Jackson 4.10 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell

Transcription

1 Jackson 4.10 Homework Probem Soution Dr. Christopher S. Baird University of Massachusetts Lowe PROBLEM: Two concentric conducting spheres of inner and outer radii a and b, respectivey, carry charges ±. The empty space between the spheres is haf-fied by a hemispherica she of dieectric (of dieectric constant ε/ε 0 ), as shown in the figure. + - b a (a) Find the eectric fied everywhere between the spheres. (b) Cacuate the surface-charge distribution on the inner sphere. (c) Cacuate the poarization-charge density induced on the surface of the dieectric at r a. SOLUTION: We can spit the region where we want to know the eectric fied into two regions, the eft hemisphere and the right hemisphere, sove for the fied in each region separatey and then appy the boundary conditions to get the fina soution. In the eft region, there are no dieectrics and no charges. So the eectric potentia obeys the Lapace equation: 2 0 If we aign the z-axis pointing to the right, the probem has azimutha symmetry. The genera soution to the Lapace equation in spherica coordinates for azimutha symmetry is: r,, A r B r 1 P cos The potentia on the surface of a conductor is aways constant: rac

2 C A a B a 1 P cos Because this equation must hod for a vaues of the independent poar variabe and because the Legendre poynomias are orthogona, a coefficients must equate independenty: C A 0 a and B A a 2 1 for > 0 This soution now becomes (the arbitrary constant C does not contain any new information, so we eave 0 constants as is): r,,a 0 r 1 A r a a r 1 P cos The other surface is aso a conductor and must have a constant potentia as we: rbd DA 0 b 1 A b a a b 1 P cos Because this equation must hod for a vaues of the independent poar variabe and because the Legendre poynomias are orthogona, a coefficients must equate independenty. A must be zero for a > 1. This eaves the potentia: r,,a 0 r This eads to a tota eectric fied of: E L We coud have guessed this form of the soution based on the symmetry of the probem, but it is often safer and more instructive to go through a the steps. The right hemisphere is a separate region and can be now soved separatey. In the region between the shes, we can te that there are no free charges and no bound charges (the bound charges, or poarization charges, wi reside aong boundaries, but not in the voume where we want to know the potentia). The tota eectric potentia then obeys the Lapace equation: 2 R 0 The same process as above is appied eading to the same form of the soution:

3 E R C 0 We can now appy boundary conditions where the dieectric materia meets free space. The tangent component of the eectric fied must be continuous across the boundary: E L E R [E L re R r ] /2 r C 0 2 r 2 C 0 which eads to: E R E L Because a materias are inear, this directy tes us that D L 0 and D R Because the free charges give rise to the D fied, the fact that D on the eft is different from D on the right means that the free charge densities on each side must be different. The ast fact that we know is that there is a tota free charge on the inner conductor surface and a tota free charge - on the outer conductor surface. We ony know the tota free charge, not the free charge density. We cannot assume the free charge spreads out uniformy over the sphere, because there is nothing in the probem to suggest that. Because we don't know the charge density, we cannot use a oca boundary condition, but instead an integrated boundary condition that makes use of the tota charge. We draw an integration sphere with radius r where a < r < b so that it competey encoses the free charge and use Gauss's aw in integra form. We coud use Gauss's aw for the E, D, or P fieds. We ony know the tota free charge, which ony gives rise to the D fied, so we use: S D rd 3 x We must be carefu here. We found above that the D fied is not the same over this integration sphere, so we must spit the integra into two pieces: 2 / D R rr 2 sin d d D L rr 2 sin d d 0 / 2

4 2 / sin d d 0 sin d d 0 /2 We now have our fina soutions: D L r r 2 and D R 0 2r r 2 E 0 2r r 2 (b) Cacuate the surface-charge distribution on the inner sphere. The distribution of free surface charges σ is inked to the D-fied according to: D We draw a Gaussian pibox stradding the surface and use the divergence theorem to find: [D 2 D 1 n ] on S Inside a conductor there are no fieds, so D 1 0. [D L r ] ra and R [D R r ] ra a r 2 and R 0 2a r 2 (c) Cacuate the poarization-charge density induced on the surface of the dieectric at r a. We use Gauss's aw on a pibox surface again, but instead ony encose the poarization charge. [P 2 P 1 n po ] on S Inside the conductor there are no fieds, so P 2 0 and the norma now points inwards, n -r [P r po ] r a po [ P ] r a

5 For inear materias, we know P 0 E po [ 0 E ] ra po a 2 Of course, this ony appies to the right side. There is no dieectric materia on the eft side and therefore no poarization charge.