chpter 24 Guss s Lw 24.1 lectric Flux 24.2 Guss s Lw 24.3 Appliction of Guss s Lw to Vrious Chrge Distributions 24.4 Conductors in lectrosttic quilibrium In Chpter 23, we showed how to clculte the electric field due to given chrge distribution by integrting over the distribution. In this chpter, we describe Guss s lw nd n lterntive procedure for clculting electric fields. Guss s lw is bsed on the inverse-squre behvior of the electric force between point chrges. Although Guss s lw is direct consequence of Coulomb s lw, it is more convenient for clculting the electric fields of highly symmetric chrge distributions nd mkes it possible to del with complicted problems using qulittive resoning. As we show in this chpter, Guss s lw is importnt in understnding nd verifying the properties of conductors in electrosttic equilibrium. In tbletop plsm bll, the colorful lines emnting from the sphere give evidence of strong electric fields. Using Guss s lw, we show in this chpter tht the electric field surrounding uniformly chrged sphere is identicl to tht of point chrge. (teve Cole/Getty Imges) Are A Figure 24.1 Field lines representing uniform electric field penetrting plne of re A perpendiculr to the field. 24.1 lectric Flux The concept of electric field lines ws described qulittively in Chpter 23. We now tret electric field lines in more quntittive wy. Consider n electric field tht is uniform in both mgnitude nd direction s shown in Figure 24.1. The field lines penetrte rectngulr surfce of re A, whose plne is oriented perpendiculr to the field. Recll from ection 23.6 tht the number of lines per unit re (in other words, the line density) is proportionl to the mgnitude of the electric field. Therefore, the totl number of lines penetrting the surfce is proportionl to the product A. This product of the mgnitude 690
24.1 lectric Flux 691 of the electric field nd surfce re A perpendiculr to the field is clled the electric flux F (uppercse Greek letter phi): F 5 A (24.1) From the I units of nd A, we see tht F hs units of newton meters squred per coulomb (N? m 2 /C). lectric flux is proportionl to the number of electric field lines penetrting some surfce. If the surfce under considertion is not perpendiculr to the field, the flux through it must be less thn tht given by qution 24.1. Consider Figure 24.2, where the norml to the surfce of re A is t n ngle u to the uniform electric field. Notice tht the number of lines tht cross this re A is equl to the number of lines tht cross the re A, which is projection of re A onto plne oriented perpendiculr to the field. Figure 24.2 shows tht the two res re relted by A 5 A cos u. Becuse the flux through A equls the flux through A, the flux through A is F 5 A 5 A cos u (24.2) From this result, we see tht the flux through surfce of fixed re A hs mximum vlue A when the surfce is perpendiculr to the field (when the norml to the surfce is prllel to the field, tht is, when u 5 08 in Fig. 24.2); the flux is zero when the surfce is prllel to the field (when the norml to the surfce is perpendiculr to the field, tht is, when u 5 908). We ssumed uniform electric field in the preceding discussion. In more generl situtions, the electric field my vry over lrge surfce. Therefore, the definition of flux given by qution 24.2 hs mening only for smll element of re over which the field is pproximtely constnt. Consider generl surfce divided into lrge number of smll elements, ech of re DA. It is convenient to define vector DA i whose mgnitude represents the re of the ith element of the lrge surfce nd whose direction is defined to be perpendiculr to the surfce element s shown in Figure 24.3. The electric field i t the loction of this element mkes n ngle u i with the vector DA i. The electric flux F,i through this element is The number of field lines tht go through the re A is the sme s the number tht go through re A. u A A A cos u Norml u Figure 24.2 Field lines representing uniform electric field penetrting n re A tht is t n ngle u to the field. The electric field mkes n ngle u i with the vector A i, defined s being norml to the surfce element. i u i A i Figure 24.3 A smll element of surfce re DA i. F,i 5 i DA i cos u i 5 i? DA i where we hve used the definition of the sclr product of two vectors (A? B ; AB cos u; see Chpter 7). umming the contributions of ll elements gives n pproximtion to the totl flux through the surfce: F < i? DA i If the re of ech element pproches zero, the number of elements pproches infinity nd the sum is replced by n integrl. Therefore, the generl definition of electric flux is F ; 3? da (24.3) Definition of electric flux surfce qution 24.3 is surfce integrl, which mens it must be evluted over the surfce in question. In generl, the vlue of F depends both on the field pttern nd on the surfce. We re often interested in evluting the flux through closed surfce, defined s surfce tht divides spce into n inside nd n outside region so tht one cnnot move from one region to the other without crossing the surfce. The surfce of sphere, for exmple, is closed surfce. Consider the closed surfce in Active Figure 24.4 (pge 692). The vectors DA i point in different directions for the vrious surfce elements, but t ech point they
692 CHAPTR 24 Guss s Lw ACTIV FIGUR 24.4 A closed surfce in n electric field. The re vectors re, by convention, norml to the surfce nd point outwrd. A 1 A 3 u n A 2 n u The electric flux through this re element is negtive. The electric flux through this re element is zero. The electric flux through this re element is positive. re norml to the surfce nd, by convention, lwys point outwrd. At the element lbeled, the field lines re crossing the surfce from the inside to the outside nd u, 908; hence, the flux F,1 5? DA 1 through this element is positive. For element, the field lines grze the surfce (perpendiculr to DA 2); therefore, u 5 908 nd the flux is zero. For elements such s, where the field lines re crossing the surfce from outside to inside, 1808. u. 908 nd the flux is negtive becuse cos u is negtive. The net flux through the surfce is proportionl to the net number of lines leving the surfce, where the net number mens the number of lines leving the surfce minus the number of lines entering the surfce. If more lines re leving thn entering, the net flux is positive. If more lines re entering thn leving, the net flux is negtive. Using the symbol r to represent n integrl over closed surfce, we cn write the net flux F through closed surfce s F 5 C? da 5 C n da (24.4) where n represents the component of the electric field norml to the surfce. Quick Quiz 24.1 uppose point chrge is locted t the center of sphericl surfce. The electric field t the surfce of the sphere nd the totl flux through the sphere re determined. Now the rdius of the sphere is hlved. Wht hppens to the flux through the sphere nd the mgnitude of the electric field t the surfce of the sphere? () The flux nd field both increse. (b) The flux nd field both decrese. (c) The flux increses, nd the field decreses. (d) The flux decreses, nd the field increses. (e) The flux remins the sme, nd the field increses. (f) The flux decreses, nd the field remins the sme.
24.2 Guss s Lw 693 xmple 24.1 Flux Through Cube Consider uniform electric field oriented in the x direction in empty spce. A cube of edge length, is plced in the field, oriented s shown in Figure 24.5. Find the net electric flux through the surfce of the cube. OLUTION Conceptulize xmine Figure 24.5 crefully. Notice tht the electric field lines pss through two fces perpendiculrly nd re prllel to four other fces of the cube. Ctegorize We evlute the flux from its definition, so we ctegorize this exmple s substitution problem. The flux through four of the fces (,, nd the unnumbered fces) is zero becuse is prllel to the four fces nd therefore perpendiculr to da on these fces. z da 1 y da 3 da 4 da 2 Figure 24.5 (xmple 24.1) A closed surfce in the shpe of cube in uniform electric field oriented prllel to the x xis. ide is the bottom of the cube, nd side is opposite side. x Write the integrls for the net flux through fces nd : For fce, is constnt nd directed inwrd but da 1 is directed outwrd (u 5 1808). Find the flux through this fce: For fce, is constnt nd outwrd nd in the sme direction s da 2 (u 5 08). Find the flux through this fce: F 5 3 1 3 1 3 2? da 1 3 2? da? da 5 3 1cos 180 2 da 52 3 da 52A 52, 2 1 1? da 5 3 1cos 0 2 da 5 3 da 51A 5, 2 2 2 Find the net flux by dding the flux over ll six fces: F 52, 2 1, 2 1 0 1 0 1 0 1 0 5 0 24.2 Guss s Lw In this section, we describe generl reltionship between the net electric flux through closed surfce (often clled gussin surfce) nd the chrge enclosed by the surfce. This reltionship, known s Guss s lw, is of fundmentl importnce in the study of electric fields. Consider positive point chrge q locted t the center of sphere of rdius r s shown in Figure 24.6. From qution 23.9, we know tht the mgnitude of the electric field everywhere on the surfce of the sphere is 5 k e q/r 2. The field lines re directed rdilly outwrd nd hence re perpendiculr to the surfce t every point on the surfce. Tht is, t ech surfce point, is prllel to the vector DA i representing locl element of re DA i surrounding the surfce point. Therefore,? DA i 5 DA i nd, from qution 24.4, we find tht the net flux through the gussin surfce is F 5 C? da 5 C da 5 C da where we hve moved outside of the integrl becuse, by symmetry, is constnt over the surfce. The vlue of is given by 5 k e q/r 2. Furthermore, becuse the When the chrge is t the center of the sphere, the electric field is everywhere norml to the surfce nd constnt in mgnitude. phericl gussin surfce r q Figure 24.6 A sphericl gussin surfce of rdius r surrounding positive point chrge q. A
694 CHAPTR 24 Guss s Lw Americn Institute of Physics Krl Friedrich Guss Germn mthemticin nd stronomer (1777 1855) Guss received doctorl degree in mthemtics from the University of Helmstedt in 1799. In ddition to his work in electromgnetism, he mde contributions to mthemtics nd science in number theory, sttistics, non-ucliden geometry, nd cometry orbitl mechnics. He ws founder of the Germn Mgnetic Union, which studies the rth s mgnetic field on continul bsis. surfce is sphericl, rda 5 A 5 4pr 2. Hence, the net flux through the gussin surfce is q F 5 k e r 14pr 2 2 5 4pk 2 e q Reclling from qution 23.3 tht k e 5 1/4pP 0, we cn write this eqution in the form F 5 q P 0 (24.5) qution 24.5 shows tht the net flux through the sphericl surfce is proportionl to the chrge inside the surfce. The flux is independent of the rdius r becuse the re of the sphericl surfce is proportionl to r 2, wheres the electric field is proportionl to 1/r 2. Therefore, in the product of re nd electric field, the dependence on r cncels. Now consider severl closed surfces surrounding chrge q s shown in Figure 24.7. urfce 1 is sphericl, but surfces 2 nd 3 re not. From qution 24.5, the flux tht psses through 1 hs the vlue q/p 0. As discussed in the preceding section, flux is proportionl to the number of electric field lines pssing through surfce. The construction shown in Figure 24.7 shows tht the number of lines through 1 is equl to the number of lines through the nonsphericl surfces 2 nd 3. Therefore, the net flux through ny closed surfce surrounding point chrge q is given by q/p 0 nd is independent of the shpe of tht surfce. Now consider point chrge locted outside closed surfce of rbitrry shpe s shown in Figure 24.8. As cn be seen from this construction, ny electric field line entering the surfce leves the surfce t nother point. The number of electric field lines entering the surfce equls the number leving the surfce. Therefore, the net electric flux through closed surfce tht surrounds no chrge is zero. Applying this result to xmple 24.1, we see tht the net flux through the cube is zero becuse there is no chrge inside the cube. Let s extend these rguments to two generlized cses: (1) tht of mny point chrges nd (2) tht of continuous distribution of chrge. We once gin use the superposition principle, which sttes tht the electric field due to mny chrges is the vector sum of the electric fields produced by the individul chrges. Therefore, the flux through ny closed surfce cn be expressed s The number of field lines entering the surfce equls the number leving the surfce. The net electric flux is the sme through ll surfces. 2 3 1 q Figure 24.7 Closed surfces of vrious shpes surrounding positive chrge. Figure 24.8 A point chrge locted outside closed surfce.
24.2 Guss s Lw 695 C? da 5 C 1 1 1 2 1 c 2? da where is the totl electric field t ny point on the surfce produced by the vector ddition of the electric fields t tht point due to the individul chrges. Consider the system of chrges shown in Active Figure 24.9. The surfce surrounds only one chrge, q 1 ; hence, the net flux through is q 1 /P 0. The flux through due to chrges q 2, q 3, nd q 4 outside it is zero becuse ech electric field line from these chrges tht enters t one point leves it t nother. The surfce 9 surrounds chrges q 2 nd q 3 ; hence, the net flux through it is (q 2 1 q 3 )/P 0. Finlly, the net flux through surfce 0 is zero becuse there is no chrge inside this surfce. Tht is, ll the electric field lines tht enter 0 t one point leve t nother. Chrge q 4 does not contribute to the net flux through ny of the surfces. The mthemticl form of Guss s lw is generliztion of wht we hve just described nd sttes tht the net flux through ny closed surfce is F 5 C? da 5 q in P 0 (24.6) where represents the electric field t ny point on the surfce nd q in represents the net chrge inside the surfce. When using qution 24.6, you should note tht lthough the chrge q in is the net chrge inside the gussin surfce, represents the totl electric field, which includes contributions from chrges both inside nd outside the surfce. In principle, Guss s lw cn be solved for to determine the electric field due to system of chrges or continuous distribution of chrge. In prctice, however, this type of solution is pplicble only in limited number of highly symmetric situtions. In the next section, we use Guss s lw to evlute the electric field for chrge distributions tht hve sphericl, cylindricl, or plnr symmetry. If one chooses the gussin surfce surrounding the chrge distribution crefully, the integrl in qution 24.6 cn be simplified nd the electric field determined. Quick Quiz 24.2 If the net flux through gussin surfce is zero, the following four sttements could be true. Which of the sttements must be true? () There re no chrges inside the surfce. (b) The net chrge inside the surfce is zero. (c) The electric field is zero everywhere on the surfce. (d) The number of electric field lines entering the surfce equls the number leving the surfce. Chrge q 4 does not contribute to the flux through ny surfce becuse it is outside ll surfces. q 1 q 4 q 3 ACTIV FIGUR 24.9 q 2 The net electric flux through ny closed surfce depends only on the chrge inside tht surfce. The net flux through surfce is q 1 /P 0, the net flux through surfce 9 is (q 2 1 q 3 )/P 0, nd the net flux through surfce 0 is zero. Pitfll Prevention 24.1 Zero Flux Is Not Zero Field In two situtions, there is zero flux through closed surfce: either (1) there re no chrged prticles enclosed by the surfce or (2) there re chrged prticles enclosed, but the net chrge inside the surfce is zero. For either sitution, it is incorrect to conclude tht the electric field on the surfce is zero. Guss s lw sttes tht the electric flux is proportionl to the enclosed chrge, not the electric field. Conceptul xmple 24.2 Flux Due to Point Chrge A sphericl gussin surfce surrounds point chrge q. Describe wht hppens to the totl flux through the surfce if (A) the chrge is tripled, (B) the rdius of the sphere is doubled, (C) the surfce is chnged to cube, nd (D) the chrge is moved to nother loction inside the surfce. OLUTION (A) The flux through the surfce is tripled becuse flux is proportionl to the mount of chrge inside the surfce. (B) The flux does not chnge becuse ll electric field lines from the chrge pss through the sphere, regrdless of its rdius. (C) The flux does not chnge when the shpe of the gussin surfce chnges becuse ll electric field lines from the chrge pss through the surfce, regrdless of its shpe. (D) The flux does not chnge when the chrge is moved to nother loction inside tht surfce becuse Guss s lw refers to the totl chrge enclosed, regrdless of where the chrge is locted inside the surfce.
696 CHAPTR 24 Guss s Lw 24.3 Appliction of Guss s Lw to Vrious Chrge Distributions Pitfll Prevention 24.2 Gussin urfces Are Not Rel A gussin surfce is n imginry surfce you construct to stisfy the conditions listed here. It does not hve to coincide with physicl surfce in the sitution. As mentioned erlier, Guss s lw is useful for determining electric fields when the chrge distribution is highly symmetric. The following exmples demonstrte wys of choosing the gussin surfce over which the surfce integrl given by qution 24.6 cn be simplified nd the electric field determined. In choosing the surfce, lwys tke dvntge of the symmetry of the chrge distribution so tht cn be removed from the integrl. The gol in this type of clcultion is to determine surfce for which ech portion of the surfce stisfies one or more of the following conditions: 1. The vlue of the electric field cn be rgued by symmetry to be constnt over the portion of the surfce. 2. The dot product in qution 24.6 cn be expressed s simple lgebric product da becuse nd da re prllel. 3. The dot product in qution 24.6 is zero becuse nd da re perpendiculr. 4. The electric field is zero over the portion of the surfce. Different portions of the gussin surfce cn stisfy different conditions s long s every portion stisfies t lest one condition. All four conditions re used in exmples throughout the reminder of this chpter nd will be identified by number. If the chrge distribution does not hve sufficient symmetry such tht gussin surfce tht stisfies these conditions cn be found, Guss s lw is not useful for determining the electric field for tht chrge distribution. xmple 24.3 A phericlly ymmetric Chrge Distribution An insulting solid sphere of rdius hs uniform volume chrge density r nd crries totl positive chrge Q (Fig. 24.10). (A) Clculte the mgnitude of the electric field t point outside the sphere. OLUTION For points outside the sphere, lrge, sphericl gussin surfce is drwn concentric with the sphere. For points inside the sphere, sphericl gussin surfce smller thn the sphere is drwn. Conceptulize Notice how this problem differs from our previous discussion of Guss s lw. The electric field due to point chrges ws discussed in ection 24.2. Now we re considering the electric field due to distribution of chrge. We found the field for vrious distributions of chrge in Chpter 23 by integrting over the distribution. This exmple demonstrtes difference from our discussions in Chpter 23. In this chpter, we find the electric field using Guss s lw. Ctegorize Becuse the chrge is distributed uniformly throughout the sphere, the chrge distribution hs sphericl symmetry nd we cn pply Guss s lw to find the electric field. r Q Gussin sphere r Gussin sphere Figure 24.10 (xmple 24.3) A uniformly chrged insulting sphere of rdius nd totl chrge Q. In digrms such s this one, the dotted line represents the intersection of the gussin surfce with the plne of the pge. b Anlyze To reflect the sphericl symmetry, let s choose sphericl gussin surfce of rdius r, concentric with the sphere, s shown in Figure 24.10. For this choice, condition (2) is stisfied everywhere on the surfce nd? da 5 da.
24.3 Appliction of Guss s Lw to Vrious Chrge Distributions 697 24.3 cont. Replce? da in Guss s lw with da: F 5 C? da 5 C da 5 Q P 0 By symmetry, is constnt everywhere on the surfce, which stisfies condition (1), so we cn remove from the integrl: C da 5 C da 5 14pr 2 2 5 Q P 0 olve for : (1) 5 Q 4pP 0 r 5 k Q 2 e r 2 1for r. 2 Finlize This field is identicl to tht for point chrge. Therefore, the electric field due to uniformly chrged sphere in the region externl to the sphere is equivlent to tht of point chrge locted t the center of the sphere. (B) Find the mgnitude of the electric field t point inside the sphere. OLUTION Anlyze In this cse, let s choose sphericl gussin surfce hving rdius r,, concentric with the insulting sphere (Fig. 24.10b). Let V 9 be the volume of this smller sphere. To pply Guss s lw in this sitution, recognize tht the chrge q in within the gussin surfce of volume V 9 is less thn Q. Clculte q in by using q in 5 rv 9: q in 5rV r 5r1 4 3pr 3 2 Notice tht conditions (1) nd (2) re stisfied everywhere on the gussin surfce in Figure 24.10b. Apply Guss s lw in the region r, : C da 5 C da 5 14pr 2 2 5 q in P 0 olve for nd substitute for q in : 5 q in 4pP 0 r 5 r14 3pr 3 2 2 4pP 0 r 5 r r 2 3P 0 ubstitute r5q / 4 3p 3 nd P 0 5 1/4pk e : (2) 5 Q /4 3p 3 311/4pk e 2 r 5 k Q e r 3 1for r, 2 Finlize This result for differs from the one obtined in prt (A). It shows tht 0 s r 0. Therefore, the result elimintes the problem tht would exist t r 5 0 if vried s 1/r 2 inside the sphere s it does outside the sphere. Tht is, if ~ 1/r 2 for r,, the field would be infinite t r 5 0, which is physiclly impossible. WHAT IF? uppose the rdil position r 5 is pproched from inside the sphere nd from outside. Do we obtin the sme vlue of the electric field from both directions? Answer qution (1) shows tht the electric field pproches vlue from the outside given by k e Q 3 r k e Q r 2 5 lim r k e Q r 2b 5 k e Q 2 r From the inside, qution (2) gives 5 lim r k e Q 3 rb 5 k e Q 3 5 k e Q 2 Therefore, the vlue of the field is the sme s the surfce is pproched from both directions. A plot of versus r is shown in Figure 24.11. Notice tht the mgnitude of the field is continuous. Figure 24.11 (xmple 24.3) A plot of versus r for uniformly chrged insulting sphere. The electric field inside the sphere (r, ) vries linerly with r. The field outside the sphere (r. ) is the sme s tht of point chrge Q locted t r 5 0.
698 CHAPTR 24 Guss s Lw xmple 24.4 A Cylindriclly ymmetric Chrge Distribution Find the electric field distnce r from line of positive chrge of infinite length nd constnt chrge per unit length l (Fig. 24.12). Gussin surfce r OLUTION Conceptulize The line of chrge is infinitely long. Therefore, the field is the sme t ll points equidistnt from the line, regrdless of the verticl position of the point in Figure 24.12. Ctegorize Becuse the chrge is distributed uniformly long the line, the chrge distribution hs cylindricl symmetry nd we cn pply Guss s lw to find the electric field. Anlyze The symmetry of the chrge distribution requires tht be perpendiculr to the line chrge nd directed outwrd s shown in Figure 24.12b. To reflect the symmetry of the chrge distribution, let s choose cylindricl gussin surfce of rdius r nd length, tht is coxil with the line chrge. For the curved prt of this surfce, is constnt in mgnitude nd perpendiculr to the surfce t ech point, stisfying conditions (1) nd (2). Furthermore, the flux through the ends of the gussin cylinder is zero becuse is prllel to these surfces. Tht is the first ppliction we hve seen of condition (3). We must tke the surfce integrl in Guss s lw over the entire gussin surfce. Becuse? da is zero for the flt ends of the cylinder, however, we restrict our ttention to only the curved surfce of the cylinder. da Figure 24.12 (xmple 24.4) () An infinite line of chrge surrounded by cylindricl gussin surfce concentric with the line. (b) An end view shows tht the electric field t the cylindricl surfce is constnt in mgnitude nd perpendiculr to the surfce. b Apply Guss s lw nd conditions (1) nd (2) for the curved surfce, noting tht the totl chrge inside our gussin surfce is l,: F 5 C? da 5 C da 5 A 5 q in P 0 5 l, P 0 ubstitute the re A 5 2pr, of the curved surfce: 12pr,2 5 l, P 0 olve for the mgnitude of the electric field: 5 l 2pP 0 r 5 2k e l r (24.7) Finlize This result shows tht the electric field due to cylindriclly symmetric chrge distribution vries s 1/r, wheres the field externl to sphericlly symmetric chrge distribution vries s 1/r 2. qution 24.7 cn lso be derived by direct integrtion over the chrge distribution. (ee Problem 37 in Chpter 23.) WHAT IF? Wht if the line segment in this exmple were not infinitely long? Answer If the line chrge in this exmple were of finite length, the electric field would not be given by qution 24.7. A finite line chrge does not possess sufficient symmetry to mke use of Guss s lw becuse the mgnitude of the electric field is no longer constnt over the surfce of the gussin cylinder: the field ner the ends of the line would be different from tht fr from the ends. Therefore, condition (1) would not be stisfied in this sitution. Furthermore, is not perpendiculr to the cylindricl surfce t ll points: the field vectors ner the ends would hve component prllel to the line. Therefore, condition (2) would not be stisfied. For points close to finite line chrge nd fr from the ends, qution 24.7 gives good pproximtion of the vlue of the field. It is left for you to show (see Problem 33) tht the electric field inside uniformly chrged rod of finite rdius nd infinite length is proportionl to r.
24.4 Conductors in lectrosttic quilibrium 699 xmple 24.5 A Plne of Chrge Find the electric field due to n infinite plne of positive chrge with uniform surfce chrge density s. OLUTION Conceptulize Notice tht the plne of chrge is infinitely lrge. Therefore, the electric field should be the sme t ll points equidistnt from the plne. Ctegorize Becuse the chrge is distributed uniformly on the plne, the chrge distribution is symmetric; hence, we cn use Guss s lw to find the electric field. Anlyze By symmetry, must be perpendiculr to the plne t ll points. The direction of is wy from positive chrges, indicting tht the direction of on one side of the plne must be opposite its direction on the other side s shown in Figure 24.13. A gussin surfce tht reflects the symmetry is smll cylinder whose xis is perpendiculr to the plne nd whose ends ech hve n re A nd re equidistnt from the plne. Becuse is prllel to the curved surfce nd therefore perpendiculr to da everywhere on the surfce condition (3) is stisfied nd there is no contribution to the surfce integrl from this surfce. For the flt ends of the cylinder, conditions (1) nd (2) re stisfied. The flux through ech end of the cylinder is A; hence, the totl flux through the entire gussin surfce is just tht through the ends, F 5 2A. Write Guss s lw for this surfce, noting tht the enclosed chrge is q in 5 sa: olve for : 5 F 5 2A 5 q in P 0 5 sa P 0 s 2P 0 (24.8) Gussin surfce Figure 24.13 (xmple 24.5) A cylindricl gussin surfce penetrting n infinite plne of chrge. The flux is A through ech end of the gussin surfce nd zero through its curved surfce. Finlize Becuse the distnce from ech flt end of the cylinder to the plne does not pper in qution 24.8, we conclude tht 5 s/2p 0 t ny distnce from the plne. Tht is, the field is uniform everywhere. WHAT IF? uppose two infinite plnes of chrge re prllel to ech other, one positively chrged nd the other negtively chrged. Both plnes hve the sme surfce chrge density. Wht does the electric field look like in this sitution? Answer The electric fields due to the two plnes dd in the region between the plnes, resulting in uniform field of mgnitude s/p 0, nd cncel elsewhere to give field of zero. This method is prcticl wy to chieve uniform electric fields with finite-sized plnes plced close to ech other. A Conceptul xmple 24.6 Don t Use Guss s Lw Here! xplin why Guss s lw cnnot be used to clculte the electric field ner n electric dipole, chrged disk, or tringle with point chrge t ech corner. OLUTION The chrge distributions of ll these configurtions do not hve sufficient symmetry to mke the use of Guss s lw prcticl. We cnnot find closed surfce surrounding ny of these distributions for which ll portions of the surfce stisfy one or more of conditions (1) through (4) listed t the beginning of this section. 24.4 Conductors in lectrosttic quilibrium As we lerned in ection 23.2, good electricl conductor contins chrges (electrons) tht re not bound to ny tom nd therefore re free to move bout within the mteril. When there is no net motion of chrge within conductor, the
700 CHAPTR 24 Guss s Lw conductor is in electrosttic equilibrium. A conductor in electrosttic equilibrium hs the following properties: Properties of conductor in electrosttic equilibrium Figure 24.14 A conducting slb in n externl electric field. The chrges induced on the two surfces of the slb produce n electric field tht opposes the externl field, giving resultnt field of zero inside the slb. Gussin surfce Figure 24.15 A conductor of rbitrry shpe. The broken line represents gussin surfce tht cn be just inside the conductor s surfce. 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolted nd crries chrge, the chrge resides on its surfce. 3. The electric field t point just outside chrged conductor is perpendiculr to the surfce of the conductor nd hs mgnitude s/p 0, where s is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest t loctions where the rdius of curvture of the surfce is smllest. We verify the first three properties in the discussion tht follows. The fourth property is presented here (but not verified until Chpter 25) to provide complete list of properties for conductors in electrosttic equilibrium. We cn understnd the first property by considering conducting slb plced in n externl field (Fig. 24.14). The electric field inside the conductor must be zero, ssuming electrosttic equilibrium exists. If the field were not zero, free electrons in the conductor would experience n electric force (F 5 q ) nd would ccelerte due to this force. This motion of electrons, however, would men tht the conductor is not in electrosttic equilibrium. Therefore, the existence of electrosttic equilibrium is consistent only with zero field in the conductor. Let s investigte how this zero field is ccomplished. Before the externl field is pplied, free electrons re uniformly distributed throughout the conductor. When the externl field is pplied, the free electrons ccelerte to the left in Figure 24.14, cusing plne of negtive chrge to ccumulte on the left surfce. The movement of electrons to the left results in plne of positive chrge on the right surfce. These plnes of chrge crete n dditionl electric field inside the conductor tht opposes the externl field. As the electrons move, the surfce chrge densities on the left nd right surfces increse until the mgnitude of the internl field equls tht of the externl field, resulting in net field of zero inside the conductor. The time it tkes good conductor to rech equilibrium is on the order of 10 216 s, which for most purposes cn be considered instntneous. If the conductor is hollow, the electric field inside the conductor is lso zero, whether we consider points in the conductor or in the cvity within the conductor. The zero vlue of the electric field in the cvity is esiest to rgue with the concept of electric potentil, so we will ddress this issue in ection 25.6. Guss s lw cn be used to verify the second property of conductor in electrosttic equilibrium. Figure 24.15 shows n rbitrrily shped conductor. A gussin surfce is drwn inside the conductor nd cn be very close to the conductor s surfce. As we hve just shown, the electric field everywhere inside the conductor is zero when it is in electrosttic equilibrium. Therefore, the electric field must be zero t every point on the gussin surfce, in ccordnce with condition (4) in ection 24.3, nd the net flux through this gussin surfce is zero. From this result nd Guss s lw, we conclude tht the net chrge inside the gussin surfce is zero. Becuse there cn be no net chrge inside the gussin surfce (which is rbitrrily close to the conductor s surfce), ny net chrge on the conductor must reside on its surfce. Guss s lw does not indicte how this excess chrge is distributed on the conductor s surfce, only tht it resides exclusively on the surfce. To verify the third property, let s begin with the perpendiculrity of the field to the surfce. If the field vector hd component prllel to the conductor s surfce, free electrons would experience n electric force nd move long the surfce; in such cse, the conductor would not be in equilibrium. Therefore, the field vector must be perpendiculr to the surfce. To determine the mgnitude of the electric field, we use Guss s lw nd drw gussin surfce in the shpe of smll cylinder whose end fces re prllel
24.4 Conductors in lectrosttic quilibrium 701 to the conductor s surfce (Fig. 24.16). Prt of the cylinder is just outside the conductor, nd prt is inside. The field is perpendiculr to the conductor s surfce from the condition of electrosttic equilibrium. Therefore, condition (3) in ection 24.3 is stisfied for the curved prt of the cylindricl gussin surfce: there is no flux through this prt of the gussin surfce becuse is prllel to the surfce. There is no flux through the flt fce of the cylinder inside the conductor becuse here 5 0, which stisfies condition (4). Hence, the net flux through the gussin surfce is equl to tht through only the flt fce outside the conductor, where the field is perpendiculr to the gussin surfce. Using conditions (1) nd (2) for this fce, the flux is A, where is the electric field just outside the conductor nd A is the re of the cylinder s fce. Applying Guss s lw to this surfce gives F 5 C da 5 A 5 q in P 0 5 sa P 0 where we hve used q in 5 sa. olving for gives for the electric field immeditely outside chrged conductor: The flux through the gussin surfce is A. A Figure 24.16 A gussin surfce in the shpe of smll cylinder is used to clculte the electric field immeditely outside chrged conductor. 5 s P 0 (24.9) Quick Quiz 24.3 Your younger brother likes to rub his feet on the crpet nd then touch you to give you shock. While you re trying to escpe the shock tretment, you discover hollow metl cylinder in your bsement, lrge enough to climb inside. In which of the following cses will you not be shocked? () You climb inside the cylinder, mking contct with the inner surfce, nd your chrged brother touches the outer metl surfce. (b) Your chrged brother is inside touching the inner metl surfce nd you re outside, touching the outer metl surfce. (c) Both of you re outside the cylinder, touching its outer metl surfce but not touching ech other directly. xmple 24.7 A phere Inside phericl hell A solid insulting sphere of rdius crries net positive chrge Q uniformly distributed throughout its volume. A conducting sphericl shell of inner rdius b nd outer rdius c is concentric with the solid sphere nd crries net chrge 22Q. Using Guss s lw, find the electric field in the regions lbeled,,, nd in Active Figure 24.17 nd the chrge distribution on the shell when the entire system is in electrosttic equilibrium. OLUTION Conceptulize Notice how this problem differs from xmple 24.3. The chrged sphere in Figure 24.10 ppers in Active Figure 24.17, but it is now surrounded by shell crrying chrge 22Q. Ctegorize The chrge is distributed uniformly throughout the sphere, nd we know tht the chrge on the conducting shell distributes itself uniformly on the surfces. Therefore, the system hs sphericl symmetry nd we cn pply Guss s lw to find the electric field in the vrious regions. r Q b Anlyze In region between the surfce of the solid sphere nd the inner surfce of the shell we construct sphericl gussin surfce of rdius r, where, r, b, noting tht the chrge inside this surfce is 1Q (the chrge on the solid sphere). Becuse of the sphericl symmetry, the electric field lines must be directed rdilly outwrd nd be constnt in mgnitude on the gussin surfce. 2Q c ACTIV FIGUR 24.17 (xmple 24.7) An insulting sphere of rdius nd crrying chrge Q surrounded by conducting sphericl shell crrying chrge 22Q. continued
702 CHAPTR 24 Guss s Lw 24.7 cont. The chrge on the conducting shell cretes zero electric field in the region r, b, so the shell hs no effect on the field due to the sphere. Therefore, write n expression for the field in region s tht due to the sphere from prt (A) of xmple 24.3: 2 5 k e Q r 2 1for, r, b2 Becuse the conducting shell cretes zero field inside itself, it lso hs no effect on the field inside the sphere. Therefore, write n expression for the field in region s tht due to the sphere from prt (B) of xmple 24.3: 1 5 k e Q 3 r 1for r, 2 In region, where r. c, construct sphericl gussin surfce; this surfce surrounds totl chrge q in 5 Q 1 (22Q) 5 2Q. Therefore, model the chrge distribution s sphere with chrge 2Q nd write n expression for the field in region from prt (A) of xmple 24.3: 4 5 2k e Q r 2 1for r. c2 In region, the electric field must be zero becuse the sphericl shell is conductor in equilibrium: 3 5 0 1for b, r, c2 Construct gussin surfce of rdius r, where b, r, c, nd note tht q in must be zero becuse 3 5 0. Find the mount of chrge q inner on the inner surfce of the shell: q in 5 q sphere 1 q inner q inner 5 q in 2 q sphere 5 0 2 Q 52Q Finlize The chrge on the inner surfce of the sphericl shell must be 2Q to cncel the chrge 1Q on the solid sphere nd give zero electric field in the mteril of the shell. Becuse the net chrge on the shell is 22Q, its outer surfce must crry chrge 2Q. WHAT IF? How would the results of this problem differ if the sphere were conducting insted of insulting? Answer The only chnge would be in region, where r,. Becuse there cn be no chrge inside conductor in electrosttic equilibrium, q in 5 0 for gussin surfce of rdius r, ; therefore, on the bsis of Guss s lw nd symmetry, 1 5 0. In regions,, nd, there would be no wy to determine from observtions of the electric field whether the sphere is conducting or insulting. Definition ummry lectric flux is proportionl to the number of electric field lines tht penetrte surfce. If the electric field is uniform nd mkes n ngle u with the norml to surfce of re A, the electric flux through the surfce is In generl, the electric flux through surfce is F 5 A cos u (24.2) F ; 3 surfce? da (24.3) continued
Objective Questions 703 Concepts nd Principles Guss s lw sys tht the net electric flux F through ny closed gussin surfce is equl to the net chrge q in inside the surfce divided by P 0 : F 5 C? da 5 q in P 0 (24.6) Using Guss s lw, you cn clculte the electric field due to vrious symmetric chrge distributions. A conductor in electrosttic equilibrium hs the following properties: 1. The electric field is zero everywhere inside the conductor, whether the conductor is solid or hollow. 2. If the conductor is isolted nd crries chrge, the chrge resides on its surfce. 3. The electric field t point just outside chrged conductor is perpendiculr to the surfce of the conductor nd hs mgnitude s/p 0, where s is the surfce chrge density t tht point. 4. On n irregulrly shped conductor, the surfce chrge density is gretest t loctions where the rdius of curvture of the surfce is smllest. Objective Questions denotes nswer vilble in tudent olutions Mnul/tudy Guide 1. Chrges of 3.00 nc, 22.00 nc, 27.00 nc, nd 1.00 nc re contined inside rectngulr box with length 1.00 m, width 2.00 m, nd height 2.50 m. Outside the box re chrges of 1.00 nc nd 4.00 nc. Wht is the electric flux through the surfce of the box? () 0 (b) 25.64 3 10 2 N? m 2 /C (c) 21.47 3 10 3 N? m 2 /C (d) 1.47 3 10 3 N? m 2 /C (e) 5.64 3 10 2 N? m 2 /C 2. A uniform electric field of 1.00 N/C is set up by uniform distribution of chrge in the xy plne. Wht is the electric field inside metl bll plced 0.500 m bove the xy plne? () 1.00 N/C (b) 21.00 N/C (c) 0 (d) 0.250 N/C (e) vries depending on the position inside the bll 3. In which of the following contexts cn Guss s lw not be redily pplied to find the electric field? () ner long, uniformly chrged wire (b) bove lrge, uniformly chrged plne (c) inside uniformly chrged bll (d) outside uniformly chrged sphere (e) Guss s lw cn be redily pplied to find the electric field in ll these contexts. 4. A prticle with chrge q is locted inside cubicl gussin surfce. No other chrges re nerby. (i) If the prticle is t the center of the cube, wht is the flux through ech one of the fces of the cube? () 0 (b) q/2p 0 (c) q/6p 0 (d) q/8p 0 (e) depends on the size of the cube (ii) If the prticle cn be moved to ny point within the cube, wht mximum vlue cn the flux through one fce pproch? Choose from the sme possibilities s in prt (i). 5. A cubicl gussin surfce surrounds long, stright, chrged filment tht psses perpendiculrly through two opposite fces. No other chrges re nerby. (i) Over how mny of the cube s fces is the electric field zero? () 0 (b) 2 (c) 4 (d) 6 (ii) Through how mny of the cube s fces is the electric flux zero? Choose from the sme possibilities s in prt (i). 6. A cubicl gussin surfce is bisected by lrge sheet of chrge, prllel to its top nd bottom fces. No other chrges re nerby. (i) Over how mny of the cube s fces is the electric field zero? () 0 (b) 2 (c) 4 (d) 6 (ii) Through how mny of the cube s fces is the electric flux zero? Choose from the sme possibilities s in prt (i). 7. Two solid spheres, both of rdius 5 cm, crry identicl totl chrges of 2 mc. phere A is good conductor. phere B is n insultor, nd its chrge is distributed uniformly throughout its volume. (i) How do the mgnitudes of the electric fields they seprtely crete t rdil distnce of 6 cm compre? () A. B 5 0 (b) A. B. 0 (c) A 5 B. 0 (d) 0, A, B (e) 0 5 A, B (ii) How do the mgnitudes of the electric fields they seprtely crete t rdius 4 cm compre? Choose from the sme possibilities s in prt (i). 8. A coxil cble consists of long, stright filment surrounded by long, coxil, cylindricl conducting shell. Assume chrge Q is on the filment, zero net chrge is on the shell, nd the electric field is 1 i^ t prticulr point P midwy between the filment nd the inner surfce of the shell. Next, you plce the cble into uniform externl field 2 i^. Wht is the x component of the electric field t P then? () 0 (b) between 0 nd 1 (c) 1 (d) between 0 nd 2 1 (e) 2 1 9. A solid insulting sphere of rdius 5 cm crries electric chrge uniformly distributed throughout its volume. Concentric with the A B C D sphere is conducting sphericl shell with no net chrge s shown in Figure OQ24.9. The inner rdius of the shell is 10 cm, nd the outer rdius is 15 cm. No other Figure OQ24.9 chrges re nerby. () Rnk the mgnitude of the electric field t points A (t rdius 4 cm), B (rdius 8 cm), C (rdius 12 cm), nd D (rdius 16 cm) from lrgest to smllest. Disply ny cses of equlity in your rnking. (b) imilrly rnk the electric flux through concentric sphericl surfces through points A, B, C, nd D.
704 CHAPTR 24 Guss s Lw 10. A lrge, metllic, sphericl shell hs no net chrge. It is supported on n insulting stnd nd hs smll hole t the top. A smll tck with chrge Q is lowered on silk thred through the hole into the interior of the shell. (i) Wht is the chrge on the inner surfce of the shell, () Q (b) Q/2 (c) 0 (d) 2Q/2 or (e) 2Q? Choose your nswers to the following questions from the sme possibilities. (ii) Wht is the chrge on the outer surfce of the shell? (iii) The tck is now llowed to touch the interior surfce of the shell. After this contct, wht is the chrge on the tck? (iv) Wht is the chrge on the inner surfce of the shell now? (v) Wht is the chrge on the outer surfce of the shell now? 11. Rnk the electric fluxes through ech gussin surfce shown in Figure OQ24.11 from lrgest to smllest. Disply ny cses of equlity in your rnking. Q Q 3Q 4Q b c d Figure OQ24.11 Conceptul Questions 1. The un is lower in the sky during the winter thn it is during the summer. () How does this chnge ffect the flux of sunlight hitting given re on the surfce of the rth? (b) How does this chnge ffect the wether? 2. If more electric field lines leve gussin surfce thn enter it, wht cn you conclude bout the net chrge enclosed by tht surfce? 3. A uniform electric field exists in region of spce contining no chrges. Wht cn you conclude bout the net electric flux through gussin surfce plced in this region of spce? 4. If the totl chrge inside closed surfce is known but the distribution of the chrge is unspecified, cn you use Guss s lw to find the electric field? xplin. 5. xplin why the electric flux through closed surfce with given enclosed chrge is independent of the size or shpe of the surfce. 6. A cubicl surfce surrounds point chrge q. Describe wht hppens to the totl flux through the surfce if () the chrge is doubled, (b) the volume of the cube is doubled, (c) the surfce is chnged to sphere, (d) the chrge is moved to nother loction inside the surfce, nd (e) the chrge is moved outside the surfce. 7. A person is plced in lrge, hollow, metllic sphere tht is insulted from ground. () If lrge chrge is plced on the sphere, will the person be hrmed upon touching the inside of the sphere? (b) xplin wht will hppen if denotes nswer vilble in tudent olutions Mnul/tudy Guide the person lso hs n initil chrge whose sign is opposite tht of the chrge on the sphere. 8. On the bsis of the repulsive nture of the force between like chrges nd the freedom of motion of chrge within conductor, explin why excess chrge on n isolted conductor must reside on its surfce. 9. A common demonstrtion involves chrging rubber blloon, which is n insultor, by rubbing it on your hir nd then touching the blloon to ceiling or wll, which is lso n insultor. Becuse of the electricl ttrction between the chrged blloon nd the neutrl wll, the blloon sticks to the wll. Imgine now tht we hve two infinitely lrge, flt sheets of insulting mteril. One is chrged, nd the other is neutrl. If these sheets re brought into contct, does n ttrctive force exist between them s there ws for the blloon nd the wll? 10. Consider two identicl conducting spheres whose surfces re seprted by smll distnce. One sphere is given lrge net positive chrge, nd the other is given smll net positive chrge. It is found tht the force between the spheres is ttrctive even though they both hve net chrges of the sme sign. xplin how this ttrction is possible. 11. Consider n electric field tht is uniform in direction throughout certin volume. Cn it be uniform in mgnitude? Must it be uniform in mgnitude? Answer these questions () ssuming the volume is filled with n insulting mteril crrying chrge described by volume chrge density nd (b) ssuming the volume is empty spce. tte resoning to prove your nswers. Problems The problems found in this chpter my be ssigned online in nhnced WebAssign 1. denotes strightforwrd problem; 2. denotes intermedite problem; 3. denotes chllenging problem 1. full solution vilble in the tudent olutions Mnul/tudy Guide 1. denotes problems most often ssigned in nhnced WebAssign; these provide students with trgeted feedbck nd either Mster It tutoril or Wtch It solution video. denotes sking for quntittive nd conceptul resoning denotes symbolic resoning problem denotes Mster It tutoril vilble in nhnced WebAssign denotes guided problem shded denotes pired problems tht develop resoning with symbols nd numericl vlues
Problems 705 ection 24.1 lectric Flux 1. A 40.0-cm-dimeter circulr loop is rotted in uniform electric field until the position of mximum electric flux is found. The flux in this position is mesured to be 5.20 3 10 5 N? m 2 /C. Wht is the mgnitude of the electric field? 2. A verticl electric field of mgnitude 2.00 3 10 4 N/C exists bove the rth s surfce on dy when thunderstorm is brewing. A cr with rectngulr size of 6.00 m by 3.00 m is trveling long dry grvel rodwy sloping downwrd t 10.08. Determine the electric flux through the bottom of the cr. 3. A flt surfce of re 3.20 m 2 is rotted in uniform electric field of mgnitude 5 6.20 3 10 5 N/C. Determine the electric flux through this re () when the electric field is perpendiculr to the surfce nd (b) when the electric field is prllel to the surfce. 4. Consider closed tringulr box resting within horizontl electric field of mgnitude 5 7.80 3 10 4 N/C s shown in Figure P24.4. Clculte the electric flux through () the verticl rectngulr surfce, (b) the slnted surfce, nd (c) the entire surfce of the box. 10.0 cm 60.0 30.0 cm Figure P24.4 5. An electric field of mgnitude 3.50 kn/c is pplied long the x xis. Clculte the electric flux through rectngulr plne 0.350 m wide nd 0.700 m long () if the plne is prllel to the yz plne, (b) if the plne is prllel to the xy plne, nd (c) if the plne contins the y xis nd its norml mkes n ngle of 40.08 with the x xis. ection 24.2 Guss s Lw 6. Find the net electric flux through the sphericl closed surfce shown in Figure P24.6. The two chrges on the right re inside the sphericl surfce. 2.00 nc Figure P24.6 1.00 nc 3.00 nc 7. An unchrged, nonconducting, hollow sphere of rdius 10.0 cm surrounds 10.0-mC chrge locted t the origin of Crtesin coordinte system. A drill with rdius of 1.00 mm is ligned long the z xis, nd hole is drilled in the sphere. Clculte the electric flux through the hole. 8. A chrge of 170 mc is t the center of cube of edge 80.0 cm. No other chrges re nerby. () Find the flux through ech fce of the cube. (b) Find the flux through the whole surfce of the cube. (c) Wht If? Would your nswers to either prt () or prt (b) chnge if the chrge were not t the center? xplin. 9. The following chrges re locted inside submrine: 5.00 mc, 29.00 mc, 27.0 mc, nd 284.0 mc. () Clculte the net electric flux through the hull of the submrine. (b) Is the number of electric field lines leving the submrine greter thn, equl to, or less thn the number entering it? 10. The electric field everywhere on the surfce of thin, sphericl shell of rdius 0.750 m is of mgnitude 890 N/C nd points rdilly towrd the center of the sphere. () Wht is the net chrge within the sphere s surfce? (b) Wht is the distribution of the chrge inside the sphericl shell? 11. Four closed surfces, 1 through 4, together with the chrges 22Q, Q, nd 2Q re sketched in Figure P24.11. (The colored lines re the intersections of the surfces with the pge.) Find the electric flux through ech surfce. 12. A prticle with chrge of 12.0 mc is plced t the center of sphericl shell Figure P24.11 of rdius 22.0 cm. Wht is the totl electric flux through () the surfce of the shell nd (b) ny hemisphericl surfce of the shell? (c) Do the results depend on the rdius? xplin. 13. In the ir over prticulr region t n ltitude of 500 m bove the ground, the electric field is 120 N/C directed downwrd. At 600 m bove the ground, the electric field is 100 N/C downwrd. Wht is the verge volume chrge density in the lyer of ir between these two elevtions? Is it positive or negtive? 14. () Find the net electric flux through the cube shown in Figure P24.14. (b) Cn you use Guss s lw to find the electric field on the surfce of this cube? xplin. 15. An infinitely long line chrge hving uniform chrge per unit length l lies distnce d from point O s shown in Figure P24.15. Determine the totl electric flux through the surfce of sphere of rdius R centered t O resulting from this line chrge. Consider both cses, where () R, d nd (b) R. d. 16. () A prticle with chrge q is locted distnce d from n infinite plne. Determine the electric flux through 4 3.00 nc 2Q Q Q Figure P24.14 l d Figure P24.15 2 1 3 8.00 nc O R