. 2 Quick Quiz 23.7 Rnk the mgnitudes of the electric field t points A, B, nd C shown in Figure (gretest mgnitude first). Quick Quiz 23.8 Which of the following sttements bout electric field lines ssocited with electric chrges is flse? () Electric field lines cn be either stright or curved. (b) Electric field lines cn form closed loops. (c) Electric field lines begin on positive chrges nd end on negtive chrges. (d) Electric field lines cn never intersect with one nother Motion of Chrged Prticles in Uniform Electric Field Active Figure The electric field lines for point chrge 2 nd second point chrge. Note tht two lines leve 2 for ever one tht termintes on. At the Active Figures link t ou cn choose the vlues nd signs for the two chrges nd observe the electric field lines for the configurtion tht ou hve chosen. When prticle of chrge nd mss m is plced in n electric field E, the electric force eerted on the chrge is E ccording to Eution If this is the onl force eerted on the prticle, it must be the net force nd cuses the prticle to ccelerte ccording to Newton s second lw. Thus, F e E m The ccelertion of the prticle is therefore E m (23.12) If E is uniform (tht is, constnt in mgnitude nd direction), then the ccelertion is constnt. If the prticle hs positive chrge, its ccelertion is in the direction of the

21 726 CHAPTER 23 Electric Fields electric field. If the prticle hs negtive chrge, its ccelertion is in the direction opposite the electric field. Emple An Accelerting Positive Chrge A positive point chrge of mss m is relesed from rest in uniform electric field E directed long the is, s shown in Figure Describe its motion. Solution The ccelertion is constnt nd is given b E/m. The motion is simple liner motion long the is. Therefore, we cn ppl the eutions of kinemtics in one dimension (see Chpter 2): v f 2 v 2 i 2( f i ) Choosing the initil position of the chrge s i 0 nd ssigning v i 0 becuse the prticle strts from rest, the position of the prticle s function of time is The speed of the prticle is given b The third kinemtic eution gives us v 2 f f i v i t 1 2 t 2 v f v i t f 1 2 t 2 E 2m t 2 v f t E m t 2 f 2 E m f from which we cn find the kinetic energ of the chrge fter it hs moved distnce f i : K 1 2 mv f m 2 E m E We cn lso obtin this result from the workkinetic energ theorem becuse the work done b the electric force is F e E nd W K. v = 0 E v Figure (Emple 23.10) A positive point chrge in uniform electric field E undergoes constnt ccelertion in the direction of the field. The electric field in the region between two oppositel chrged flt metllic pltes is pproimtel uniform (Fig ). Suppose n electron of chrge e is projected horizontll into this field from the origin with n initil velocit v i î t time t 0. Becuse the electric field E in Figure is in the positive direction, the ccelertion of the electron is in the negtive direction. Tht is, e E m e ĵ (23.13) Becuse the ccelertion is constnt, we cn ppl the eutions of kinemtics in two dimensions (see Chpter 4) with v i v i nd v i 0. After the electron hs been in the At the Active Figures link t ou cn choose the strength of the electric field nd the mss nd chrge of the projected prticle. v i î (0, 0) (,) E v Active Figure An electron is projected horizontll into uniform electric field produced b two chrged pltes. The electron undergoes downwrd ccelertion (opposite E), nd its motion is prbolic while it is between the pltes.

22 SECTION 23.7 Motion of Chrged Prticles in Uniform Electric Field 727 electric field for time intervl, the components of its velocit t time t re v v i constnt v t ee m e t (23.14) (23.15) Its position coordintes t time t re f v i t f 1 2 t e E m e t 2 (23.16) (23.17) Substituting the vlue t f /v i from Eution into Eution 23.17, we see tht f is proportionl to f 2. Hence, the trjector is prbol. This should not be surprise consider the nlogous sitution of throwing bll horizontll in uniform grvittionl field (Chpter 4). After the electron leves the field, the electric force vnishes nd the electron continues to move in stright line in the direction of v in Figure with speed v v i. Note tht we hve neglected the grvittionl force cting on the electron. This is good pproimtion when we re deling with tomic prticles. For n electric field of 10 4 N/C, the rtio of the mgnitude of the electric force ee to the mgnitude of the grvittionl force mg is on the order of for n electron nd on the order of for proton. PITFALL PREVENTION 23.4 Just Another Force Electric forces nd fields m seem bstrct to ou. However, once F e is evluted, it cuses prticle to move ccording to our well-estblished understnding of forces nd motion from Chpters 5 nd 6. Keeping this link with the pst in mind will help ou solve problems in this chpter. Emple An Accelerted Electron Interctive An electron enters the region of uniform electric field s shown in Figure 23.26, with v i m/s nd E 200 N/C. The horizontl length of the pltes is m. (A) Find the ccelertion of the electron while it is in the electric field. Solution The chrge on the electron hs n bsolute vlue of C, nd m e kg. Therefore, Eution gives e E m e ĵ ( C)(200 N/C) kg ĵ m/s2 (B) If the electron enters the field t time t 0, find the time t which it leves the field. ĵ Solution The horizontl distnce cross the field is m. Using Eution with f, we find tht the time t which the electron eits the electric field is t v i (C) If the verticl position of the electron s it enters the field is i 0, wht is its verticl position when it leves the field? Solution Using Eution nd the results from prts (A) nd (B), we find tht f 1 2 t ( m/s 2 )( s) m m m/s 1.95 cm s If the electron enters just below the negtive plte in Figure nd the seprtion between the pltes is less thn the vlue we hve just clculted, the electron will strike the positive plte. At the Interctive Worked Emple link t ou cn predict the reuired initil velocit for the eiting electron to just miss the right edge of the lower plte, for rndom vlues of the electric field. The Cthode R Tube The emple we just worked describes portion of cthode r tube (CRT). This tube, illustrted in Figure 23.27, is commonl used to obtin visul displ of electronic informtion in oscilloscopes, rdr sstems, television receivers, nd computer monitors. The CRT is vcuum tube in which bem of electrons is ccelerted nd deflected under the influence of electric or mgnetic fields. The electron bem is

23 728 CHAPTER 23 Electric Fields Electron gun C A Verticl deflection pltes Verticl input Horizontl deflection pltes Electron bem Horizontl input Fluorescent screen Figure Schemtic digrm of cthode r tube. Electrons leving the cthode C re ccelerted to the node A. In ddition to ccelerting electrons, the electron gun is lso used to focus the bem of electrons, nd the pltes deflect the bem. produced b n ssembl clled n electron gun locted in the neck of the tube. These electrons, if left undisturbed, trvel in stright-line pth until the strike the front of the CRT, the screen, which is coted with mteril tht emits visible light when bombrded with electrons. In n oscilloscope, the electrons re deflected in vrious directions b two sets of pltes plced t right ngles to ech other in the neck of the tube. (A television CRT steers the bem with mgnetic field, s discussed in Chpter 29.) An eternl electric circuit is used to control the mount of chrge present on the pltes. The plcing of positive chrge on one horizontl plte nd negtive chrge on the other cretes n electric field between the pltes nd llows the bem to be steered from side to side. The verticl deflection pltes ct in the sme w, ecept tht chnging the chrge on them deflects the bem verticll. SUMMARY Tke prctice test for this chpter b clicking on the Prctice Test link t Electric chrges hve the following importnt properties: Chrges of opposite sign ttrct one nother nd chrges of the sme sign repel one nother. Totl chrge in n isolted sstem is conserved. Chrge is untized. Conductors re mterils in which electrons move freel. Insultors re mterils in which electrons do not move freel. Coulomb s lw sttes tht the electric force eerted b chrge 1 on second chrge 2 is F 12 k e 1 2 r 2 rˆ (23.6) where r is the distnce between the two chrges nd rˆ is unit vector directed from 1 towrd 2. The constnt k e, which is clled the Coulomb constnt, hs the vlue k e N m 2 /C 2. The smllest unit of free chrge e known to eist in nture is the chrge on n electron ( e) or proton ( e), where e C. The electric field E t some point in spce is defined s the electric force F e tht cts on smll positive test chrge plced t tht point divided b the mgnitude 0 of the test chrge: E F e 0 (23.7) Thus, the electric force on chrge plced in n electric field E is given b F e E (23.8)

24 Questions 729 At distnce r from point chrge, the electric field due to the chrge is given b E k e (23.9) where rˆ is unit vector directed from the chrge towrd the point in uestion. The electric field is directed rdill outwrd from positive chrge nd rdill inwrd towrd negtive chrge. The electric field due to group of point chrges cn be obtined b using the superposition principle. Tht is, the totl electric field t some point euls the vector sum of the electric fields of ll the chrges: E k e i r 2 rˆ i r i 2 rˆi The electric field t some point due to continuous chrge distribution is E k e d r 2 rˆ (23.10) (23.11) where d is the chrge on one element of the chrge distribution nd r is the distnce from the element to the point in uestion. Electric field lines describe n electric field in n region of spce. The number of lines per unit re through surfce perpendiculr to the lines is proportionl to the mgnitude of E in tht region. A chrged prticle of mss m nd chrge moving in n electric field E hs n ccelertion E (23.12) m QUESTIONS 1. Eplin wht is ment b the term neutrl tom. Eplin wht negtivel chrged tom mens. 2. A chrged comb often ttrcts smll bits of dr pper tht then fl w when the touch the comb. Eplin. 3. Sprks re often seen or herd on dr d when fbrics re removed from clothes drer in dim light. Eplin. 4. Hospitl personnel must wer specil conducting shoes while working round ogen in n operting room. Wh? Contrst with wht might hppen if people wore rubbersoled shoes. 5. Eplin from n tomic viewpoint wh chrge is usull trnsferred b electrons. 6. A light, unchrged metllic sphere suspended from thred is ttrcted to chrged rubber rod. After it touches the rod, the sphere is repelled b the rod. Eplin. 7. A foreign student who grew up in tropicl countr but is studing in the United Sttes m hve hd no eperience with sttic electricit sprks or shocks until he or she first eperiences n Americn winter. Eplin. 8. Eplin the similrities nd differences between Newton s lw of universl grvittion nd Coulomb s lw. 9. A blloon is negtivel chrged b rubbing nd then clings to wll. Does this men tht the wll is positivel chrged? Wh does the blloon eventull fll? 10. A light strip of luminum foil is drped over horizontl wooden pencil. When rod crring positive chrge is brought close to the foil, the two prts of the foil stnd prt. Wh? Wht kind of chrge is on the foil? 11. When defining the electric field, wh is it necessr to specif tht the mgnitude of the test chrge be ver smll? 12. How could ou eperimentll distinguish n electric field from grvittionl field? 13. A lrge metllic sphere insulted from ground is chrged with n electrosttic genertor while student stnding on n insulting stool holds the sphere. Wh is it sfe to do this? Wh would it not be sfe for nother person to touch the sphere fter it hd been chrged? 14. Is it possible for n electric field to eist in empt spce? Eplin. Consider point A in Figure 23.23(). Does chrge eist t this point? Does force eist t this point? Does field eist t this point? 15. When is it vlid to pproimte chrge distribution b point chrge? 16. Eplin wh electric field lines never cross. Suggestion: Begin b eplining wh the electric field t prticulr point must hve onl one direction. 17. Figures nd show three electric field vectors t the sme point. With little etrpoltion, Figure

25 730 CHAPTER 23 Electric Fields would show mn electric field lines t the sme point. Is it rell true tht no two field lines cn cross? Are the digrms drwn correctl? Eplin our nswers. 18. A free electron nd free proton re relesed in identicl electric fields. Compre the electric forces on the two prticles. Compre their ccelertions. 19. Eplin wht hppens to the mgnitude of the electric field creted b point chrge s r pproches zero. 20. An object with negtive chrge is plced in region of spce where the electric field is directed verticll upwrd. Wht is the direction of the electric force eerted on this chrge? 21. A chrge 4 is t distnce r from chrge. Compre the number of electric field lines leving the chrge 4 with the number entering the chrge. Where do the etr lines beginning on 4 end? 22. Consider two eul point chrges seprted b some distnce d. At wht point (other thn ) would third test chrge eperience no net force? 23. Eplin the differences between liner, surfce, nd volume chrge densities, nd give emples of when ech would be used. 24. If the electron in Figure is projected into the electric field with n rbitrr velocit v i (t n rbitrr ngle to E), will its trjector still be prbolic? Eplin. 25. Would life be different if the electron were positivel chrged nd the proton were negtivel chrged? Does the choice of signs hve n bering on phsicl nd chemicl interctions? Eplin. 26. Wh should ground wire be connected to the metl support rod for television ntenn? 27. Suppose someone proposes the ide tht people re bound to the Erth b electric forces rther thn b grvit. How could ou prove this ide is wrong? 28. Consider two electric dipoles in empt spce. Ech dipole hs zero net chrge. Does n electric force eist between the dipoles tht is, cn two objects with zero net chrge eert electric forces on ech other? If so, is the force one of ttrction or of repulsion? PROBLEMS 1, 2, 3 = strightforwrd, intermedite, chllenging = full solution vilble in the Student Solutions Mnul nd Stud Guide = coched solution with hints vilble t = computer useful in solving problem = pired numericl nd smbolic problems Section 23.1 Properties of Electric Chrges 1. () Find to three significnt digits the chrge nd the mss of n ionized hdrogen tom, represented s H. Suggestion: Begin b looking up the mss of neutrl tom on the periodic tble of the elements. (b) Find the chrge nd the mss of N, singl ionized sodium tom. (c) Find the chrge nd the verge mss of chloride ion Cl tht joins with the N to mke one molecule of tble slt. (d) Find the chrge nd the mss of C C 2, doubl ionized clcium tom. (e) You cn model the center of n mmoni molecule s n N 3 ion. Find its chrge nd mss. (f) The plsm in hot str contins udrupl ionized nitrogen toms, N 4. Find their chrge nd mss. (g) Find the chrge nd the mss of the nucleus of nitrogen tom. (h) Find the chrge nd the mss of the moleculr ion H 2 O. 2. () Clculte the number of electrons in smll, electricll neutrl silver pin tht hs mss of 10.0 g. Silver hs 47 electrons per tom, nd its molr mss is g/mol. (b) Electrons re dded to the pin until the net negtive chrge is 1.00 mc. How mn electrons re dded for ever 10 9 electrons lred present? Section 23.2 Section 23.3 Chrging Objects b Induction Coulomb s Lw 3. The Nobel lurete Richrd Fenmn once sid tht if two persons stood t rm s length from ech other nd ech person hd 1% more electrons thn protons, the force of repulsion between them would be enough to lift weight eul to tht of the entire Erth. Crr out n order-of-mgnitude clcultion to substntite this ssertion. 4. Two protons in n tomic nucleus re tpicll seprted b distnce of m. The electric repulsion force between the protons is huge, but the ttrctive nucler force is even stronger nd keeps the nucleus from bursting prt. Wht is the mgnitude of the electric force between two protons seprted b m? 5. () Two protons in molecule re seprted b m. Find the electric force eerted b one proton on the other. (b) How does the mgnitude of this force compre to the mgnitude of the grvittionl force between the two protons? (c) Wht If? Wht must be the chrgeto-mss rtio of prticle if the mgnitude of the grvittionl force between two of these prticles euls the mgnitude of electric force between them? 6. Two smll silver spheres, ech with mss of 10.0 g, re seprted b 1.00 m. Clculte the frction of the electrons in one sphere tht must be trnsferred to the other in order to produce n ttrctive force of N (bout 1 ton) between the spheres. (The number of electrons per tom of silver is 47, nd the number of toms per grm is Avogdro s number divided b the molr mss of silver, g/mol.) 7. Three point chrges re locted t the corners of n euilterl tringle s shown in Figure P23.7. Clculte the resultnt electric force on the C chrge.

26 Problems µ C 0.50 m µ C 4.00 µ C d/2 d/2 Q Figure P23.7 Problems 7 nd Suppose tht 1.00 g of hdrogen is seprted into electrons nd protons. Suppose lso tht the protons re plced t the Erth s north pole nd the electrons re plced t the south pole. Wht is the resulting compressionl force on the Erth? 9. Two identicl conducting smll spheres re plced with their centers m prt. One is given chrge of 12.0 nc nd the other chrge of 18.0 nc. () Find the electric force eerted b one sphere on the other. (b) Wht If? The spheres re connected b conducting wire. Find the electric force between the two fter the hve come to euilibrium. 10. Two smll beds hving positive chrges 3 nd re fied t the opposite ends of horizontl, insulting rod, etending from the origin to the point d. As shown in Figure P23.10, third smll chrged bed is free to slide on the rod. At wht position is the third bed in euilibrium? Cn it be in stble euilibrium? Section Figure P23.12 The Electric Field Wht re the mgnitude nd direction of the electric field tht will blnce the weight of () n electron nd (b) proton? (Use the dt in Tble 23.1.) 14. An object hving net chrge of 24.0 C is plced in uniform electric field of 610 N/C directed verticll. Wht is the mss of this object if it flots in the field? 15. In Figure P23.15, determine the point (other thn infinit) t which the electric field is zero m 2.50 µc µ 6.00 µc µ 3 Figure P23.15 d Figure P Review problem. In the Bohr theor of the hdrogen tom, n electron moves in circulr orbit bout proton, where the rdius of the orbit is m. () Find the electric force between the two. (b) If this force cuses the centripetl ccelertion of the electron, wht is the speed of the electron? 12. Review problem. Two identicl prticles, ech hving chrge, re fied in spce nd seprted b distnce d. A third point chrge Q is free to move nd lies initill t rest on the perpendiculr bisector of the two fied chrges distnce from the midpoint between the two fied chrges (Fig. P23.12). () Show tht if is smll compred with d, the motion of Q will be simple hrmonic long the perpendiculr bisector. Determine the period of tht motion. (b) How fst will the chrge Q be moving when it is t the midpoint between the two fied chrges, if initill it is relesed t distnce d from the midpoint? 16. An irplne is fling through thundercloud t height of m. (This is ver dngerous thing to do becuse of updrfts, turbulence, nd the possibilit of electric dischrge.) If chrge concentrtion of 40.0 C is bove the plne t height of m within the cloud nd chrge concentrtion of 40.0 C is t height m, wht is the electric field t the ircrft? 17. Two point chrges re locted on the is. The first is chrge Q t. The second is n unknown chrge locted t 3. The net electric field these chrges produce t the origin hs mgnitude of 2k e Q / 2. Wht re the two possible vlues of the unknown chrge? 18. Three chrges re t the corners of n euilterl tringle s shown in Figure P23.7. () Clculte the electric field t the position of the C chrge due to the C nd C chrges. (b) Use our nswer to prt () to determine the force on the C chrge. 19. Three point chrges re rrnged s shown in Figure P () Find the vector electric field tht the 6.00-nC nd 3.00-nC chrges together crete t the origin. (b) Find the vector force on the 5.00-nC chrge.

27 732 CHAPTER 23 Electric Fields 5.00 nc m m 6.00 nc 24. Consider n infinite number of identicl chrges (ech of chrge ) plced long the is t distnces, 2, 3, 4,..., from the origin. Wht is the electric field t the origin due to this distribution? Suggestion: Use the fct tht nc 20. Two C point chrges re locted on the is. One is t 1.00 m, nd the other is t 1.00 m. () Determine the electric field on the is t m. (b) Clculte the electric force on C chrge plced on the is t m. 21. Four point chrges re t the corners of sure of side s shown in Figure P () Determine the mgnitude nd direction of the electric field t the loction of chrge. (b) Wht is the resultnt force on? 2 3 Figure P23.19 Figure P Consider the electric dipole shown in Figure P Show tht the electric field t distnt point on the is is E 4k e / 3. 2 Figure P Consider n eul positive point chrges ech of mgnitude Q/n plced smmetricll round circle of rdius R. () Clculte the mgnitude of the electric field t point distnce on the line pssing through the center of the circle nd perpendiculr to the plne of the circle. (b) Eplin wh this result is identicl to tht of the clcultion done in Emple Section 23.5 Electric Field of Continuous Chrge Distribution 25. A rod 14.0 cm long is uniforml chrged nd hs totl chrge of 22.0 C. Determine the mgnitude nd direction of the electric field long the is of the rod t point 36.0 cm from its center. 26. A continuous line of chrge lies long the is, etending from 0 to positive infinit. The line crries chrge with uniform liner chrge densit 0. Wht re the mgnitude nd direction of the electric field t the origin? 27. A uniforml chrged ring of rdius 10.0 cm hs totl chrge of 75.0 C. Find the electric field on the is of the ring t () 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, nd (d) 100 cm from the center of the ring. 28. A line of chrge strts t 0 nd etends to positive infinit. The liner chrge densit is 0 0 /. Determine the electric field t the origin. 29. Show tht the mimum mgnitude E m of the electric field long the is of uniforml chrged ring occurs t / 2 (see Fig ) nd hs the vlue Q /( ). 30. A uniforml chrged disk of rdius 35.0 cm crries chrge with densit of C/m 2. Clculte the electric field on the is of the disk t () 5.00 cm, (b) 10.0 cm, (c) 50.0 cm, nd (d) 200 cm from the center of the disk. 31. Emple 23.9 derives the ect epression for the electric field t point on the is of uniforml chrged disk. Consider disk, of rdius R 3.00 cm, hving uniforml distributed chrge of 5.20 C. () Using the result of Emple 23.9, compute the electric field t point on the is nd 3.00 mm from the center. Wht If? Compre this nswer with the field computed from the ner-field pproimtion E /2 0. (b) Using the result of Emple 23.9, compute the electric field t point on the is nd 30.0 cm from the center of the disk. Wht If? Compre this with the electric field obtined b treting the disk s C point chrge t distnce of 30.0 cm. 32. The electric field long the is of uniforml chrged disk of rdius R nd totl chrge Q ws clculted in Emple Show tht the electric field t distnces tht re lrge compred with R pproches tht of point chrge Q R 2. (Suggestion: First show tht /( 2 R 2 ) 1/2 (1 R 2 / 2 ) 1/2 nd use the binomil epnsion (1 ) n 1 n when 1.) 33. A uniforml chrged insulting rod of length 14.0 cm is bent into the shpe of semicircle s shown in Figure P The rod hs totl chrge of 7.50 C. Find the mgnitude nd direction of the electric field t O, the center of the semicircle.

28 Problems () Consider uniforml chrged thin-wlled right circulr clindricl shell hving totl chrge Q, rdius R, nd height h. Determine the electric field t point distnce d from the right side of the clinder s shown in Figure P (Suggestion: Use the result of Emple 23.8 nd tret the clinder s collection of ring chrges.) (b) Wht If? Consider now solid clinder with the sme dimensions nd crring the sme chrge, uniforml distributed through its volume. Use the result of Emple 23.9 to find the field it cretes t the sme point. O Figure P Three solid plstic clinders ll hve rdius 2.50 cm nd length 6.00 cm. One () crries chrge with uniform densit 15.0 nc/m 2 everwhere on its surfce. Another (b) crries chrge with the sme uniform densit on its curved lterl surfce onl. The third (c) crries chrge with uniform densit 500 nc/m 3 throughout the plstic. Find the chrge of ech clinder. 37. Eight solid plstic cubes, ech 3.00 cm on ech edge, re glued together to form ech one of the objects (i, ii, iii, nd iv) shown in Figure P () Assuming ech object crries chrge with uniform densit 400 nc/m 3 throughout its volume, find the chrge of ech object. (b) Assuming ech object crries chrge with uniform densit 15.0 nc/m 2 everwhere on its eposed surfce, find the chrge on ech object. (c) Assuming chrge is plced onl on the edges where perpendiculr surfces meet, with uniform densit 80.0 pc/m, find the chrge of ech object. h R d d Figure P23.34 (i) (ii) (iii) (iv) Figure P A thin rod of length nd uniform chrge per unit length lies long the is, s shown in Figure P () Show tht the electric field t P, distnce from the rod long its perpendiculr bisector, hs no component nd is given b E 2k e sin 0 /. (b) Wht If? Using our result to prt (), show tht the field of rod of infinite length is E 2k e /. (Suggestion: First clculte the field t P due to n element of length d, which hs chrge d. Then chnge vribles from to, using the reltionships tn nd d sec 2 d, nd integrte over.) Section 23.6 Electric Field Lines 38. A positivel chrged disk hs uniform chrge per unit re s described in Emple Sketch the electric field lines in plne perpendiculr to the plne of the disk pssing through its center. 39. A negtivel chrged rod of finite length crries chrge with uniform chrge per unit length. Sketch the electric field lines in plne contining the rod. 40. Figure P23.40 shows the electric field lines for two point chrges seprted b smll distnce. () Determine the rtio 1 / 2. (b) Wht re the signs of 1 nd 2? P 0 2 O d Figure P Figure P23.40

29 734 CHAPTER 23 Electric Fields 41. Three eul positive chrges re t the corners of n euilterl tringle of side s shown in Figure P () Assume tht the three chrges together crete n electric field. Sketch the field lines in the plne of the chrges. Find the loction of point (other thn ) where the electric nd with positive chrge of C leves the center of the bottom negtive plte with n initil speed of m/s t n ngle of 37.0 bove the horizontl. Describe the trjector of the prticle. Which plte does it strike? Where does it strike, reltive to its strting point? field is zero. (b) Wht re the mgnitude nd direction of 49. Protons re projected with n initil speed v i 9.55 the electric field t P due to the two chrges t the bse? 10 3 m/s into region where uniform electric field E 720 ĵ N/C is present, s shown in Figure P The P protons re to hit trget tht lies t horizontl distnce of 1.27 mm from the point where the protons cross the plne nd enter the electric field in Figure P Find () the two projection ngles tht will result in hit nd (b) the totl time of flight (the time intervl during which the proton is bove the plne in Figure P23.49) for ech trjector. Figure P23.41 E = (720j) ˆ N/C Section 23.7 Motion of Chrged Prticles in Uniform Electric Field 42. An electron nd proton re ech plced t rest in n electric field of 520 N/C. Clculte the speed of ech prticle 48.0 ns fter being relesed. 43. A proton ccelertes from rest in uniform electric field of 640 N/C. At some lter time, its speed is m/s (nonreltivistic, becuse v is much less thn the speed of light). () Find the ccelertion of the proton. (b) How long does it tke the proton to rech this speed? (c) How fr hs it moved in this time? (d) Wht is its kinetic energ t this time? 44. A proton is projected in the positive direction into region of uniform electric field E î N/C t t 0. The proton trvels 7.00 cm before coming to rest. Determine () the ccelertion of the proton, (b) its initil speed, nd (c) the time t which the proton comes to rest. 45. The electrons in prticle bem ech hve kinetic energ K. Wht re the mgnitude nd direction of the electric field tht will stop these electrons in distnce d? 46. A positivel chrged bed hving mss of 1.00 g flls from rest in vcuum from height of 5.00 m in uniform verticl electric field with mgnitude of N/C. The bed hits the ground t speed of 21.0 m/s. Determine () the direction of the electric field (up or down), nd (b) the chrge on the bed. 47. A proton moves t m/s in the horizontl direction. It enters uniform verticl electric field with mgnitude of N/C. Ignoring n grvittionl effects, find () the time intervl reuired for the proton to trvel 5.00 cm horizontll, (b) its verticl displcement during the time intervl in which it trvels 5.00 cm horizontll, nd (c) the horizontl nd verticl components of its velocit fter it hs trveled 5.00 cm horizontll. 48. Two horizontl metl pltes, ech 100 mm sure, re ligned 10.0 mm prt, with one bove the other. The re given eul-mgnitude chrges of opposite sign so tht uniform downwrd electric field of N/C eists in the region between them. A prticle of mss kg 4.00 nc Proton bem m v i Additionl Problems 1.27 mm Trget Figure P Two known chrges, 12.0 C nd 45.0 C, nd n unknown chrge re locted on the is. The chrge 12.0 C is t the origin, nd the chrge 45.0 C is t 15.0 cm. The unknown chrge is to be plced so tht ech chrge is in euilibrium under the ction of the electric forces eerted b the other two chrges. Is this sitution possible? Is it possible in more thn one w? Find the reuired loction, mgnitude, nd sign of the unknown chrge. 51. A uniform electric field of mgnitude 640 N/C eists between two prllel pltes tht re 4.00 cm prt. A proton is relesed from the positive plte t the sme instnt tht n electron is relesed from the negtive plte. () Determine the distnce from the positive plte t which the two pss ech other. (Ignore the electricl ttrction between the proton nd electron.) (b) Wht If? Repet prt () for sodium ion (N ) nd chloride ion (Cl ). 52. Three point chrges re ligned long the is s shown in Figure P Find the electric field t () the position (2.00, 0) nd (b) the position (0, 2.00) nc m Figure P nc

30 Problems A resercher studing the properties of ions in the upper tmosphere wishes to construct n pprtus with the following chrcteristics: Using n electric field, bem of ions, ech hving chrge, mss m, nd initil velocit vî, is turned through n ngle of 90 s ech ion undergoes displcement Rî R ĵ. The ions enter chmber s shown in Figure P23.53, nd leve through the eit port with the sme speed the hd when the entered the chmber. The electric field cting on the ions is to hve constnt mgnitude. () Suppose the electric field is produced b two concentric clindricl electrodes not shown in the digrm, nd hence is rdil. Wht mgnitude should the field hve? Wht If? (b) If the field is produced b two flt pltes nd is uniform in direction, wht vlue should the field hve in this cse? the ngle. Find () the chrge on the bll nd (b) the tension in the string. E Figure P23.55 Problems 55 nd 56. v v R R 57. Four identicl point chrges ( 10.0 C) re locted on the corners of rectngle s shown in Figure P The dimensions of the rectngle re L 60.0 cm nd W 15.0 cm. Clculte the mgnitude nd direction of the resultnt electric force eerted on the chrge t the lower left corner b the other three chrges. Figure P A smll, 2.00-g plstic bll is suspended b 20.0-cm-long string in uniform electric field s shown in Figure P If the bll is in euilibrium when the string mkes 15.0 ngle with the verticl, wht is the net chrge on the bll? L Figure P23.57 W E = ˆi N/C 58. Inez is putting up decortions for her sister s uinceñer (fifteenth birthd prt). She ties three light silk ribbons together to the top of gtew nd hngs rubber blloon from ech ribbon (Fig. P23.58). To include the 20.0 cm 15.0 m = 2.00 g Figure P A chrged cork bll of mss 1.00 g is suspended on light string in the presence of uniform electric field s shown in Figure P When E (3.00î 5.00ĵ) 10 5 N/C, the bll is in euilibrium t Find () the chrge on the bll nd (b) the tension in the string. 56. A chrged cork bll of mss m is suspended on light string in the presence of uniform electric field s shown in Figure P When E (Aî Bĵ) N/C, where A nd B re positive numbers, the bll is in euilibrium t Figure P23.58