) 4n+2 sin[(4n + 2)φ] n=0. a n ρ n sin(nφ + α n ) + b n ρ n sin(nφ + β n ) n=1. n=1. [A k ρ k cos(kφ) + B k ρ k sin(kφ)] (1) 2 + k=1

Transcription

1 Physics 505 Fll 2007 Homework Assignment #3 Solutions Textbook problems: Ch. 2: 2.4, 2.5, 2.22, A vrint of the preceeding two-dimensionl problem is long hollow conducting cylinder of rdius b tht is divided into equl qurters, lternte segments being held t potentil +V nd V. Solve by mens of the series solution (2.7 nd show tht the potentil inside the cylinder is Φ(ρ, φ = 4V ( ρ 4n+2 sin[(4n + 2φ] π b 2n + n=0 The generl series solution for the two-dimensionl problem in polr coordintes is given by (2.7 Φ(ρ, φ = 0 + b 0 log ρ + n ρ n sin(nφ + α n + b n ρ n sin(nφ + β n Since we re interested in the interior solution, we demnd tht the potentil remins finite t ρ = 0. This indictes tht the b n coefficients must ll vnish. We re thus left with Φ(ρ, φ = 0 + n ρ n sin(nφ + α n which we my choose to rewrite s Φ(ρ, φ = A [A k ρ k cos(kφ + B k ρ k sin(kφ] ( k= This form of the series is supposed to be reminiscent of Fourier series. The boundry condition for this problem is tht the potentil t ρ = b is either +V or V, depending on which qudrnt we re in V b V V V

2 This cn be plotted s function of φ Φ(b, ϕ V π V π ϕ It should be obvious tht Φ(b, φ is n odd function of φ. As result, we immeditely deduce tht the A k Fourier coefficients in ( must vnish, leving us with Φ(ρ, φ = B k ρ k sin(kφ (2 k= On the interior surfce of the conducting cylinder, this reds Φ(b, φ = B k b k sin(kφ k= where Φ(b, φ is given by the figure bove. In prticulr, we see tht the quntities B k b k re explicitly the Fourier expnsion coefficients of squre wve with period π (which is hlf the usul 2π period. As result, we my simply look up the stndrd Fourier expnsion of the squre wve nd mp it to this present problem. Alterntively, it is strightforwrd to clculte the coefficients directly π B k b k = Φ(b, φ sin(kφdφ π π ( = V π/2 0 π/2 π + sin(kφdφ π π π/2 0 π/2 = V ( cos(kφ π/2 + cos(kφ 0 cos(kφ kπ π π/2 = 2V ( ( kπ 2 cos + cos(kπ kπ 2 = 8V k = 2, 6, 0, 4,... (ie k = 4n + 2 kπ π/2 0 cos(kφ Substituting B k = 8V/kπb k into (2 nd usiing k = 4n + 2 then gives π π/2 Φ(ρ, φ = 4V π ( ρ 4n+2 sin[(4n + 2φ] b 2n + n=0 (3

3 b Sum the series nd show tht Φ(ρ, φ = 2V ( 2ρ 2 b 2 sin 2φ π tn b 4 ρ 4 This series is esy to sum if we work with complex vribles. Since sin θ is the imginry prt of e iθ, we write (3 s Φ(ρ, φ = 4V π I (ρ/b 4n+2 e (4n+2iφ 2n + n=0 = 4V π I n=0 z 2n+ 2n + = 4V π I ( z + 3 z3 + 5 z5 + (4 where z ρ2 b 2 e2iφ (5 Now recll tht the Tylor series expnsion for log( + z is given by log( + z = ( k+ z k = z k 2 z2 + 3 z3 4 z4 + 5 z5 k= We my eliminte the even powers of z by tking the difference between log(+z nd log( z. This llows us to derive the series expression Substituting this into (4 gives z + 3 z3 + 5 z5 + = 2 log + z z Φ(ρ, φ = 2V π I log + z z = 2V π rg + z z Since rg(x + iy = tn (y/x, bit of lgebr gives Φ(ρ, φ = 2V ( 2Iz π tn z 2 Using the expression for z given in (5, we finlly obtin Φ(ρ, φ = 2V ( 2(ρ 2 /b 2 sin 2φ π tn (ρ 2 /b 2 2 = 2V ( 2ρ 2 b 2 sin 2φ π tn b 4 ρ 4 (6

4 c Sketch the field lines nd equipotentils. The equipotentils correspond to φ(ρ, φ = Φ 0. To see wht this looks like, we my invert (6 to solve for b s function of φ t fixed Φ 0. The result is 2ρ 2 b 2 sin 2φ b 4 ρ 4 ( πφ0 = tn 2V (ρ/b 2 sin 2φ = tn(πφ 0 /2V + + A plot of the equipotentils is given by sin 2 2φ tn 2 (πφ 0 /2V V V V V where we hve lso shown the electric field lines (the curves with rrows. 2.5 Show tht the Green function G(x, y; x, y pproprite for Dirichlet boundry conditions for squre two-dimensionl region, 0 x, 0 y, hs n expnsion G(x, y; x, y = 2 g n (y, y sin(nπx sin(nπx where g n (y, y stisfies ( 2 y 2 n2 π 2 g n (y, y = 4πδ(y y nd g n (y, 0 = g n (y, = 0 We strt by reclling the the Green s function is defined by ( 2 x + 2 y G(x, y; x, y = 4πδ(x xδ(y y (7 Although this is symmetric in x nd y, the problem suggests tht we begin by expnding in x (nd lso x. This of course breks the symmetry in the expnded form of the Green s function by treting x somewht differently. Nevertheless G(x, y; x, y is unique for the given boundry conditions; it just my dmit

5 different expnsions, nd we re free to choose whtever expnsion is the most convenient. Given the boundry condition tht G vnishes for x = 0 nd x =, this suggests n expnsion in Fourier sine series G(x, y; x, y = Substituting this into (7 then gives f n (x, y; y sin(nπx ( y 2 n2 π 2 f n (x, y; y sin(nπx = 4πδ(x xδ(y y (8 However this is not prticulrly useful (yet, since the δ(x x on the right hnd side does not mtch with the Fourier sine series on the left. We cn get round this by invoking the completeness reltion for the sine series sin(nπx sin(nπx = 2 δ(x x By replcing the delt function in (8 by this sum, we end up with ( y 2 n2 π 2 f n (x, y; y sin(nπx = 8πδ(y y sin(nπx sin(nπx (9 Mtching left nd right sides of the Fourier sine series indictes tht the x behvior of f n (x, y; y must be given by sin(nπx. Putting in fctor of two for convenience f n (x, y; y = 2g n (y, y sin(nπx finlly motivtes the expnsion G(x, y; x, y = 2 g n (y, y sin(nπx sin(nπx When this is inserted into (9, we mtch the x nd x behvior perfectly, nd we re left with n eqution in y ( 2 y n2 π 2 g n (y, y = 4πδ(y y (0 The boundry conditions re tht G vnishes t y = 0 nd y =. Hence we must lso demnd g n (y, 0 = g n (y, = 0. b Tking for g n (y, y pproprite liner combintions of sinh(nπy nd cosh(nπy in the two regions, y < y nd y > y, in ccord with the boundry conditions

6 nd the discontinuity in slope required by the source delt function, show tht the explicit form of G is G(x, y; x, y = 8 n sinh(nπ sin(nπx sin(nπx sinh(nπy < sinh[nπ( y > ] where y < (y > is the smller (lrger of y nd y. To find the Green s function for (0, we begin with the solution to the homogeneous eqution ( 2 y n2 π 2 g n (y, y = 0. This clerly hs exponentil solutions e ±nπy, or equivlently sinh(nπy nd cosh(nπy. As result, we cn write the Green s function s g n (y, y = { g< < sinh(nπy + b < cosh(nπy y < y g > > sinh(nπy + b > cosh(nπy y > y ( We wish to solve for the four constnts <, b <, >, b > given the boundry conditions g n (y, 0 = 0, g n (y, = 0 nd the continuity nd jump conditions g > = g < y g > = y g < 4π when y = y We strt with the boundry conditions. For g < to vnish t y = 0 we must tke the sinh solution, while for g > to vnish t y = we end up with > sinh(nπ + b > cosh(nπ = 0 or b > = > tnh(nπ. Thus g n (y, y = { < sinh(nπy y < y > [sinh(nπy tnh(nπ cosh(nπy ] y > y (2 The continuity nd jump conditions yield the system of equtions ( ( sinh(nπy sinh(nπy + tnh(nπ cosh(nπy < = cosh(nπy cosh(nπy + tnh(nπ sinh(nπy > ( 0 4/n which is solved by ( ( < 4 sinh(nπy tnh(nπ cosh(nπy = > n tnh(nπ sinh(nπy ( 4 cosh(nπ sinh(nπy sinh(nπ cosh(nπy = n sinh(nπ cosh(nπ sinh(nπy Inserting this into (2 gives g n (y, y 4 = n sinh(nπ { sinh(nπy [sinh(nπ cosh(nπy cosh(nπ sinh(nπy] y < y sinh(nπy[sinh(nπ cosh(nπy cosh(nπ sinh(nπy ] y > y

7 This is simplified by noting sinh[nπ( y] = sinh(nπ cosh(nπy cosh(nπ sinh(nπy nd by using the definition y < = min(y, y nd y > = mx(y, y. The result is which yields g n (y, y = 4 n sinh(nπ sinh(nπy < sinh[nπ( y > ] G(x, y; x, y = n 8 n sinh(nπ sin(nπx sin(nπx sinh(nπy < sinh[nπ( y > ] Alterntively, insted of using (, note tht we cn utomticlly solve the boundry conditions g n (y, 0 = g n (y, = 0 by writing { g n (y, y g< = < sinh(nπy y < y g > > sinh[nπ( y ] y > y Solving the continuity nd jump conditions then gives directly so tht < = 4 n g n (y, y = sinh[nπ( y], > = 4 sinh(nπy sinh(nπ n sinh(nπ 4 n sinh(nπ { sinh[nπ( y] sinh(nπy y < y sinh(nπy sinh[nπ( y ] y > y which is the sme result s bove. Finlly, we note tht the one-dimensionl Green s function g n (y, y cn lso be obtined through Sturm-Liouville theory s g n (y, y = A u(y <v(y > where u(y nd v(y re solutions to the homogeneous eqution stisfying boundry conditions t y = 0 nd y =, respectively. Here A is constnt given by W (u, v = A/p where W is the Wronskin, nd the self-djoint differentil opertor is L = d dy p(y d dy + q(y 2.22 For the exmple of oppositely chrged conducting hemisphericl shells seprted by tiny gp, s shown in Figure 2.8, show tht the interior potentil (r < on the z xis is Φ in (z = V [ (2 z 2 ] z 2 + z 2

8 Find the first few terms of the expnsion in powers of z nd show tht they gree with (2.27 with the pproprite substitutions. As we hve seen, the Green s function for the interior conducting sphere problem is equivlent to tht for the exterior problem. The only difference we need to ccount for is tht, for the interior problem, the outwrd pointing norml indeed points wy from the center of the sphere. This indictes tht the interior potentil my be expressed s where Φ(r.Ω = 4π Φ(, Ω ( 2 r 2 (r r cos γ cos γ = cos θ cos θ + sin θ sin θ cos(φ φ 3/2 dω In fct, introducing the bsolute vlue 2 r 2, it is esy to see tht the expression Φ(r, Ω = 2 r 2 4π Φ(, Ω (r dω 2r cos γ 3/2 is vlid for both the interior nd the exterior problem. For the oppositely chrged hemisphere problem, this integrl tkes the form 2π Φ(r, Ω = V 2 r 2 dφ d(cos θ 4π 0 0 [(r r cos γ 3/2 (r r cos γ 3/2] This simplifies on the z xis, where θ = 0 implies cos γ = cos θ. We find 2π Φ(z = V 2 z 2 dφ d(cos θ 4π 0 0 [(z z cos θ 3/2 (z z cos θ 3/2] = V 2 z 2 2z = V 2 z 2 = V z [(z z cos θ /2 (z z cos θ /2] ( z + z + 2 z z ( mx(, z 2 z 2 z2 + 2 For the interior, z <, this my be rewritten s Φ in (z = V z ( 2 z z 2 0 (3 (4

9 while for the exterior, z >, this becomes Φ out (z = V ( z2 2 z 2 + z 2 The interior solution, (4, my be expnded for z 0. The result is Φ in (z = 3V 2 ( z [ 7 2 ( z 2 ( z 4 + ] 24 (5 This my be compred with the exterior solution (2.27 Φ out (r, θ = 3V 2 Φ out (z = 3V 2 ( r 2 [ P (cos θ 7 ( z 2 [ 7 2 ( z ( 2 r 2 + ] 2 P3 (cos θ + ] This demonstrtes tht the expnsion coefficients gree, nd tht in fct the interior nd exterior expressions re identicl up to the substitution in: ( r l out: ( r l+ b From the result of prt nd (2.22, show tht the rdil electric field on the positive z xis is V 2 E r (z = (3 + 2 (z /2 z 2 for z >, nd E r (z = V [ 3 + (/z 2 ] 2 ( + (z/ 2 3/2 z 2 for z <. Show tht the second form is well behved t the origin, with the vlue, E r (0 = 3V/2. Show tht t z = (north pole inside it hs the vlue ( 2 V/. Show tht the rdil field t the north pole outside hs the vlue 2V/. The rdil electric field on the positive z xis is given by Rewriting the potentil Φ(z in (3 s Φ(z = V E r (z = z Φ(z ( mx(/z, 2 z 2 z z 2 + 2

10 we find ( V z E r (z = ( 2 + 3z 2 ( z 2 (z /2 V 2 ( 2 + 3z 2 z 2 (z /2 This my be simplified to red z < z > V ( 3 + (/z 2 2 ( + (z/ E r (z = 2 3/2 z 2 z < V 3 + (/z 2 z > ( + (z/ 2 3/2 (6 which is the desired result. As z 0, we my Tylor expnd the interior solution to obtin E r (z = 3V [ 7 ( z 2 55 ( z 4 + ] Hence E r (0 = 3V/2. Note tht this result could hve been obtined directly by differentiting (5. Finlly, the vlue of the rdil electric field t z = (immeditely inside nd z = + (immeditely outside my be obtined from (6 V E r ( ± = ( 2 z = V 2 z = + c Mke sketch of the electric field lines both inside nd outside the conducting hemispheres, with directions indicted. Mke plot of the rdil electric field long the z xis from z = 2 to z = +2. A rough sketch of the electric field lines is s follows

11 Note tht the field lines re not necessrily continuous from the inside to the outside of the hemispheres. The z component of the electric field long the z xis is given by (6 /( V / E z z/ Note tht E z is positive (pointed upwrds outside the sphere, nd negtive (pointed downwrds inside the sphere. By symmetry, E z is the only nonvnishing component of the electric field long the xis. The rdil or r component of the electric field, E r, is the sme s E z on the +z xis, but hs the opposite sign on the z xis /( V / E r z/ A hollow cube hs conducting wlls defined by six plnes x = 0, y = 0, z = 0, nd x =, y =, z =. The wlls z = 0 nd z = re held t constnt potentil V. The other four sides re t zero potentil. Find the potentil Φ(x, y, z t ny point inside the cube. The potentil my be obtined by superposition Φ = Φ top + Φ bottom where Φ top (Φ bottom is the solution for hollow cube with the top (bottom held t constnt potentil V nd ll other sides t zero potentil. As we hve seen, the series solution for Φ top is given by Φ top = ( nπx ( mπx ( n A n,m sin sin sinh 2 + n 2 πz n,m where A n,m = 4 2 sinh( n 2 + m 2 π 0 dx dy V sin 0 ( nπx ( mπx sin

12 Noting tht 0 ( nπx sin dx = ( nπx nπ cos = 0 nπ ( ( n = 2 nπ for n odd we hve nd hence Φ top = 6V π 2 A n,m = n,m odd 6V nmπ 2 sinh( n 2 + m 2 π nm sinh( n 2 + m 2 π n, m odd ( nπx ( mπx ( n sin sin sinh 2 + n 2 πz To obtin Φ bottom, it is sufficient to relize tht symmetry llows us to tke z z. More precisely As result Φ bottom (x, y, z = Φ top (x, y, z Φ = Φ top + Φ bottom = 6V ( nπx ( mπx π 2 nm sinh( n n,m odd 2 + m 2 π sin sin [ ( n sinh 2 + n 2 πz ( n + sinh 2 + m 2 π( z ] Note tht this my be simplified using to red Φ = 6V π 2 sinh ζ + sinh(α ζ = 2 sinh(α/2 cosh(ζ α/2 n,m odd nm cosh( n 2 + m 2 π/2 ( nπx ( mπx ( n sin sin cosh 2 + m 2 π(z /2 (7 b Evlute the potentil t the center of the cube numericlly, ccurte to three significnt figures. How mny terms in the series is it necessry to keep in order to ttin this ccurcy? Compre your numericl result with the verge vlue of the potentil on the wlls. See Problem 2.28.

13 At the center of the cube, (x, y, z = (/2, /2, /2, the potentil from (7 reds Φ(center = 6V π 2 = 6V π 2 sin(nπ/2 sin(mπ/2 n,m odd i,j=0 nm cosh( n 2 + m 2 π/2 ( i+j (2i + (2j + cosh( (2i (2j + 2 π/2 Numericlly, the first few terms in this series re given by n m Φ n,m /V running totl This tble indictes tht we need to keep t lest the first four terms to chieve ccurcy to three significnt figures. To this level of ccurcy, we hve Φ(center.333V If we went to higher orders, it ppers tht the potentil t the center is precisely Φ(center = 3 V which is the verge vlue of the potentil on the wlls. In fct, we cn prove (s in Problem 2.28 tht the potentil t the center of regulr polyhedron is equl to the verge of the potentil on the wlls. Hence this vlue of V/3 is indeed exct. c Find the surfce-chrge density on the surfce z =. For the surfce-chrge density on the inside top surfce (z =, we use Φ σ = ɛ 0 Φ n = ɛ 0 S z z= where the norml pointing wy from the top conductor is ˆn = ẑ. This is wht ccounts for the sign flip in the bove. Substituting in (7 gives σ = 6ɛ 0V π n2 + m 2 n,m odd nm tnh( ( nπx n 2 + m 2 π/2 sin sin ( mπy