Msschusetts Institute of Technology Deprtment of Physics Physics 8.07 Fll 2005 Problem Set 3 Solutions Problem 1: Cylindricl Cpcitor Griffiths Problems 2.39: Let the totl chrge per unit length on the inner conductor be λ. By symmetry, the electric field points in the direction e s where s is the distnce from the centrl xis. To determine the electric field we use Guss s lw with cylinder of rdius s nd length L contining (for < s < b chrge λl, giving E s = λl 2πɛ 0 Ls = λ 2πɛ 0 s. The potentil difference between the two cylinders is thus V = b E s ds = (λ/2πɛ 0 ln(b/, giving cpcitnce per unit length dc dl = λ V = 2πɛ 0 ln(b/. b The electric energy density is 1ɛ 2 0E 2 nd the volume element in cylindricl coordintes is 2πLs ds, giving n energy per unit length du dl = ɛ b ( 2 ( 0 λ 2πs ds = λ2 b ln. 2 2πɛ 0 s 4πɛ 0 Problem 2: Men Vlue Theorem Griffiths Problems 3.1: The derivtion prllels tht in Section 3.1.4 of Griffiths. With µ cos θ, V ve = = = = 1 q(2πr 2 (z 2 + R 2 2zRµ 1/2 dµ (4πɛ 0 (4πR 2 1 q [ 1 z2 + R 8πɛ 0 zr 2 2zRµ] 1 q ( z R z + R 8πɛ 0 zr q 1/z, z > R (chrge outside sphere 4πɛ 0 1/R, z < R (chrge inside sphere. 1
As on p. 114 of Griffiths, chrges outside the sphere, distnce z from the center, contribute to V ve exctly their potentil V (0 = q/(4πɛ 0 z V center ; if there is more thn one chrge the potentils dd linerly. Chrges inside the sphere contribute totl mount Q enc /(4πɛ 0 R. Thus, V ve = V center + Q enc 4πɛ 0 R where V center is the potentil t the center due to ll the externl chrges, nd Q enc is the totl enclosed chrge. b Griffiths Problems 3.2: The men vlue theorem sttes tht wy from chrges, the potentil hs no extrem except on boundries, nd therefore cn provide no stble equilibrium points (which would require the potentil to hve locl minumum or mximum. For the configurtion of equl chrges t the corners of cube, clerly i V = 0 t the center of the cube (i.e. there is no net electric field, the forces blnce. Moreover, clcultion (which you need not present shows tht i j V = 0, i.e. the center of the cube is sddle point. However, this is not good enough. Chrges cn lek out through the middle of ny of the 6 fces. Problem 3: Method of Imges From Exmple 3.2 of Griffiths, the sphere r = R will be t potentil V = 0 if we dd n pproprite imge chrge q t distnce b long the sme line s the chrge q nd the center of the sphere (tken to be the z-xis. The potentil inside the sphere is thus V ( x = 1 4πɛ 0 ( q x e z + q x b e z for r < R, b R2, q R q, where b nd q re chosen so tht V = 0 for x 2 + y 2 + z 2 = R 2. The formul given bove does not provide the potentil outside the sphere, becuse there is no ctul chrge t r = b. Insted, we get the externl potentil from solving the boundry vlue problem 2 V = 0 for r > R with boundry condition V = 0 t r = R. One solution nd, from the uniqueness theorem, the only solution is V = 0 for r > R. b In sphericl coordintes, for r R we hve [ ] V (r,θ = q 1 4πɛ 0 r2 + 2 2r cos θ 1 R2 + (r/r 2 2r cos θ. 2
Now, becuse E = 0 inside the conductor t r = R, the surfce chrge density t r = R is ( V σ = ɛ 0 [E r ] = ɛ 0 E r (r = R = ɛ 0. r r=r This gives ( q (1 ã 2 σ(θ = 4πR 2 (1 2ã cos θ + ã 2, ã 3/2 R. c Integrting the induced surfce chrge over the sphere gives Q ind = σ R 2 dω = q 1 2 (1 dx ã2 = q. (1 2ãx + ã 3/2 As expected, the induced chrge on the inner surfce exctly cncels the chrge in the interior of the sphere. For grounded sphere there is no chrge on the outer surfce. The force on the test chrge is obtined most simply from the imge chrge: F = qq 4πɛ 0 (b 2 = 1 q 2 R 4πɛ 0 (R 2 2 2 nd it is ttrctive (rdilly outwrd on the chrge q. In the limit R 0, F q 2 16πɛ 0 (R 2. This is the sme s the force on test chrge the distnce R from grounded infinite plne. d To mke the totl chrge on the conductor zero, we simply dd chrge q to the sphere. It ll goes to the outer surfce, dding q/(4πɛ 0 R to the potentil for r < R nd q/(4πɛ 0 r for r > R. Thus, for r R, V (r,θ = q 4πɛ 0 nd for r > R, [ 1 R + 1 r2 + 2 2r cos θ 1 R2 + (r/r 2 2r cos θ V (r,θ = 1 4πɛ 0 r. The induced chrge distribution on the inner surfce is unchnged, nd the chrge dded to the outer surfce produces no force inside the sphere. Thus the force on the test chrge is unchnged from prt c. ] 3
Problem 4: Seprtion of Vribles Using seprtion of vribles, substituting V (x, y, z = X(xY (yz(z into 2 V = 0 gives 1 d 2 X X dx = 2 1, 1 Y d 2 Y dy 2 = 2, 1 d 2 Z Z dz = 2 3, 1 + 2 + 3 = 0. Imposing the boundry conditions V = 0 for x = 0 nd x = requires ( mπx X(x = sin where m = 1, 2, 3,... Imposing the boundry conditions V = 0 for y = 0 nd y = b requires Y (y = sin where n = 1, 2, 3,... b Therefore 3 = ( 1 + 2 = κ 2 mn ( mπ 2 ( nπ + b 2 > 0. The solutions of d 2 Z/dz 2 = κ 2 mnz re exp(±κ mn z, giving ( mπx V mn (1 = sin sin exp(κ mn z b nd V (2 mn = sin ( mπx sin exp( κ mn z, b where m nd n re nturl numbers 1, 2, 3,... (One my use ny independent liner combintions of the exponentils. b From prt, it follows tht inside the box we my write V (x,y,z = m=1 n=1 [ C (1 mnv (1 mn + C (2 mnv (2 mn]. Imposing boundry conditions tht E z = V/ z hs the sme vlue t z = 0 nd z = selects out one prticulr liner combintion of exp(±k mn z, V (x,y,z = m=1 n=1 ( mπx C mn sin sin Now requiring V/ z = E 0 t z = 0 nd z = c gives E 0 = m=1 n=1 ( mπx C mn sin sin 4 b sinh κ mn (z c/2. κ mn cosh( 1 2 b κ mnc.
We cn solve for the C mn using the orthonormlity of Fourier series, 2 ( mπx ( m πx sin sin dx = δ mm. 0 Simply multiply the expression for E 0 given bove by sin(mπx/ sin(nπy/b nd integrte over x nd y to get ( ( 2 2 C mn κ mn cosh( 1κ 2 mnc = E 0 [1 cos(mπ][1 cos(nπ] mπ nπ or C mn κ mn cosh( 1 2 κ mnc = where κ 2 mn = (mπ/ 2 + (nπ/b 2. (16E 0 /(mnπ 2, m nd n both odd 0, otherwise, c Evluting E z in the middle of cube with = b = c nd writing m = 2k+1 nd n = 2l + 1, E z (/2,/2,/2 = k=0 = 16E 0 π 2 C mn κ mn ( 1 k+l l=0 k=0 ( 1 k+l mn cosh( π m2 + n 2 2. I don t know how to evlute this sum nlyticlly. Numericlly, it evlutes to E z /E 0 = 1 to mchine precision, s determined using the following C 3 progrm. /* Sum the series in Problem 4c of 8.07 Problem Set 3. Revised by EB September 27, 2005. */ #include <mth.h> #include <stdio.h> #define pi 3.1415926535898 int min(void l=0 double x, cosh, prtilsum, sum, tolernce = 0.5e-13; int k, l, kplusl, sign, m, n; 5
sum = 0.0; sign = 1; kplusl = 0; /* Sum over k nd l by zig-zgging long digonls of constnt k+l, becuse the contributions to the sum decrese stedily with incresing k+l. */ do prtilsum = 0.0; for (k=0; k <= kplusl; k++ l = kplusl-k; m = 2*k+1; n = 2*l+1; x = 0.5*pi*sqrt((double(m*m+n*n; cosh = 0.5*(exp(x+exp(-x; prtilsum += 1.0/(m*n*cosh; } sum += 16.*sign*prtilsum/(pi*pi; printf("(k+l_mx, E_z/E_0 = %d %15.13g\n",kplusl, sum; kplusl++; sign *= -1; } while (prtilsum > tolernce; exit(0; } Problem 5: Legendre series First we rewrite the potentil on the sphere in terms of Legendre functions using P 0 = 1, P 1 = cos θ, P 2 = 1 2 (3 cos2 θ 1 nd cos 2θ = 2 cos 2 θ 1 = 4 3 P 2 1 3 P 0: V (R,θ = (V 0 1 3 V 2P 0 + V 1 P 1 + 4 3 V 2P 2. We know tht with xisymmetry, in sphericl coordintes the generl solution of 2 V = 0 is V (r,θ = (A l r l + B l r (l+1 P l (cos θ l=0 where A l nd B l re constnts. Becuse there re no chrges inside the sphere B l = 0 for r < R nd becuse there re no chrges t infinity A l = 0 for r > R. Therefore we conclude (V 0 1 3 V (r,θ = V r 2 + V 1 cos θ + 4V R 3 2 r2 P R 2 2 (cos θ, r < R (V 0 1V 3 2 R + V r 1 R2 cos θ + 4V r 2 3 2 R3 P r 3 2 (cos θ, r > R. 6
b The electric field components re V 1 R cos θ 8V 3 2 r nd E r = V r = E θ = 1 r P R 2 2 (cosθ, r < R (V 0 1V 3 2 R R + 2V 2 R r 2 1 cos θ + 4V 3 r 3 2 P r 4 2 (cosθ, r > R V θ = ( V 1 r R + 4V 2 cos θ sin θ, r < R R 2 ( V 1R 2 R + 4V 3 r 3 2 cos θ sin θ, r > R. r 4 The tngentil jump condition sttes tht E θ is continuous cross r = R. This follows t once from the result given here. c The norml jump condition is [E r ] = σ/ɛ 0 which gives σ(θ = ɛ 0 [E r (r = R + E r (r = R ] = (V 0 1V 3 2 ɛ 0 R + 3V ɛ 0 20 1 cos θ + R 3 V ɛ 0 2 R P 2(cosθ. 7