2.2 This is the Nearest One Head (Optional) Experimental Verification of Gauss s Law and Coulomb s Law

Transcription

1 2.2 This is the Neest One Hed 743 P U Z Z L R Some ilwy compnies e plnning to cot the windows of thei commute tins with vey thin lye of metl. (The coting is so thin you cn see though it.) They e doing this in esponse to ide complints bout othe pssenges tlking loudly on cellul telephones. How cn metllic coting tht is only few hunded nnometes thick ovecome this poblem? (Athu Tilley/FPG Intentionl) c h p t e Guss s Lw Chpte Outline 24.1 lectic Flux 24.2 Guss s Lw 24.3 Appliction of Guss s Lw to Chged Insultos 24.4 Conductos in lectosttic uilibium 24.5 (Optionl) xpeimentl Veifiction of Guss s Lw nd Coulomb s Lw 24.6 (Optionl) Foml Deivtion of Guss s Lw 743

2 744 CHAPTR 24 Guss s Lw Ae = A In the peceding chpte we showed how to use Coulomb s lw 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 invese-sue 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 we e deling with complicted poblems. Figue 24.1 Field lines epesenting unifom electic field 11.6 penetting plne of e A pependicul to the field. The electic flux though this e is eul to A LCTRIC FLUX The concept of electic field lines is descibed ulittively in Chpte 23. We now use the concept of electic flux to 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, which is 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): A (24.1) Fom the SI units of nd A, we see tht hs units of newton metes sued pe coulomb (Nm 2 /C). lectic flux is popotionl to the numbe of electic field lines penetting some sufce. XAMPL 24.1 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 given by ution 23.4, 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 xecise A ( N/C)(12.6 m 2 ) Nm 2 /C Wht would be the () electic field nd (b) flux though the sphee if it hd dius of m? Answe () N/C; (b) Nm 2 /C. 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, in which 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 ligned pependicul to the field. Fom Figue 24.2 we see tht the two es e elted by A Acos. Becuse the flux though A euls the flux though A, we

3 24.1 lectic Flux 745 A = A cos θ θ A Noml 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 A cos. QuickLb Shine desk lmp onto plying cd nd notice how the size of the shdow on you desk depends on the oienttion of the cd with espect to the bem of light. Could fomul like ution 24.2 be used to descibe how much light ws being blocked by the cd? conclude tht the flux though A is A A cos (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 (in othe wods, when the noml to the sufce is pllel to the field, tht is, in Figue 24.2); the flux is zeo when the sufce is pllel to the field (in othe wods, when the noml to the sufce is pependicul to the field, tht is, 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 flux though this element is whee we hve used the definition of the scl poduct of two vectos (A B AB cos ). By summing the contibutions of ll elements, we obtin the totl flux though the sufce. 1 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 i A i cos i A i lim i A i A i :0 sufce da 0 90). (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. 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 Figue 24.3 A i θ Definition of electic flux i A smll element of sufce e A i. The electic field mkes n ngle 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. 1 It is impotnt to note tht dwings with field lines hve thei inccucies becuse smll e element (depending on its loction) my hppen to hve too mny o too few field lines penetting it. We stess tht the bsic definition of electic flux is da. The use of lines is only n id fo visulizing the concept.

4 746 CHAPTR 24 Guss s Lw A i θ A i A i θ 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 ). Kl Fiedich Guss Gemn mthemticin nd stonome ( ) 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 hence, the flux Ai though this element is positive. Fo element, the field lines gze the sufce (pependicul to the vecto A i ); thus, nd the flux is zeo. Fo elements such s, whee the field lines e cossing the sufce fom outside to inside, 180 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 90 da n da 90; 90 (24.4) whee n epesents the component of the electic field noml to the sufce. vluting the net flux though closed sufce cn be vey cumbesome. Howeve, if the field is noml to the sufce t ech point nd constnt in mgnitude, the clcultion is stightfowd, s it ws in xmple The next exmple lso illusttes this point. XAMPL 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 edges, 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 da on these fces. The net flux though fces nd is 1 da da 2

5 24.2 Guss s Lw 747 z Figue 24.5 da 1 y da 3 da 4 da 2 A closed sufce in the shpe of cube in unifom electic field oiented pllel to the x xis. The net flux though the closed sufce is zeo. Side is the bottom of the cube, nd side is opposite side. x Fo, is constnt nd diected inwd but da 1 is diected outwd thus, the flux though this fce is becuse the e of ech fce is A 2. Fo, is constnt nd outwd nd in the sme diection s da 2 ( 0 ); hence, the flux though this fce is 2 ( 180); 1 da 1 (cos 180)dA 1 da A 2 da (cos 0)dA da A Theefoe, the net flux ove ll six fces is GAUSS S LAW 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 23.4 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 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 da da da Figue 24.6 Gussin sufce dai 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. 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 k e 2 (4 2 ) 4k e Reclling fom Section 23.3 tht k e 1/(40), 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 2 /Nm 2 ) Nm 2 ( C) /C. 0

6 748 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. Figue 24.8 A point chge locted outside closed sufce. The numbe of lines enteing the sufce euls the numbe leving the sufce. The net electic flux though closed sufce is zeo if thee is no chge inside 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 pevious 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 is independent of the shpe of tht sufce. The net flux though ny closed sufce suounding point chge is given by / 0. 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. Quick Quiz 24.1 Suppose tht the chge in xmple 24.1 is just outside the sphee, 1.01 m fom its cente. Wht is the totl flux though the sphee? 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 da ( 1 2 ) da 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.

7 24.2 Guss s Lw 749 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 nd 3 outside it is zeo becuse ech electic field line tht entes S t one point leves it t 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. Guss s lw, which is geneliztion of wht we hve just descibed, sttes tht the net flux though ny closed sufce is da 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. As we shll see in the next section, Guss s lw cn be used 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. You should lso note tht gussin sufce is mthemticl constuction nd need not coincide with ny el physicl sufce. Quick Quiz 24.2 Fo gussin sufce though which the net flux 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. Guss s lw is useful fo evluting when the chge distibution hs high symmety S 1 2 S Figue S 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. CONCPTUAL XAMPL 24.3 A spheicl gussin sufce suounds point chge. Descibe wht hppens to the totl flux though the sufce if () 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 () 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.

8 750 CHAPTR 24 Guss s Lw 24.3 APPLICATION OF GAUSS S LAW TO CHARGD INSULATORS 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: 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 da becuse nd da e pllel. 3. The dot poduct in ution 24.6 is zeo becuse nd da 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. XAMPL 24.4 The lectic Field Due to Point Chge Stting with Guss s lw, clculte the electic field due to n isolted point chge. whee we hve used the fct tht the sufce e of sphee is 4 2. Now, we solve fo the electic field: 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. 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 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 da nd Guss s lw gives da da By symmety, is constnt eveywhee on the sufce, which stisfies condition (1), so it cn be emoved fom the integl. Theefoe, da da (4 2 ) 0 0 Figue This is the fmili electic field due to point chge tht we developed fom Coulomb s lw in Chpte 23. k e 2 Gussin sufce 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 11.6 XAMPL 24.5 A Spheiclly Symmetic Chge Distibution An insulting solid sphee of dius hs unifom volume Solution Becuse the chge distibution is spheiclly chge density nd cies totl positive chge Q (Fig. symmetic, we gin select spheicl gussin sufce of dius 24.11). () Clculte the mgnitude of the electic field t, concentic with the sphee, s shown in Figue point outside the sphee. Fo this choice, conditions (1) nd (2) e stisfied, s they

9 24.3 Appliction of Guss s Lw to Chged Insultos 751 wee fo the point chge in xmple Following the line of esoning given in xmple 24.4, we find tht (fo ) 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 k e Q 2 In this cse we select spheicl gussin sufce hving dius, concentic with the insulted 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 Q. To clculte in, we use the fct tht in V : in V ( ) 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 Becuse by definition nd since k e 1/(40), this expession fo cn be witten s da da (4 2 ) in Q / in 40 2 Q 40 3 (fo ) Note tht this esult fo diffes fom the one we obtined in pt (). 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. Note lso tht the expessions fo pts () nd (b) mtch when. A plot of vesus is shown in Figue k e Q Figue () Gussin sphee (b) Gussin sphee A unifomly chged insulting sphee of dius nd totl chge Q. () The mgnitude of the electic field t point exteio to the sphee is k e Q / 2. (b) The mgnitude of the electic field inside the insulting sphee is due only to the chge within the gussin sphee defined by the dshed cicle nd is k e Q / 3. Figue = k e Q 2 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 Q locted t 0. XAMPL 24.6 The lectic Field Due to Thin Spheicl Shell A thin spheicl shell of dius hs totl chge Q distibuted unifomly ove its sufce (Fig ). Find the electic field t points () outside nd (b) inside the shell. the shell is euivlent to tht due to point chge Q locted t the cente: Solution () 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 Q. Theefoe, the field t point outside k e Q 2 (fo ) (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

10 752 CHAPTR 24 Guss s Lw 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 23.6 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 sufce Gussin sufce in = 0 Figue () () The electic field inside unifomly chged spheicl shell is zeo. The field outside is the sme s tht due to point chge Q locted t the cente of the shell. (b) Gussin sufce fo. (c) Gussin sufce fo. (b) (c) XAMPL 24.7 A Cylindiclly Symmetic Chge Distibution 11.7 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 Gussin sufce () da da da A in 0 0 Figue () 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. (b)

11 24.3 Appliction of Guss s Lw to Chged Insultos 753 The e of the cuved sufce is A 2; theefoe, (2) 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 in Chpte 23 (see Poblem 35[b]), by integtion of the field of point chge. 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 0 2k e 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. When thee is insufficient symmety in the chge distibution, s in this sitution, it is necessy to use ution 23.6 to clculte. 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. XAMPL 24.8 A Nonconducting Plne of Chge Find the electic field due to nonconducting, 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 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 2A in 20 0 A 0 (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. An impotnt chge configution elted to this exmple consists of two pllel plnes, one positively chged nd the othe negtively chged, nd ech with sufce chge density (see Poblem 58). In this sitution, the electic fields due to the two plnes dd in the egion between the plnes, esulting in field of mgnitude / 0, nd cncel elsewhee to give field of zeo. Figue Gussin cylinde 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 CONCPTUAL XAMPL 24.9 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.

12 754 CHAPTR 24 Guss s Lw 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. 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. Figue A conducto of bity shpe. The boken line epesents gussin sufce just inside the conducto. Gussin sufce 24.4 CONDUCTORS IN LCTROSTATIC QUILIBRIUM 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. As we shll see, conducto in electosttic euilibium hs the following popeties: 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 without futhe discussion so tht we hve complete list of popeties fo conductos in electosttic euilibium. We cn undestnd the fist popety by consideing conducting slb plced in n extenl field (Fig ). We cn gue tht the electic field inside the conducto must be zeo unde the ssumption tht we hve electosttic euilibium. If the field wee not zeo, fee chges in the conducto would ccelete unde the ction of the field. 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 density inceses until the mgnitude of the intenl field euls tht of the extenl field, nd the net esult is net field of zeo inside the conducto. The time it tkes good conducto to ech euilibium is of the ode of s, which fo most puposes cn be consideed instntneous. 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 su-

13 24.4 Conductos in lectosttic uilibium 755 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). fce 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. We cn lso use Guss s lw to veify the thid popety. 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 noml to the conducto s sufce fom the condition of electosttic euilibium. (If hd component pllel to the conducto s sufce, the fee chges would move long the sufce; in such cse, the conducto would not be in 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 stisfction of 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 whee we hve used the fct tht in A. Solving fo gives 0 0 A 0 (24.9) n A Figue lectic field just outside chged conducto 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 n A. Remembe tht is zeo inside the conducto. XAMPL A Sphee Inside Spheicl Shell A solid conducting sphee of dius cies net positive chge 2Q. A conducting spheicl shell of inne dius b nd oute dius c is concentic with the solid sphee nd cies net chge Q. 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

14 756 CHAPTR 24 Guss s Lw Q 2Q b c 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 2 A 2 (4 2 ) in 2 2Q k e Q 2 2Q 0 (fo b) Figue A solid conducting sphee of dius nd cying chge 2Q suounded by conducting spheicl shell cying chge Q. 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 2Q (the chge on the solid sphee). Becuse of the spheicl symmety, the electic field In egion, whee c, the spheicl gussin sufce we constuct suounds totl chge of in 2Q (Q) Q. Theefoe, ppliction of Guss s lw to this sufce gives 4 k e Q 2 (fo c) In egion, the electic field must be zeo becuse the spheicl shell is lso conducto in euilibium. 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 2Q to cncel the chge 2Q on the solid sphee. Becuse the net chge on the shell is Q, we conclude tht its oute sufce must cy chge Q. Quick Quiz 24.3 How would the electic flux though gussin sufce suounding the shell in xmple chnge if the solid sphee wee off-cente but still inside the shell? Optionl Section 24.5 XPRIMNTAL VRIFICATION OF GAUSS S LAW AND COULOMB S LAW When net chge is plced on conducto, the chge distibutes itself on the sufce in such wy tht the electic field inside the conducto is zeo. Guss s lw shows tht thee cn be no net chge inside the conducto in this sitution. In this section, we investigte n expeimentl veifiction of the bsence of this chge. We hve seen tht Guss s lw is euivlent to ution 23.6, the expession fo the electic field of distibution of chge. Becuse this eution ises fom Coulomb s lw, we cn clim theoeticlly tht Guss s lw nd Coulomb s lw e euivlent. Hence, it is possible to test the vlidity of both lws by ttempting to detect net chge inside conducto o, euivlently, nonzeo electic field inside the conducto. If nonzeo field is detected within the conducto, Guss s lw nd Coulomb s lw e invlid. Mny expeiments, including

15 24.5 xpeimentl Veifiction of Guss s Lw nd Coulomb s Lw 757 ely wok by Fdy, Cvendish, nd Mxwell, hve been pefomed to detect the field inside conducto. In ll epoted cses, no electic field could be detected inside conducto. Hee is one of the expeiments tht cn be pefomed. 2 A positively chged metl bll t the end of silk thed is loweed though smll opening into n unchged hollow conducto tht is insulted fom gound (Fig ). The positively chged bll induces negtive chge on the inne wll of the hollow conducto, leving n eul positive chge on the oute wll (Fig b). The pesence of positive chge on the oute wll is indicted by the deflection of the needle of n electomete ( device used to mesue chge nd tht mesues chge only on the oute sufce of the conducto). The bll is then loweed nd llowed to touch the inne sufce of the hollow conducto (Fig c). Chge is tnsfeed between the bll nd the inne sufce so tht neithe is chged fte contct is mde. The needle deflection emins unchnged while this hppens, indicting tht the chge on the oute sufce is unffected. When the bll is emoved, the electomete eding emins the sme (Fig d). Futhemoe, the bll is found to be unchged; this veifies tht chge ws tnsfeed between the bll nd the inne sufce of the hollow conducto. The ovell effect is tht the chge tht ws oiginlly on the bll now ppes on the hollow conducto. The fct tht the deflection of the needle on the electomete mesuing the chge on the oute sufce emined unchnged egdless of wht ws hppening inside the hollow conducto indictes tht the net chge on the system lwys esided on the oute sufce of the conducto. If we now pply nothe positive chge to the metl bll nd plce it ne the outside of the conducto, it is epelled by the conducto. This demonsttes tht 0 outside the conducto, finding consistent with the fct tht the conducto cies net chge. If the chged metl bll is now loweed into the inteio of the chged hollow conducto, it exhibits no evidence of n electic foce. This shows tht 0 inside the hollow conducto. This expeiment veifies the pedictions of Guss s lw nd theefoe veifies Coulomb s lw. The euivlence of Guss s lw nd Coulomb s lw is due to the invese-sue behvio of the electic foce. Thus, we cn intepet this expeiment s veifying the exponent of 2 in the 1/ 2 behvio of the electic foce. xpeiments by Willims, Flle, nd Hill in 1971 showed tht the exponent of in Coulomb s lw is (2 ), whee In the expeiment we hve descibed, the chged bll hnging in the hollow conducto would show no deflection even in the cse in which n extenl electic field is pplied to the entie system. The field inside the conducto is still zeo. This bility of conductos to block extenl electic fields is utilized in mny plces, fom electomgnetic shielding fo compute components to thin metl cotings on the glss in ipot contol towes to keep d oiginting outside the towe fom disupting the electonics inside. Cellul telephone uses iding tins like the one pictued t the beginning of the chpte hve to spek loudly to be hed bove the noise of the tin. In esponse to complints fom othe pssenges, the tin compnies e consideing coting the windows with thin metllic conducto. This coting, combined with the metl fme of the tin c, blocks cellul telephone tnsmissions into nd out of the tin. ( ) 10 16! Hollow conducto () (b) (c) (d) Figue An expeiment showing tht ny chge tnsfeed to conducto esides on its sufce in electosttic euilibium. The hollow conducto is insulted fom gound, nd the smll metl bll is suppoted by n insulting thed. QuickLb Wp dio o codless telephone in luminum foil nd see if it still woks. Does it mtte if the foil touches the ntenn? The expeiment is often efeed to s Fdy s ice-pil expeiment becuse Fdy, the fist to pefom it, used n ice pil fo the hollow conducto.

16 758 CHAPTR 24 Guss s Lw Figue Ω θ A A closed sufce of bity shpe suounds point chge. The net electic flux though the sufce is independent of the shpe of the sufce. Optionl Section 24.6 FORMAL DRIVATION OF GAUSS S LAW 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 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 A cos 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 24.22, 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 A θ θ Ω A cos θ A Figue The e element A subtends solid ngle (A cos )/ 2 t the chge.

17 Summy 759 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) In genel, the electic flux though sufce is da (24.3) sufce You need to 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 inside the sufce divided by 0 : da 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. 0 TABL 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 Q Thin spheicl shell of dius R nd totl chge Q Line chge of infinite length nd chge pe unit length Nonconducting, infinite chged plne hving sufce chge density Conducto hving sufce chge density k Q e 2 k Q e R 3 k e Q 2 2k e R R R R Outside the line veywhee outside the plne Just outside the conducto Inside the conducto

18 760 CHAPTR 24 Guss s Lw 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. Poblem-Solving Hints Guss s lw, s we hve seen, is vey poweful in solving poblems involving highly symmetic chge distibutions. In this chpte, you encounteed thee kinds of symmety: pln, cylindicl, nd spheicl. It is impotnt to eview xmples 24.4 though nd to dhee to the following pocedue when using Guss s lw: Select gussin sufce tht hs symmety to mtch tht of the chge distibution nd stisfies one o moe of the conditions listed in Section Fo point chges o spheiclly symmetic chge distibutions, the gussin sufce should be sphee centeed on the chge s in xmples 24.4, 24.5, 24.6, nd Fo unifom line chges o unifomly chged cylindes, you gussin sufce should be cylindicl sufce tht is coxil with the line chge o cylinde s in xmple Fo plnes of chge, useful choice is cylindicl gussin sufce tht stddles the plne, s shown in xmple These choices enble you to simplify the sufce integl tht ppes in Guss s lw nd epesents the totl electic flux though tht sufce. vlute the in / 0 tem in Guss s lw, which mounts to clculting the totl electic chge in inside the gussin sufce. If the chge density is unifom (tht is, if,, o is constnt), simply multiply tht chge density by the length, e, o volume enclosed by the gussin sufce. If the chge distibution is nonunifom, integte the chge density ove the egion enclosed by the gussin sufce. Fo exmple, if the chge is distibuted long line, integte the expession d whee d is the chge on n infinitesiml length element dx. Fo plne of chge, integte d da, whee da is n infinitesiml element of e. Fo volume of chge, integte d whee dv is n infinitesiml element of volume. Once the tems in Guss s lw hve been evluted, solve fo the electic field on the gussin sufce if the chge distibution is given in the poblem. Convesely, if the electic field is known, clculte the chge distibution tht poduces the field. dv, dx, QUSTIONS 1. The Sun is lowe in the sky duing the winte 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 no electic chges e in tht egion? xplin. 3. If moe electic field lines e leving gussin sufce thn enteing, 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?

19 Poblems 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. (Hint: Chnge the size of the gussin sufce.) 9. On the bsis of the epulsive ntue of the foce between like chges nd the feedom of motion of chge within the 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. How would the obsevtions descibed in Figue diffe if the hollow conducto wee gounded? How would they diffe if the smll chged bll wee n insulto the thn conducto? 12. Wht othe expeiment might be pefomed on the bll in Figue to show tht its chge ws tnsfeed to the hollow conducto? 13. Wht would hppen to the electomete eding if the chged bll in Figue touched the inne wll of the conducto? the oute wll? 14. You my hve hed tht one of the sfe plces to be duing lightning stom is inside c. Why would this be the cse? 15. Two solid sphees, both of dius R, cy identicl totl chges Q. 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, how do the electic fields outside these two sphees compe? Ae the fields identicl inside the two sphees? PROBLMS 1, 2, 3 = stightfowd, intemedite, chllenging = full solution vilble in the Student Solutions Mnul nd Study Guide WB = solution posted t = Compute useful in solving poblem = Intective Physics = pied numeicl/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 if () the plne is pllel to the yz plne; (b) the plne is pllel to the xy plne; nd (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 ppoximtely 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 40.0-cm-dimete 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 N m 2 /C. Wht is the mgnitude of the electic field? 4. A spheicl shell is plced in unifom electic field. Find the totl electic flux though the shell. 5. Conside closed tingul box esting within hoizontl electic field of mgnitude N/C, s shown in Figue P24.5. Clculte the electic flux though () the veticl ectngul sufce, (b) the slnted sufce, nd (c) the entie sufce of the box cm 30.0 cm A unifom electic field i bj 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? 7. A point chge is locted t the cente of unifom ing hving line chge density nd dius, s shown in Figue P24.7. Detemine the totl electic flux λ Figue P24.5 R Figue P24.7

20 762 CHAPTR 24 Guss s Lw though sphee centeed t the point chge nd hving dius R, whee R. 8. A pymid with 6.00-m-sue bse 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. 9. 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 P24.9. Detemine the electic flux tht entes the left-hnd side of the cone. 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. 14. Clculte the totl electic flux though the pboloidl sufce due to constnt electic field of mgnitude 0 in the diection shown in Figue P Section 24.2 Guss s Lw 10. The electic field eveywhee on the sufce of thin spheicl shell of dius m is mesued to be eul to 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. The following chges e locted inside submine: 5.00 C, 9.00 C, 27.0 C, nd 84.0 C. () Clculte the net electic flux though the submine. (b) Is the numbe of electic field lines leving the submine gete thn, eul to, o less thn the numbe enteing it? 12. Fou closed sufces, S 1 though S 4, togethe with the chges 2Q, Q, nd Q e sketched in Figue P Find the electic flux though ech sufce. R h Figue P24.9 WB A point chge Q 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? δ Figue P d Q R Figue P24.15 S 4 2Q Figue P () A point chge is locted distnce d fom n infinite plne. Detemine the electic flux though the plne due to the point chge. (b) A point chge is Q Q S 2 S 1 S 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. 17. A point chge of C is inside pymid. Detemine the totl electic flux though the sufce of the pymid. 18. 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. (Hint: Conside both cses: when R d, nd when R d.)