Homework Assignment 5 Solution Set


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1 Homework Assignment 5 Solution Set PHYCS 44 3 Februry, 4 Problem Griffiths 3.8 The first imge chrge gurntees potentil of zero on the surfce. The secon imge chrge won t chnge the contribution to the potentil from the first, so we shoul just put chrge t the center of the sphere to mke the potentil some uniform constnt over the sphere s surfce. For potentil of V we shoul use chrge of q imge = 4πɛ V R where R is the rius of the sphere. The force between the rel chrge q n the conucting sphere will be the sme s the force felt by the rel chrge s ue to the two imge chrges. If the sphere is neutrl, then ll the imge chrges insie shoul to zero see the text for convincing rgument of this lmost obvious fct. Thus, F q r = q q imge 4πɛ r q imge r R ˆr r = q qr 4πɛ r 3 qr r ˆr r R r = q 3 R R r 4πɛ r R r ˆr where r is the istnce from the center of the sphere to the physicl chrge outsie. Since r > R the force of ttrction is F = q 3 R r R 4πɛ r r R Problem Griffiths optionl The chrge ensity etermines the fiel just outsie the strip. Thus, V σy, x = = ɛ n = ɛ V x
2 4V = ɛ x π = 4ɛ V π = 4ɛ V on on on nπx e sin nπy n nπ nπx e sin nπy n sin nπy x= x= Problem 3 Griffiths 3.7b Fortuntely we know the nswer before we strt, since uniform surfce chrge istribution is exctly wht we get from chrge conucting sphere. Outsie we shoul hve V r = σr ɛ r n insie we shoul get constnt V = σr ɛ. However, the book wnts us to use the exmple n solve it s bounry vlue problem. We know from Exmple 3.9 tht the generl solution for fixe, sphericl surfce chrge ensity is V r, θ insie = l V r, θ outsie = l A l r l P l cos θ B l r l+ P l cos θ with A l = R l B l = A l R l+. π σp l cos θ sin θθ In this cse, though, σ is constnt. Thus, the orthogonlity of the Legenre polynomils insures us tht A = σr ɛ n ll other A l vnish. Thus, B = σr ɛ n ll other B l vnish. Therefore, our potentil is V r, θ = σr ɛ ; r < R V r, θ = σr ɛ r ; r > R which is just s we suppose to begin with.
3 Problem 4 Griffiths 3.8 Agin, the generl solution is V r, θ = l= with the following bounry conitions: A l r l + B l r l+ P l cos θ V r V r = finite V R, θ outsie = V R, θ insie = k cos 3θ. 3 From conitions n we hve n V r, θ outsie = V r, θ insie = B l r l+ P l cos θ l= A l r l P l cos θ. l= From conition 3 n the results bove we see tht n V R, θ = A l R l = B l R l+ A l = B l R l+ B l R l+ R l P l cos θ = k cos 3θ. l= However, we cn write cos 3θ in terms of the Legenre polynomils s cos 3θ = 4 cos 3 θ 3 cos θ = 8 5 P 3cos θ 3 5 P cos θ. Therefore, ll coefficients of orers other thn n 3 must vnish n we hve B R P cos θ + B 3 R 4 P 3cos θ = 3k 5 P cos θ + 8k 5 P 3cos θ. from which we cn get ll of our coefficients becuse of the reltionship between the A l n B l we foun before. Therefore our potentil is V r, θ insie = 4k r3 R 3 cos3 θ k 5 V r, θ outsie = 4k R4 r 4 cos3 θ k 5 3 R 3 cos θ k 3 r 5 R cos θ R 4 r 4 cos θ k 3 R cos θ. 5 r r 3
4 To fin the surfce chrge ensity we just nee to fin the iscontinuity in the grient of the potentil. We know tht V out r + V in r = σ. r=r ɛ So, from eq in the book we get σθ = ɛ l + A l R l P l cos θ l = 9kɛ 5R cos θ + 8kɛ R cos3 θ 84ɛ 5R = kɛ 4 cos 3 θ 93 cos θ. 5R Problem 5 Griffiths 3.  optionl With n rbitrry chrge on the sphere the problem iffers very little from Exmple 3.8 in the text. There is just one more bounry conition. The integrl of the surfce chrge ensity over the totl re of the sphere shoul return Q. You my be ble to convince yourself tht this mounts to chnging the surfce chrge ensity in the exmple σ = 3ɛ E cos θ by constnt, since the istribution of the extr chrge shoul be the sme s the istribution in the neutrl cse. After ll, Q= is just prticulr cse of the generl solution. So, we cn only bounry conition tht ffects the l = term in the potentil outsie the sphere. The only such ition tht integrtes to give us Q is V = 4πɛ Q. Therefore, the new potentil outsie the sphere is V r, θ = E r R3 r cos θ + Q 4πɛ r. It s s though we e the chrge s point chrge t the center of the sphere! This shoul mke sense if you i the imge chrge problem correctly bove Griffiths 3.8. Problem 6 Griffiths 4.3 We coul try to o this one by solving Lplce s eqution in cylinricl coorintes using seprtion of vribles, but tht seems little out of the scope of wht you shoul be expecte to know t this point. However, we cn tke hint from Exmple 4.3 in the text n tret the uniformly polrize cyliner s two superimpose cyliners of equl n opposite uniform chrge istribution seprte by some vector whose size is smll. Ech cyliner gives us fiel insie of E s i in = ρ i s i n fiel outsie of E s i out = ρ i 4 s i ŝ i
5 where i enotes which of the two cyliners we re consiering. Let the origin be irectly between the centers of the two overlpping cyliners. Thus, the totl fiel insie is Outsie the sphere we get E s = ρ s ρ s = ρ s s = ρ. E s = ρ ŝ ρ ŝ s s ŝ = ρ ŝ s s We like to put this in terms of the vector s from the origin. We hve n Thus, E s = ρ s = ρ s = s + s = s. s s s s s s + s +. Here we nee some trickery I got stuck little myself. We my write n similrly s + = s + + s 4 s + s s s s s s s + s s 5
6 Here we hve ignore nonliner terms of becuse it is so smll, n we hve lso expne binomil expnsion in s s n truncte the expnsion for the sme reson. Finlly, combining everything together we hve E s = ρ s s s s s + s + s = ρ s s s s + = s ŝŝ P P becuse s I chose it, for whtever reson points from the positively chrge xis to the negtively chrge xis. Problem 7 Griffiths 4.6b & c b The ens of such long cvity will not contribute very much to the fiel insie. Therefore, the fiel insie the neeleshpe cvity will be ominte by the bounry conitions long the long bounry, which is prllel to P. The electric fiel t this bounry is continuous becuse it is prllel to the surfce, so the fiel in the cvity is just E. The isplcement fiel, however, must be ifferent. It becomes D = ɛ E. c In this cse, the ens ominte n the wlls prllel to E re insignificnt. Therefore, E is most certinly not continuous, but D is. Thus, D = D n E = ɛ D. ALTERNATE METHOD Following the hint in the problem, tret the cvities s superpositions of oppositely polrize ielectrics. Then, the neele like cvity resembles single ipole, cncelling the isplcement but leving E. The isclike cvity resembles cpcitor with chrge ensity given by the polriztion. In tht cse D is the sme, but E is not. You get the sme results either wy you look t it. Problem 8 Griffiths 4.8, & e The free chrge ensity on the pltes is fixe, n there re no free chrges in the ielectrics. Thus, D is uniform throughout both ielectrics n is given by D A = Q free enclose DA = σ free A D = σâ 6
7 where â is the vector pointing from the positively chrge plte to the negtively chrge plte ẑ. We know how V is relte to E, n we know how E is relte to D. So, we cn just use the result of prt to show tht V = = = 7 6ɛ σ D z ɛ σz + ɛ D z.5ɛ σz e In slb we hve D = ɛ E + P P = σẑ. Therefore, insie slb we hve ρ b = P = n t the lower n upper interfces respectively we get σ b = P ẑ = ± σ. In slb we hve D = ɛ E + P P = 3 σẑ. So, ρ b = P = n t the lower n upper interfces respectively we get σ b = P ẑ = ± 3 σ. Problem 9 Griffiths 4. Plce n rbitrry liner chrge ensity λ on the inner wire n clculte the potentil ifference between the wire n the outer shell. This free chrge gives rise to uniform D everywhere between the two conuctors s seen in numerous exmples: D r = λ πr ˆr. 7
8 The potentil ifference between the conuctors is then V = b = λ πɛ λ πɛ r r + b ln So, the cpcitnce per unit length is c b + ɛ r ln λ πɛ ɛ r r r c. b C l = λ V λ = πɛ ln b + ɛ r ln c b 8
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