Question

2. (40 points) A conducting sphere of radius R1 has potential V. The per- mittivity is co for 0<T< R2, and it is e1 for R2 <T R3. A spherical conducting shell at radius R3 is at potential 0.. Assume Ri < R2 < R3 Find, at all points in space, the following: V, D, E, P, Po, and ob, where the last two are the volume bound charge density and the surface bound charge density. Be sure to find on all relevant surfaces.

Answer 1

Electric field inside a conductor is zero Hence for 0 < r < R_1 the electric field is zero. Now E = 0 The polarization is given as P = epsilon_0 chi E P = 0 The Displacement current is given as D = (1 + chi) epsilon_0 E = 0 D = 0 V = integral E .dr = 0 V = 0 Bound surface change density id given as sigma_b = Q/A = Q/4 pi R_1^2 sigma_b = Q/4 pi R^2_1 ...........(1) From gauss�s law integral E dA = Q/epsilon_0 rightarrow Q = 0 putting in (1) Sigma_b = 0 The volume change density is given as S_b = -nabla P S_b = 0 \r\n For R_1 < r < R_2 Using Gauss�s law integral E. dA = Q/epsilon_0 E = Q/epsilon_0 times 1/4 pi (R_2 - R_1)^2 P = chi epsilon_0 E = Q chi/4 pi (R_2 - R_1)^2 D = (1 + chi) epsilon_0 E = (1 + chi)