Chapter 24. Gauss s Law


 Allison Henry
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1 Chpte 24 Guss s Lw CHAPTR OUTLIN 24.1 lectic Flux 24.2 Guss s Lw 24.3 Appliction of Guss s Lw to Vious Chge Distibutions 24.4 Conductos in lectosttic uilibium 24.5 Foml Deivtion of Guss s Lw In tbletop plsm bll, the coloful lines emnting fom the sphee give evidence of stong electic fields. Using Guss s lw, we show in this chpte tht the electic field suounding chged sphee is identicl to tht of point chge. (Getty Imges) 739
2 In the peceding chpte we showed how to clculte the electic field geneted by given chge distibution. In this chpte, we descibe Guss s lw nd n ltentive pocedue fo clculting electic fields. The lw is bsed on the fct tht the fundmentl electosttic foce between point chges exhibits n invesesue behvio. Although conseuence of Coulomb s lw, Guss s lw is moe convenient fo clculting the electic fields of highly symmetic chge distibutions nd mkes possible useful ulittive esoning when deling with complicted poblems. Ae = A 24.1 lectic Flux The concept of electic field lines ws descibed ulittively in Chpte 23. We now tet electic field lines in moe untittive wy. Conside n electic field tht is unifom in both mgnitude nd diection, s shown in Figue The field lines penette ectngul sufce of e A, whose plne is oiented pependicul to the field. Recll fom Section 23.6 tht the numbe of lines pe unit e (in othe wods, the line density) is popotionl to the mgnitude of the electic field. Theefoe, the totl numbe of lines penetting the sufce is popotionl to the poduct A. This poduct of the mgnitude of the electic field nd sufce e A pependicul to the field is clled the electic flux (uppecse Geek phi): Figue 24.1 Field lines epesenting unifom electic field penetting plne of e A pependicul to the field. The electic flux though this e is eul to A. A (24.1) Fom the SI units of nd A, we see tht hs units of newtonmetes sued pe coulomb (N m 2 /C.) lectic flux is popotionl to the numbe of electic field lines penetting some sufce. xmple 24.1 lectic Flux Though Sphee Wht is the electic flux though sphee tht hs dius of 1.00 m nd cies chge of 1.00 C t its cente? Solution The mgnitude of the electic field 1.00 m fom this chge is found using ution 23.9: k e 2 ( Nm 2 /C 2 ) C (1.00 m) N/C The field points dilly outwd nd is theefoe eveywhee pependicul to the sufce of the sphee. The flux though the sphee (whose sufce e A m 2 ) is thus A ( N/C)(12.6 m 2 ) Nm 2 /C 740
3 SCTION 24.1 lectic Flux 741 A Noml θ A = A cos θ θ Figue 24.2 Field lines epesenting unifom electic field penetting n e A tht is t n ngle to the field. Becuse the numbe of lines tht go though the e A is the sme s the numbe tht go though A, the flux though A is eul to the flux though A nd is given by Acos. If the sufce unde considetion is not pependicul to the field, the flux though it must be less thn tht given by ution We cn undestnd this by consideing Figue 24.2, whee the noml to the sufce of e A is t n ngle to the unifom electic field. Note tht the numbe of lines tht coss this e A is eul to the numbe tht coss the e A, which is pojection of e A onto plne oiented pependicul to the field. Fom Figue 24.2 we see tht the two es e elted by A A cos. Becuse the flux though A euls the flux though A, we conclude tht the flux though A is (24.2) Fom this esult, we see tht the flux though sufce of fixed e A hs mximum vlue A when the sufce is pependicul to the field (when the noml to the sufce is pllel to the field, tht is, 0 in Figue 24.2); the flux is zeo when the sufce is pllel to the field (when the noml to the sufce is pependicul to the field, tht is, 90 ). We ssumed unifom electic field in the peceding discussion. In moe genel situtions, the electic field my vy ove sufce. Theefoe, ou definition of flux given by ution 24.2 hs mening only ove smll element of e. Conside genel sufce divided up into lge numbe of smll elements, ech of e A. The vition in the electic field ove one element cn be neglected if the element is sufficiently smll. It is convenient to define vecto A i whose mgnitude epesents the e of the ith element of the sufce nd whose diection is defined to be pependicul to the sufce element, s shown in Figue The electic field i t the loction of this element mkes n ngle i with the vecto A i. The electic flux though this element is whee we hve used the definition of the scl poduct (o dot poduct; see Chpte 7) of two vectos (A B AB cos ). By summing the contibutions of ll elements, we obtin the totl flux though the sufce. If we let the e of ech element ppoch zeo, then the numbe of elements ppoches infinity nd the sum is eplced by n integl. Theefoe, the genel definition of electic flux is 1 A A cos i A i cos i i A i lim A i i A i : 0 (24.3) ution 24.3 is sufce integl, which mens it must be evluted ove the sufce in uestion. In genel, the vlue of depends both on the field ptten nd on the sufce. sufce d A A i i Figue 24.3 A smll element of sufce e A i. The electic field mkes n ngle i with the vecto A i, defined s being noml to the sufce element, nd the flux though the element is eul to i A i cos i. θ i Definition of electic flux 1 Dwings with field lines hve thei inccucies becuse limited numbe of field lines e typiclly dwn in digm. Conseuently, smll e element dwn on digm (depending on its loction) my hppen to hve too few field lines penetting it to epesent the flux ccutely. We stess tht the bsic definition of electic flux is ution The use of lines is only n id fo visulizing the concept.
4 742 CHAPTR 24 Guss s Lw At the Active Figues link t you cn select ny segment on the sufce nd see the eltionship between the electic field vecto nd the e vecto A i. A 3 θ n A 2 A 1 n θ Active Figue 24.4 A closed sufce in n electic field. The e vectos A i e, by convention, noml to the sufce nd point outwd. The flux though n e element cn be positive (element ), zeo (element ), o negtive (element ). We e often inteested in evluting the flux though closed sufce, which is defined s one tht divides spce into n inside nd n outside egion, so tht one cnnot move fom one egion to the othe without cossing the sufce. The sufce of sphee, fo exmple, is closed sufce. Conside the closed sufce in Figue The vectos A i point in diffeent diections fo the vious sufce elements, but t ech point they e noml to the sufce nd, by convention, lwys point outwd. At the element lbeled, the field lines e cossing the sufce fom the inside to the outside nd 90 ; hence, the flux A 1 though this element is positive. Fo element, the field lines gze the sufce (pependicul to the vecto A 2 ); thus, 90 nd the flux is zeo. Fo elements such s, whee the field lines e cossing the sufce fom outside to inside, nd the flux is negtive becuse cos is negtive. The net flux though the sufce is popotionl to the net numbe of lines leving the sufce, whee the net numbe mens the numbe leving the sufce minus the numbe enteing the sufce. If moe lines e leving thn enteing, the net flux is positive. If moe lines e enteing thn leving, the net flux is negtive. Using the symbol to epesent n integl ove closed sufce, we cn wite the net flux though closed sufce s Kl Fiedich Guss Gemn mthemticin nd stonome ( ) Guss eceived doctol degee in mthemtics fom the Univesity of Helmstedt in In ddition to his wok in electomgnetism, he mde contibutions to mthemtics nd science in numbe theoy, sttistics, non ucliden geomety, nd comety obitl mechnics. He ws founde of the Gemn Mgnetic Union, which studies the th s mgnetic field on continul bsis. d A n da (24.4) whee n epesents the component of the electic field noml to the sufce. If the field is noml to the sufce t ech point nd constnt in mgnitude, the clcultion is stightfowd, s it ws in xmple xmple 24.2 lso illusttes this point. uick uiz 24.1 Suppose the dius of the sphee in xmple 24.1 is chnged to m. Wht hppens to the flux though the sphee nd the mgnitude of the electic field t the sufce of the sphee? () The flux nd field both incese. (b) The flux nd field both decese. (c) The flux inceses nd the field deceses. (d) The flux deceses nd the field inceses. (e) The flux emins the sme nd the field inceses. (f) The flux deceses nd the field emins the sme.
5 SCTION 24.2 Guss s Lw 743 uick uiz 24.2 In chgefee egion of spce, closed contine is plced in n electic field. A euiement fo the totl electic flux though the sufce of the contine to be zeo is tht () the field must be unifom, (b) the contine must be symmetic, (c) the contine must be oiented in cetin wy, o (d) the euiement does not exist the totl electic flux is zeo no mtte wht. xmple 24.2 Flux Though Cube Conside unifom electic field oiented in the x diection. Find the net electic flux though the sufce of cube of edge length, oiented s shown in Figue Solution The net flux is the sum of the fluxes though ll fces of the cube. Fist, note tht the flux though fou of the fces (,, nd the unnumbeed ones) is zeo becuse is pependicul to d A on these fces. The net flux though fces nd is d A d A 1 2 z da 1 y da 3 da 4 da 2 x Fo fce, is constnt nd diected inwd but da 1 is diected outwd ( 180 ); thus, the flux though this fce is A (cos 180) d A d A A 1d becuse the e of ech fce is A 2. Fo fce, is constnt nd outwd nd in the sme diection s da 2 ( 0 ); hence, the flux though this fce is d A (cos 0) da da A Figue 24.5 (xmple 24.2) A closed sufce in the shpe of cube in unifom electic field oiented pllel to the x xis. Side is the bottom of the cube, nd side is opposite side. Theefoe, the net flux ove ll six fces is Guss s Lw In this section we descibe genel eltionship between the net electic flux though closed sufce (often clled gussin sufce) nd the chge enclosed by the sufce. This eltionship, known s Guss s lw, is of fundmentl impotnce in the study of electic fields. Let us gin conside positive point chge locted t the cente of sphee of dius, s shown in Figue Fom ution 23.9 we know tht the mgnitude of the electic field eveywhee on the sufce of the sphee is k e / 2. As noted in xmple 24.1, the field lines e diected dilly outwd nd hence e pependicul to the sufce t evey point on the sufce. Tht is, t ech sufce point, is pllel to the vecto A i epesenting locl element of e A i suounding the sufce point. Theefoe, A i A i nd fom ution 24.4 we find tht the net flux though the gussin sufce is d A da da Gussin sufce dai Figue 24.6 A spheicl gussin sufce of dius suounding point chge. When the chge is t the cente of the sphee, the electic field is eveywhee noml to the sufce nd constnt in mgnitude.
6 744 CHAPTR 24 Guss s Lw S 3 S 2 S 1 Figue 24.7 Closed sufces of vious shpes suounding chge. The net electic flux is the sme though ll sufces. whee we hve moved outside of the integl becuse, by symmety, is constnt ove the sufce nd given by k e / 2. Futhemoe, becuse the sufce is spheicl, da A 4 2. Hence, the net flux though the gussin sufce is Figue 24.8 A point chge locted outside closed sufce. The numbe of lines enteing the sufce euls the numbe leving the sufce. S 1 2 S 4 3 S Active Figue 24.9 The net electic flux though ny closed sufce depends only on the chge inside tht sufce. The net flux though sufce S is 1 / 0, the net flux though sufce S is ( 2 3 )/ 0, nd the net flux though sufce S is zeo. Chge 4 does not contibute to the flux though ny sufce becuse it is outside ll sufces. At the Active Figues link t you cn chnge the size nd shpe of closed sufce nd see the effect on the electic flux of suounding combintions of chge with tht sufce. k e 2 (4 2 ) 4k e Reclling fom Section 23.3 tht k e 1/4 0, we cn wite this eution in the fom (24.5) We cn veify tht this expession fo the net flux gives the sme esult s xmple 24.1: ( C)/( C 2 /N m 2 ) N m 2 /C. Note fom ution 24.5 tht the net flux though the spheicl sufce is popotionl to the chge inside. The flux is independent of the dius becuse the e of the spheicl sufce is popotionl to 2, whees the electic field is popotionl to 1/ 2. Thus, in the poduct of e nd electic field, the dependence on cncels. Now conside sevel closed sufces suounding chge, s shown in Figue Sufce S 1 is spheicl, but sufces S 2 nd S 3 e not. Fom ution 24.5, the flux tht psses though S 1 hs the vlue / 0. As we discussed in the peceding section, flux is popotionl to the numbe of electic field lines pssing though sufce. The constuction shown in Figue 24.7 shows tht the numbe of lines though S 1 is eul to the numbe of lines though the nonspheicl sufces S 2 nd S 3. Theefoe, we conclude tht the net flux though ny closed sufce suounding point chge is given by / 0 nd is independent of the shpe of tht sufce. Now conside point chge locted outside closed sufce of bity shpe, s shown in Figue As you cn see fom this constuction, ny electic field line tht entes the sufce leves the sufce t nothe point. The numbe of electic field lines enteing the sufce euls the numbe leving the sufce. Theefoe, we conclude tht the net electic flux though closed sufce tht suounds no chge is zeo. If we pply this esult to xmple 24.2, we cn esily see tht the net flux though the cube is zeo becuse thee is no chge inside the cube. Let us extend these guments to two genelized cses: (1) tht of mny point chges nd (2) tht of continuous distibution of chge. We once gin use the supeposition pinciple, which sttes tht the electic field due to mny chges is the vecto sum of the electic fields poduced by the individul chges. Theefoe, we cn expess the flux though ny closed sufce s d A ( 1 2 )d A whee is the totl electic field t ny point on the sufce poduced by the vecto ddition of the electic fields t tht point due to the individul chges. Conside the system of chges shown in Figue The sufce S suounds only one chge, 1 ; hence, the net flux though S is 1 / 0. The flux though S due to chges 2, 3, nd 4 outside it is zeo becuse ech electic field line tht entes S t one point leves it t 0
7 SCTION 24.2 Guss s Lw 745 nothe. The sufce S suounds chges 2 nd 3 ; hence, the net flux though it is ( 2 3 )/ 0. Finlly, the net flux though sufce S is zeo becuse thee is no chge inside this sufce. Tht is, ll the electic field lines tht ente S t one point leve t nothe. Notice tht chge 4 does not contibute to the net flux though ny of the sufces becuse it is outside ll of the sufces. Guss s lw, which is geneliztion of wht we hve just descibed, sttes tht the net flux though ny closed sufce is d A in 0 (24.6) Guss s lw whee in epesents the net chge inside the sufce nd epesents the electic field t ny point on the sufce. A foml poof of Guss s lw is pesented in Section When using ution 24.6, you should note tht lthough the chge in is the net chge inside the gussin sufce, epesents the totl electic field, which includes contibutions fom chges both inside nd outside the sufce. In pinciple, Guss s lw cn be solved fo to detemine the electic field due to system of chges o continuous distibution of chge. In pctice, howeve, this type of solution is pplicble only in limited numbe of highly symmetic situtions. In the next section we use Guss s lw to evlute the electic field fo chge distibutions tht hve spheicl, cylindicl, o pln symmety. If one chooses the gussin sufce suounding the chge distibution cefully, the integl in ution 24.6 cn be simplified. uick uiz 24.3 If the net flux though gussin sufce is zeo, the following fou sttements could be tue. Which of the sttements must be tue? () Thee e no chges inside the sufce. (b) The net chge inside the sufce is zeo. (c) The electic field is zeo eveywhee on the sufce. (d) The numbe of electic field lines enteing the sufce euls the numbe leving the sufce. PITFALL PRVNTION 24.1 Zeo Flux is not Zeo Field We see two situtions in which thee is zeo flux though closed sufce eithe thee e no chged pticles enclosed by the sufce o thee e chged pticles enclosed, but the net chge inside the sufce is zeo. Fo eithe sitution, it is incoect to conclude tht the electic field on the sufce is zeo. Guss s lw sttes tht the electic flux is popotionl to the enclosed chge, not the electic field. uick uiz 24.4 Conside the chge distibution shown in Figue The chges contibuting to the totl electic flux though sufce S e () 1 only (b) 4 only (c) 2 nd 3 (d) ll fou chges (e) none of the chges. uick uiz 24.5 Agin conside the chge distibution shown in Figue The chges contibuting to the totl electic field t chosen point on the sufce S e () 1 only (b) 4 only (c) 2 nd 3 (d) ll fou chges (e) none of the chges. Conceptul xmple 24.3 Flux Due to Point Chge A spheicl gussin sufce suounds point chge. Descibe wht hppens to the totl flux though the sufce if (A) the chge is tipled, (B) the dius of the sphee is doubled, (C) the sufce is chnged to cube, nd (D) the chge is moved to nothe loction inside the sufce. Solution (A) The flux though the sufce is tipled becuse flux is popotionl to the mount of chge inside the sufce. (B) The flux does not chnge becuse ll electic field lines fom the chge pss though the sphee, egdless of its dius. (C) The flux does not chnge when the shpe of the gussin sufce chnges becuse ll electic field lines fom the chge pss though the sufce, egdless of its shpe. (D) The flux does not chnge when the chge is moved to nothe loction inside tht sufce becuse Guss s lw efes to the totl chge enclosed, egdless of whee the chge is locted inside the sufce.
8 746 CHAPTR 24 Guss s Lw 24.3 Appliction of Guss s Lw to Vious Chge Distibutions As mentioned elie, Guss s lw is useful in detemining electic fields when the chge distibution is chcteized by high degee of symmety. The following exmples demonstte wys of choosing the gussin sufce ove which the sufce integl given by ution 24.6 cn be simplified nd the electic field detemined. In choosing the sufce, we should lwys tke dvntge of the symmety of the chge distibution so tht we cn emove fom the integl nd solve fo it. The gol in this type of clcultion is to detemine sufce tht stisfies one o moe of the following conditions: PITFALL PRVNTION 24.2 Gussin Sufces e not Rel A gussin sufce is n imginy sufce tht you choose to stisfy the conditions listed hee. It does not hve to coincide with physicl sufce in the sitution. 1. The vlue of the electic field cn be gued by symmety to be constnt ove the sufce. 2. The dot poduct in ution 24.6 cn be expessed s simple lgebic poduct dabecuse nd d A e pllel. 3. The dot poduct in ution 24.6 is zeo becuse nd d A e pependicul. 4. The field cn be gued to be zeo ove the sufce. All fou of these conditions e used in exmples thoughout the eminde of this chpte. xmple 24.4 The lectic Field Due to Point Chge Stting with Guss s lw, clculte the electic field due to n isolted point chge. Solution A single chge epesents the simplest possible chge distibution, nd we use this fmili cse to show how to solve fo the electic field with Guss s lw. Figue nd ou discussion of the electic field due to point chge in Chpte 23 help us to conceptulize the physicl sitution. Becuse the spce ound the single chge hs spheicl symmety, we ctegoize this poblem s one in which thee is enough symmety to pply Guss s lw. To nlyze ny Guss s lw poblem, we conside the detils of the electic field nd choose gussin sufce tht stisfies some o ll of the conditions tht we hve listed bove. We choose spheicl gussin sufce of dius centeed on the point chge, s shown in Figue The electic field due to positive point chge is diected dilly outwd by Gussin sufce Figue (xmple 24.4) The point chge is t the cente of the spheicl gussin sufce, nd is pllel to d A t evey point on the sufce. da symmety nd is theefoe noml to the sufce t evey point. Thus, s in condition (2), is pllel to da t ech point. Theefoe, da dand Guss s lw gives d A da By symmety, is constnt eveywhee on the sufce, which stisfies condition (1), so it cn be emoved fom the integl. Theefoe, da d A (4 2 ) whee we hve used the fct tht the sufce e of sphee is 4 2. Now, we solve fo the electic field: 40 2 To finlize this poblem, note tht this is the fmili electic field due to point chge tht we developed fom Coulomb s lw in Chpte 23. Wht If? Wht if the chge in Figue wee not t the cente of the spheicl gussin sufce? Answe In this cse, while Guss s lw would still be vlid, the sitution would not possess enough symmety to evlute the electic field. Becuse the chge is not t the cente, the mgnitude of would vy ove the sufce of the sphee nd the vecto would not be eveywhee pependicul to the sufce. k e 2 0 0
9 SCTION 24.3 Appliction of Guss s Lw to Vious Chge Distibutions 747 xmple 24.5 A Spheiclly Symmetic Chge Distibution Intective An insulting solid sphee of dius hs unifom volume chge density nd cies totl positive chge (Fig ). (A) Clculte the mgnitude of the electic field t point outside the sphee. Solution Becuse the chge distibution is spheiclly symmetic, we gin select spheicl gussin sufce of dius, concentic with the sphee, s shown in Figue Fo this choice, conditions (1) nd (2) e stisfied, s they wee fo the point chge in xmple Following the line of esoning given in xmple 24.4, we find tht (1) k e 2 Note tht this esult is identicl to the one we obtined fo point chge. Theefoe, we conclude tht, fo unifomly chged sphee, the field in the egion extenl to the sphee is euivlent to tht of point chge locted t the cente of the sphee. (B) Find the mgnitude of the electic field t point inside the sphee. Solution In this cse we select spheicl gussin sufce hving dius, concentic with the insulting sphee (Fig b). Let us denote the volume of this smlle sphee by V. To pply Guss s lw in this sitution, it is impotnt to ecognize tht the chge in within the gussin sufce of volume V is less thn. To clculte in, we use the fct tht in V : in V ( ) (fo ) By symmety, the mgnitude of the electic field is constnt eveywhee on the spheicl gussin sufce nd is noml to the sufce t ech point both conditions (1) nd (2) e stisfied. Theefoe, Guss s lw in the egion gives Solving fo gives da da (4 2 ) in 0 / Becuse by definition nd becuse k e 1/4 0, this expession fo cn be witten s Note tht this esult fo diffes fom the one we obtined in pt (A). It shows tht : 0 s : 0. Theefoe, the esult elimintes the poblem tht would exist t 0 if vied s 1/ 2 inside the sphee s it does outside the sphee. Tht is, if 1/ 2 fo, the field would be infinite t 0, which is physiclly impossible. Wht If? Suppose we ppoch the dil position fom inside the sphee nd fom outside. Do we mesue the sme vlue of the electic field fom both diections? Answe Fom ution (1), we see tht the field ppoches vlue fom the outside given by Fom the inside, ution (2) gives us lim : k e in 40 2 (2) 40 3 lim : k e 2 k e 3 k e ( ) k e (fo ) 2 Thus, the vlue of the field is the sme s we ppoch the sufce fom both diections. A plot of vesus is shown in Figue Note tht the mgnitude of the field is continuous, but the deivtive of the field mgnitude is not. k e 3 30 Gussin sphee () Gussin sphee Figue (xmple 24.5) A unifomly chged insulting sphee of dius nd totl chge. () Fo points outside the sphee, lge spheicl gussin sufce is dwn concentic with the sphee. In digms such s this, the dotted line epesents the intesection of the gussin sufce with the plne of the pge. (b) Fo points inside the sphee, spheicl gussin sufce smlle thn the sphee is dwn. (b) = k e 3 = k e 2 Figue (xmple 24.5) A plot of vesus fo unifomly chged insulting sphee. The electic field inside the sphee ( ) vies linely with. The field outside the sphee ( ) is the sme s tht of point chge locted t 0. At the Intective Woked xmple link t you cn investigte the electic field inside nd outside the sphee.
10 748 CHAPTR 24 Guss s Lw xmple 24.6 The lectic Field Due to Thin Spheicl Shell A thin spheicl shell of dius hs totl chge distibuted unifomly ove its sufce (Fig ). Find the electic field t points k e 2 (fo ) (A) outside nd (B) inside the shell. Solution (A) The clcultion fo the field outside the shell is identicl to tht fo the solid sphee shown in xmple If we constuct spheicl gussin sufce of dius concentic with the shell (Fig b), the chge inside this sufce is. Theefoe, the field t point outside the shell is euivlent to tht due to point chge locted t the cente: (B) The electic field inside the spheicl shell is zeo. This follows fom Guss s lw pplied to spheicl sufce of dius concentic with the shell (Fig c). Becuse of the spheicl symmety of the chge distibution nd becuse the net chge inside the sufce is zeo stisfction of conditions (1) nd (2) gin ppliction of Guss s lw shows tht 0 in the egion. We obtin the sme esults using ution nd integting ove the chge distibution. This clcultion is the complicted. Guss s lw llows us to detemine these esults in much simple wy. Gussin sphee Gussin sphee in = 0 () (b) (c) Figue (xmple 24.6) () The electic field inside unifomly chged spheicl shell is zeo. The field outside is the sme s tht due to point chge locted t the cente of the shell. (b) Gussin sufce fo. (c) Gussin sufce fo. xmple 24.7 A Cylindiclly Symmetic Chge Distibution Find the electic field distnce fom line of positive chge of infinite length nd constnt chge pe unit length (Fig ). Solution The symmety of the chge distibution euies tht be pependicul to the line chge nd diected outwd, s shown in Figue nd b. To eflect the symmety of the chge distibution, we select cylindicl gussin sufce of dius nd length tht is coxil with the line chge. Fo the cuved pt of this sufce, is constnt in mgnitude nd pependicul to the sufce t ech point stisfction of conditions (1) nd (2). Futhemoe, the flux though the ends of the gussin cylinde is zeo becuse is pllel to these sufces the fist ppliction we hve seen of condition (3). We tke the sufce integl in Guss s lw ove the entie gussin sufce. Becuse of the zeo vlue of da fo the ends of the cylinde, howeve, we cn estict ou ttention to only the cuved sufce of the cylinde. The totl chge inside ou gussin sufce is. Applying Guss s lw nd conditions (1) nd (2), we find tht fo the cuved sufce d A da A in The e of the cuved sufce is A 2; theefoe, (2 ) 0 20 (24.7) Thus, we see tht the electic field due to cylindiclly symmetic chge distibution vies s 1/, whees the field extenl to spheiclly symmetic chge distibution vies s 1/ 2. ution 24.7 ws lso deived by integtion of the field of point chge. (See Poblem 35 in Chpte 23.) 2k e 0 0
11 SCTION 24.3 Appliction of Guss s Lw to Vious Chge Distibutions 749 Gussin sufce () da Wht If? Wht if the line segment in this exmple wee not infinitely long? Answe If the line chge in this exmple wee of finite length, the esult fo would not be tht given by ution A finite line chge does not possess sufficient symmety fo us to mke use of Guss s lw. This is becuse the mgnitude of the electic field is no longe constnt ove the sufce of the gussin cylinde the field ne the ends of the line would be diffeent fom tht f fom the ends. Thus, condition (1) would not be stisfied in this sitution. Futhemoe, is not pependicul to the cylindicl sufce t ll points the field vectos ne the ends would hve component pllel to the line. Thus, condition (2) would not be stisfied. Fo points close to finite line chge nd f fom the ends, ution 24.7 gives good ppoximtion of the vlue of the field. It is left fo you to show (see Poblem 29) tht the electic field inside unifomly chged od of finite dius nd infinite length is popotionl to. (b) Figue (xmple 24.7) () An infinite line of chge suounded by cylindicl gussin sufce concentic with the line. (b) An end view shows tht the electic field t the cylindicl sufce is constnt in mgnitude nd pependicul to the sufce. xmple 24.8 A Plne of Chge Find the electic field due to n infinite plne of positive chge with unifom sufce chge density. Solution By symmety, must be pependicul to the plne nd must hve the sme mgnitude t ll points euidistnt fom the plne. The fct tht the diection of is wy fom positive chges indictes tht the diection of on one side of the plne must be opposite its diection on the othe side, s shown in Figue A gussin sufce tht eflects the symmety is smll cylinde whose xis is pependicul to the plne nd whose ends Gussin sufce Figue (xmple 24.8) A cylindicl gussin sufce penetting n infinite plne of chge. The flux is A though ech end of the gussin sufce nd zeo though its cuved sufce. A ech hve n e A nd e euidistnt fom the plne. Becuse is pllel to the cuved sufce nd, theefoe, pependicul to da eveywhee on the sufce condition (3) is stisfied nd thee is no contibution to the sufce integl fom this sufce. Fo the flt ends of the cylinde, conditions (1) nd (2) e stisfied. The flux though ech end of the cylinde is A; hence, the totl flux though the entie gussin sufce is just tht though the ends, 2A. Noting tht the totl chge inside the sufce is in A, we use Guss s lw nd find tht the totl flux though the gussin sufce is leding to 2 A in (24.8) Becuse the distnce fom ech flt end of the cylinde to the plne does not ppe in ution 24.8, we conclude tht /2 0 t ny distnce fom the plne. Tht is, the field is unifom eveywhee. Wht If? Suppose we plce two infinite plnes of chge pllel to ech othe, one positively chged nd the othe negtively chged. Both plnes hve the sme sufce chge density. Wht does the electic field look like now? 0 20 A 0
12 750 CHAPTR 24 Guss s Lw Answe In this sitution, the electic fields due to the two plnes dd in the egion between the plnes, esulting in unifom field of mgnitude / 0, nd cncel elsewhee to give field of zeo. This is pcticl wy to chieve unifom electic fields, such s those needed in the CRT tube discussed in Section Conceptul xmple 24.9 Don t Use Guss s Lw Hee! xplin why Guss s lw cnnot be used to clculte the electic field ne n electic dipole, chged disk, o tingle with point chge t ech cone. Solution The chge distibutions of ll these configutions do not hve sufficient symmety to mke the use of Guss s lw pcticl. We cnnot find closed sufce suounding ny of these distibutions tht stisfies one o moe of conditions (1) though (4) listed t the beginning of this section Conductos in lectosttic uilibium As we lened in Section 23.2, good electicl conducto contins chges (electons) tht e not bound to ny tom nd theefoe e fee to move bout within the mteil. When thee is no net motion of chge within conducto, the conducto is in electosttic euilibium. A conducto in electosttic euilibium hs the following popeties: Popeties of conducto in electosttic euilibium Figue A conducting slb in n extenl electic field. The chges induced on the two sufces of the slb poduce n electic field tht opposes the extenl field, giving esultnt field of zeo inside the slb. 1. The electic field is zeo eveywhee inside the conducto. 2. If n isolted conducto cies chge, the chge esides on its sufce. 3. The electic field just outside chged conducto is pependicul to the sufce of the conducto nd hs mgnitude / 0, whee is the sufce chge density t tht point. 4. On n iegully shped conducto, the sufce chge density is getest t loctions whee the dius of cuvtue of the sufce is smllest. We veify the fist thee popeties in the discussion tht follows. The fouth popety is pesented hee so tht we hve complete list of popeties fo conductos in electosttic euilibium, but cnnot be veified until Chpte 25. We cn undestnd the fist popety by consideing conducting slb plced in n extenl field (Fig ). The electic field inside the conducto must be zeo unde the ssumption tht we hve electosttic euilibium. If the field wee not zeo, fee electons in the conducto would expeience n electic foce (F ) nd would ccelete due to this foce. This motion of electons, howeve, would men tht the conducto is not in electosttic euilibium. Thus, the existence of electosttic euilibium is consistent only with zeo field in the conducto. Let us investigte how this zeo field is ccomplished. Befoe the extenl field is pplied, fee electons e unifomly distibuted thoughout the conducto. When the extenl field is pplied, the fee electons ccelete to the left in Figue 24.16, cusing plne of negtive chge to be pesent on the left sufce. The movement of electons to the left esults in plne of positive chge on the ight sufce. These plnes of chge cete n dditionl electic field inside the conducto tht opposes the extenl field. As the electons move, the sufce chge densities on the left nd ight sufces incese until the mgnitude of the intenl field euls tht of the extenl field, esulting in net field of zeo inside the conducto. The time it tkes good conducto to ech euilibium is on the ode of s, which fo most puposes cn be consideed instntneous.
13 SCTION 24.4 Conductos in lectosttic uilibium 751 We cn use Guss s lw to veify the second popety of conducto in electosttic euilibium. Figue shows n bitily shped conducto. A gussin sufce is dwn inside the conducto nd cn be s close to the conducto s sufce s we wish. As we hve just shown, the electic field eveywhee inside the conducto is zeo when it is in electosttic euilibium. Theefoe, the electic field must be zeo t evey point on the gussin sufce, in ccodnce with condition (4) in Section Thus, the net flux though this gussin sufce is zeo. Fom this esult nd Guss s lw, we conclude tht the net chge inside the gussin sufce is zeo. Becuse thee cn be no net chge inside the gussin sufce (which is bitily close to the conducto s sufce), ny net chge on the conducto must eside on its sufce. Guss s lw does not indicte how this excess chge is distibuted on the conducto s sufce, only tht it esides exclusively on the sufce. We cn lso use Guss s lw to veify the thid popety. Fist, note tht if the field vecto hd component pllel to the conducto s sufce, fee electons would expeience n electic foce nd move long the sufce; in such cse, the conducto would not be in euilibium. Thus, the field vecto must be pependicul to the sufce. To detemine the mgnitude of the electic field, we dw gussin sufce in the shpe of smll cylinde whose end fces e pllel to the sufce of the conducto (Fig ). Pt of the cylinde is just outside the conducto, nd pt is inside. The field is pependicul to the conducto s sufce fom the condition of electosttic euilibium. Thus, we stisfy condition (3) in Section 24.3 fo the cuved pt of the cylindicl gussin sufce thee is no flux though this pt of the gussin sufce becuse is pllel to the sufce. Thee is no flux though the flt fce of the cylinde inside the conducto becuse hee 0; this stisfies condition (4). Hence, the net flux though the gussin sufce is tht though only the flt fce outside the conducto, whee the field is pependicul to the gussin sufce. Using conditions (1) nd (2) fo this fce, the flux is A, whee is the electic field just outside the conducto nd A is the e of the cylinde s fce. Applying Guss s lw to this sufce, we obtin da A in 0 A 0 Figue A conducto of bity shpe. The boken line epesents gussin sufce tht cn be s close to the sufce of the conducto s we wish. A Gussin sufce Figue A gussin sufce in the shpe of smll cylinde is used to clculte the electic field just outside chged conducto. The flux though the gussin sufce is A. Remembe tht is zeo inside the conducto. whee we hve used the fct tht in A. Solving fo gives fo the electic field just outside chged conducto 0 (24.9) Figue shows electic field lines mde visible by pieces of thed floting in oil. Notice tht the field lines e pependicul to both the cylindicl conducting sufce nd the stight conducting sufce. uick uiz 24.6 You little bothe likes to ub his feet on the cpet nd then touch you to give you shock. While you e tying to escpe the shock tetment, you discove hollow metl cylinde in you bsement, lge enough to climb inside. In which of the following cses will you not be shocked? () You climb inside the cylinde, mking contct with the inne sufce, nd you chged bothe touches the oute metl sufce. (b) You chged bothe is inside touching the inne metl sufce nd you e outside, touching the oute metl sufce. (c) Both of you e outside the cylinde, touching its oute metl sufce but not touching ech othe diectly. Coutesy of Hold M. Wge, Pinceton Univesity Figue lectic field ptten suounding chged conducting plte plced ne n oppositely chged conducting cylinde. Smll pieces of thed suspended in oil lign with the electic field lines. Note tht (1) the field lines e pependicul to both conductos nd (2) thee e no lines inside the cylinde ( 0).
14 752 CHAPTR 24 Guss s Lw xmple A Sphee Inside Spheicl Shell Intective A solid conducting sphee of dius cies net positive chge 2. A conducting spheicl shell of inne dius b nd oute dius c is concentic with the solid sphee nd cies net chge. Using Guss s lw, find the electic field in the egions lbeled,,, nd in Figue nd the chge distibution on the shell when the entie system is in electosttic euilibium. Solution Fist note tht the chge distibutions on both the sphee nd the shell e chcteized by spheicl symmety ound thei common cente. To detemine the electic field t vious distnces fom this cente, we constuct spheicl gussin sufce fo ech of the fou egions of inteest. Such sufce fo egion is shown in Figue To find inside the solid sphee (egion ), conside gussin sufce of dius. Becuse thee cn be no chge inside conducto in electosttic euilibium, we see tht in 0; thus, on the bsis of Guss s lw nd symmety, 1 0 fo. In egion between the sufce of the solid sphee nd the inne sufce of the shell we constuct spheicl gussin sufce of dius whee b nd note tht the chge inside this sufce is 2 (the chge on the solid sphee). Becuse of the spheicl symmety, the electic field lines must be diected dilly outwd nd be constnt in mgnitude on the gussin sufce. Following xmple 24.4 nd using Guss s lw, we find tht 2A 2(4 2 ) in In egion, whee c, the spheicl gussin sufce we constuct suounds totl chge of in 2 (). Theefoe, ppliction of Guss s lw to this sufce gives 4 0 k e 2 2k e (fo b) (fo c) In egion, the electic field must be zeo becuse the spheicl shell is lso conducto in euilibium. Figue shows gphicl epesenttion of the vition of electic field with. If we constuct gussin sufce of dius whee b c, we see tht in must be zeo becuse 3 0. Fom this gument, we conclude tht the chge on the inne sufce of the spheicl shell must be 2 to cncel the chge 2 on the solid sphee. Becuse the net chge on the shell is, we conclude tht its oute sufce must cy chge. 2 b = 2k e 2 Figue (xmple 24.10) A solid conducting sphee of dius nd cying chge 2 suounded by conducting spheicl shell cying chge. c b Figue (xmple 24.10) A plot of vesus fo the twoconducto system shown in Figue c = k e 2 xploe the electic field of the system in Figue t the Intective Woked xmple link t Foml Deivtion of Guss s Lw One wy of deiving Guss s lw involves solid ngles. Conside spheicl sufce of dius contining n e element A. The solid ngle (: uppecse Geek omeg) subtended t the cente of the sphee by this element is defined to be A 2 Fom this eution, we see tht hs no dimensions becuse A nd 2 both hve dimensions L 2. The dimensionless unit of solid ngle is the stedin. (You my wnt to compe this eution to ution 10.1b, the definition of the din.) Becuse the
15 Summy 753 θ A Ω A Ω θ A cos θ θ A Figue A closed sufce of bity shpe suounds point chge. The net electic flux though the sufce is independent of the shpe of the sufce. Figue The e element A subtends solid ngle (A cos )/ 2 t the chge. sufce e of sphee is 4 2, the totl solid ngle subtended by the sphee is stedins Now conside point chge suounded by closed sufce of bity shpe (Fig ). The totl electic flux though this sufce cn be obtined by evluting A fo ech smll e element A nd summing ove ll elements. The flux though ech element is A ( cos )A k e A cos 2 whee is the distnce fom the chge to the e element, is the ngle between the electic field nd A fo the element, nd k e / 2 fo point chge. In Figue 24.23, we see tht the pojection of the e element pependicul to the dius vecto is A cos. Thus, the untity (A cos )/ 2 is eul to the solid ngle tht the sufce element A subtends t the chge. We lso see tht is eul to the solid ngle subtended by the e element of spheicl sufce of dius. Becuse the totl solid ngle t point is 4 stedins, the totl flux though the closed sufce is k e da cos 2 k e d 4k e Thus we hve deived Guss s lw, ution Note tht this esult is independent of the shpe of the closed sufce nd independent of the position of the chge within the sufce. 0 SUMMARY lectic flux is popotionl to the numbe of electic field lines tht penette sufce. If the electic field is unifom nd mkes n ngle with the noml to sufce of e A, the electic flux though the sufce is A cos (24.2) Tke pctice test fo this chpte by clicking on the Pctice Test link t In genel, the electic flux though sufce is sufce d A (24.3)
16 754 CHAPTR 24 Guss s Lw Tble 24.1 Typicl lectic Field Clcultions Using Guss s Lw Chge Distibution lectic Field Loction Insulting sphee of dius R, unifom chge density, nd totl chge k e k e 2 R 2 R R Thin spheicl shell of dius R nd totl chge 0 k e 2 R R Line chge of infinite length nd chge pe unit length 2k e Outside the line Infinite chged plne hving sufce chge density Conducto hving sufce chge density veywhee outside the plne Just outside the conducto Inside the conducto You should be ble to pply utions 24.2 nd 24.3 in viety of situtions, pticully those in which symmety simplifies the clcultion. Guss s lw sys tht the net electic flux though ny closed gussin sufce is eul to the net chge in inside the sufce divided by 0 : d A in (24.6) Using Guss s lw, you cn clculte the electic field due to vious symmetic chge distibutions. Tble 24.1 lists some typicl esults. A conducto in electosttic euilibium hs the following popeties: 1. The electic field is zeo eveywhee inside the conducto. 2. Any net chge on the conducto esides entiely on its sufce. 3. The electic field just outside the conducto is pependicul to its sufce nd hs mgnitude / 0, whee is the sufce chge density t tht point. 4. On n iegully shped conducto, the sufce chge density is getest whee the dius of cuvtue of the sufce is the smllest. 0 USTIONS 1. The Sun is lowe in the sky duing the winte months thn it is in the summe. How does this chnge the flux of sunlight hitting given e on the sufce of the th? How does this ffect the wethe? 2. If the electic field in egion of spce is zeo, cn you conclude tht no electic chges e in tht egion? xplin. 3. If moe electic field lines leve gussin sufce thn ente it, wht cn you conclude bout the net chge enclosed by tht sufce? 4. A unifom electic field exists in egion of spce in which thee e no chges. Wht cn you conclude bout the net electic flux though gussin sufce plced in this egion of spce? 5. If the totl chge inside closed sufce is known but the distibution of the chge is unspecified, cn you use Guss s lw to find the electic field? xplin. 6. xplin why the electic flux though closed sufce with given enclosed chge is independent of the size o shpe of the sufce. 7. Conside the electic field due to nonconducting infinite plne hving unifom chge density. xplin why the electic field does not depend on the distnce fom the plne, in tems of the spcing of the electic field lines. 8. Use Guss s lw to explin why electic field lines must begin o end on electic chges. (Suggestion: Chnge the size of the gussin sufce.)
17 Poblems On the bsis of the epulsive ntue of the foce between like chges nd the feedom of motion of chge within conducto, explin why excess chge on n isolted conducto must eside on its sufce. 10. A peson is plced in lge hollow metllic sphee tht is insulted fom gound. If lge chge is plced on the sphee, will the peson be hmed upon touching the inside of the sphee? xplin wht will hppen if the peson lso hs n initil chge whose sign is opposite tht of the chge on the sphee. 11. Two solid sphees, both of dius R, cy identicl totl chges,. One sphee is good conducto while the othe is n insulto. If the chge on the insulting sphee is unifomly distibuted thoughout its inteio volume, 12. how do the electic fields outside these two sphees compe? Ae the fields identicl inside the two sphees? A common demonsttion involves chging ubbe blloon, which is n insulto, by ubbing it on you hi, nd touching the blloon to ceiling o wll, which is lso n insulto. The electicl ttction between the chged blloon nd the neutl wll esults in the blloon sticking to the wll. Imgine now tht we hve two infinitely lge flt sheets of insulting mteil. One is chged nd the othe is neutl. If these e bought into contct, will n ttctive foce exist between them, s thee ws fo the blloon nd the wll? 13. You my hve hed tht one of the sfe plces to be duing lightning stom is inside c. Why would this be the cse? PROBLMS 1, 2, 3 stightfowd, intemedite, chllenging full solution vilble in the Student Solutions Mnul nd Study Guide coched solution with hints vilble t compute useful in solving poblem pied numeicl nd symbolic poblems Section 24.1 lectic Flux 1. An electic field with mgnitude of 3.50 kn/c is pplied long the x xis. Clculte the electic flux though ectngul plne m wide nd m long ssuming tht () the plne is pllel to the yz plne; (b) the plne is pllel to the xy plne; (c) the plne contins the y xis, nd its noml mkes n ngle of 40.0 with the x xis. 2. A veticl electic field of mgnitude N/C exists bove the th s sufce on dy when thundestom is bewing. A c with ectngul size of 6.00 m by 3.00 m is tveling long odwy sloping downwd t Detemine the electic flux though the bottom of the c. 3. A 40.0cmdimete loop is otted in unifom electic field until the position of mximum electic flux is found. The flux in this position is mesued to be Nm 2 /C. Wht is the mgnitude of the electic field? 4. Conside closed tingul box esting within hoizontl electic field of mgnitude N/C s shown in Figue P24.4. Clculte the electic flux though () the veticl ectngul sufce, (b) the slnted sufce, nd (c) the entie sufce of the box. 6. A point chge is locted t the cente of unifom ing hving line chge density nd dius, s shown in Figue P24.6. Detemine the totl electic flux though sphee centeed t the point chge nd hving dius R, whee R. λ 7. A pymid with hoizontl sue bse, 6.00 m on ech side, nd height of 4.00 m is plced in veticl electic field of 52.0 N/C. Clculte the totl electic flux though the pymid s fou slnted sufces. 8. A cone with bse dius R nd height h is locted on hoizontl tble. A hoizontl unifom field penettes the cone, s shown in Figue P24.8. Detemine the electic flux tht entes the lefthnd side of the cone. R Figue P cm h 10 cm 60 R Figue P24.4 Figue P A unifom electic field î bĵ intesects sufce of e A. Wht is the flux though this e if the sufce lies () in the yz plne? (b) in the xz plne? (c) in the xy plne? Section 24.2 Guss s Lw 9. The following chges e locted inside submine: 5.00 C, 9.00 C, 27.0 C, nd 84.0 C. () Clculte
18 756 CHAPTR 24 Guss s Lw the net electic flux though the hull of the submine. (b) Is the numbe of electic field lines leving the submine gete thn, eul to, o less thn the numbe enteing it? 10. The electic field eveywhee on the sufce of thin spheicl shell of dius m is mesued to be 890 N/C nd points dilly towd the cente of the sphee. () Wht is the net chge within the sphee s sufce? (b) Wht cn you conclude bout the ntue nd distibution of the chge inside the spheicl shell? 11. Fou closed sufces, S 1 though S 4, togethe with the chges 2,, nd e sketched in Figue P (The coloed lines e the intesections of the sufces with the pge.) Find the electic flux though ech sufce. S 1 δ In the i ove pticul egion t n ltitude of 500 m bove the gound the electic field is 120 N/C diected downwd. At 600 m bove the gound the electic field is 100 N/C downwd. Wht is the vege volume chge density in the lye of i between these two elevtions? Is it positive o negtive? R Figue P24.15 S 4 2 S A point chge 5.00 C is locted t the cente of cube of edge L m. In ddition, six othe identicl point chges hving 1.00 C e positioned symmeticlly ound s shown in Figue P Detemine the electic flux though one fce of the cube. S 2 Figue P24.11 L 12. () A point chge is locted distnce d fom n infinite plne. Detemine the electic flux though the plne due to the point chge. (b) Wht If? A point chge is locted vey smll distnce fom the cente of vey lge sue on the line pependicul to the sue nd going though its cente. Detemine the ppoximte electic flux though the sue due to the point chge. (c) xplin why the nswes to pts () nd (b) e identicl. 13. Clculte the totl electic flux though the pboloidl sufce due to unifom electic field of mgnitude 0 in the diection shown in Figue P L L Figue P24.17 Poblems 17 nd 18. d 18. A positive point chge is locted t the cente of cube of edge L. In ddition, six othe identicl negtive point chges e positioned symmeticlly ound s shown in Figue P Detemine the electic flux though one fce of the cube. 0 Figue P A point chge of 12.0 C is plced t the cente of spheicl shell of dius 22.0 cm. Wht is the totl electic flux though () the sufce of the shell nd (b) ny hemispheicl sufce of the shell? (c) Do the esults depend on the dius? xplin. 15. A point chge is locted just bove the cente of the flt fce of hemisphee of dius R s shown in Figue P Wht is the electic flux () though the cuved sufce nd (b) though the flt fce? λ d Figue P24.19 O R
19 Poblems An infinitely long line chge hving unifom chge pe unit length lies distnce d fom point O s shown in Figue P Detemine the totl electic flux though the sufce of sphee of dius R centeed t O esulting fom this line chge. Conside both cses, whee R d nd R d. 20. An unchged nonconducting hollow sphee of dius 10.0 cm suounds 10.0C chge locted t the oigin of ctesin coodinte system. A dill with dius of 1.00 mm is ligned long the z xis, nd hole is dilled in the sphee. Clculte the electic flux though the hole. 21. A chge of 170 C is t the cente of cube of edge 80.0 cm. () Find the totl flux though ech fce of the cube. (b) Find the flux though the whole sufce of the cube. (c) Wht If? Would you nswes to pts () o (b) chnge if the chge wee not t the cente? xplin. 22. The line g in Figue P24.22 is digonl of cube. A point chge is locted on the extension of line g, vey close to vetex of the cube. Detemine the electic flux though ech of the sides of the cube which meet t the point. Section 24.3 e d h Figue P24.22 Appliction of Guss s Lw to Vious Chge Distibutions 23. Detemine the mgnitude of the electic field t the sufce of led208 nucleus, which contins 82 potons nd 126 neutons. Assume the led nucleus hs volume 208 times tht of one poton, nd conside poton to be sphee of dius m. 24. A solid sphee of dius 40.0 cm hs totl positive chge of 26.0 C unifomly distibuted thoughout its volume. Clculte the mgnitude of the electic field () 0 cm, (b) 10.0 cm, (c) 40.0 cm, nd (d) 60.0 cm fom the cente of the sphee. 25. A 10.0g piece of Styofom cies net chge of C nd flots bove the cente of lge hoizontl sheet of plstic tht hs unifom chge density on its sufce. Wht is the chge pe unit e on the plstic sheet? 26. A cylindicl shell of dius 7.00 cm nd length 240 cm hs its chge unifomly distibuted on its cuved sufce. The mgnitude of the electic field t point 19.0 cm dilly outwd fom its xis (mesued fom the midpoint of the shell) is 36.0 kn/c. Find () the net chge on the shell nd (b) the electic field t point 4.00 cm fom the xis, mesued dilly outwd fom the midpoint of the shell. b f c g 27. A pticle with chge of 60.0 nc is plced t the cente of nonconducting spheicl shell of inne dius 20.0 cm nd oute dius 25.0 cm. The spheicl shell cies chge with unifom density of 1.33 C/m 3. A poton moves in cicul obit just outside the spheicl shell. Clculte the speed of the poton. 28. A nonconducting wll cies unifom chge density of 8.60 C/cm 2. Wht is the electic field 7.00 cm in font of the wll? Does you esult chnge s the distnce fom the wll is vied? 29. Conside long cylindicl chge distibution of dius R with unifom chge density. Find the electic field t distnce fom the xis whee R. 30. A solid plstic sphee of dius 10.0 cm hs chge with unifom density thoughout its volume. The electic field 5.00 cm fom the cente is 86.0 kn/c dilly inwd. Find the mgnitude of the electic field 15.0 cm fom the cente. 31. Conside thin spheicl shell of dius 14.0 cm with totl chge of 32.0 C distibuted unifomly on its sufce. Find the electic field () 10.0 cm nd (b) 20.0 cm fom the cente of the chge distibution. 32. In nucle fission, nucleus of unium238, which contins 92 potons, cn divide into two smlle sphees, ech hving 46 potons nd dius of m. Wht is the mgnitude of the epulsive electic foce pushing the two sphees pt? 33. Fill two ubbe blloons with i. Suspend both of them fom the sme point nd let them hng down on stings of eul length. Rub ech with wool o on you hi, so tht they hng pt with noticeble seption fom ech othe. Mke odeofmgnitude estimtes of () the foce on ech, (b) the chge on ech, (c) the field ech cetes t the cente of the othe, nd (d) the totl flux of electic field ceted by ech blloon. In you solution stte the untities you tke s dt nd the vlues you mesue o estimte fo them. 34. An insulting solid sphee of dius hs unifom volume chge density nd cies totl positive chge. A spheicl gussin sufce of dius, which shes common cente with the insulting sphee, is inflted stting fom 0. () Find n expession fo the electic flux pssing though the sufce of the gussin sphee s function of fo. (b) Find n expession fo the electic flux fo. (c) Plot the flux vesus. 35. A unifomly chged, stight filment 7.00 m in length hs totl positive chge of 2.00 C. An unchged cdbod cylinde 2.00 cm in length nd 10.0 cm in dius suounds the filment t its cente, with the filment s the xis of the cylinde. Using esonble ppoximtions, find () the electic field t the sufce of the cylinde nd (b) the totl electic flux though the cylinde. 36. An insulting sphee is 8.00 cm in dimete nd cies 5.70C chge unifomly distibuted thoughout its inteio volume. Clculte the chge enclosed by concentic spheicl sufce with dius () 2.00 cm nd (b) 6.00 cm. 37. A lge flt hoizontl sheet of chge hs chge pe unit e of 9.00 C/m 2. Find the electic field just bove the middle of the sheet.
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