A spherical non-conductor thin shell of radius a is given a fixed surface charge

Question

A spherical non-conductor thin shell of radius a is given a fixed surface charge density o(0) OSS 0 cos 0), where oo is a constant; Figure 3(a). There is no charge elsewhere in space a. Find the potential V(r, 0) in all space (including inside the shell). b. Suppose that the spherical non-conducting shell of part (a) is place at the center of a spherical conductor cavity or radius b a; Figure 3(b). The surface charge o(0) at a is not affected by the presence of the conducting shell. Find the potential everywhere for r K b c. Compute the surface charge density on the wall of the cavity of part (b) (that is, at r b).

Explanation / Answer

GIven surface charge density, sigma = sigma-o ( cos(theta) + cos^2(theta))
a. Inside the shell, the potential is constant, and Electric field is zero
this potential is equal to the potential at the surface of the shell
potential at surface of the shell = potential inside the shell = kq/r
here r is the radius of the shell
and q is net charge on the shell
now, charge on a incremental ring of the shell at anfle theta
dq = sigma*dA = sigma-o ( cos(theta) + cos^2(theta)) * 2*pi*rcos(theta) * d(theta) * r
dq = sigma-o ( cos^2(theta) + cos^3(theta)) * 2*pi*r^2 * d(theta)
integrating from theta = -pi/2 to pi/2
q = [sigma - o (pi/4 + sin(pi)/4 + sin(pi/2) - sin^3(pi/2)/3)] - [sigma - o (-pi/4 + sin(-pi)/4 + sin(-pi/2) - sin^3(-pi/2)/3)]
q = [sigma - o (pi/4 + 1 - 1/3)] - [sigma - o (-pi/4 - 1 + 1/3)] = sigma - o (pi/4 + 2/3)] - [sigma - o (-pi/4 - 2/3)]
q = sigma - o (pi/4 + 2/3 + pi/4 + 2/3) = sigma-o(pi/2 + 4/3)
V = potential at surface of the shell = potential inside the shell = kq/r = k*sigma-o(pi/2 + 4/3)/R ( where R is the radisu of the shell)

POtential at distance r from the centre of the shell, outside the shell = kq/r = k*sigma-o(pi/2 + 4/3)/r

b. Electric field inside a conductor is 0
so for r = < a the answer is same as in the previous problem
V = k*sigma-o(pi/2 + 4/3)/r for r < = a

for r > a , r < b
V = k*sigma-o(pi/2 + 4/3)/r ( same as the previous problem)

for r = b
V is constant ( as E is zero)
V = k*sigma-o(pi/2 + 4/3)/b

c. Net charge developed inside the cavity = q = -sigma-o(pi/2 + 4/3) [ because this charge would then cancel out the other charge already present in the shell and make the net field inside the conductor = 0]

Charge density ( average ) = q/A = sigma-o(pi/2 + 4/3)/4*pi*b^2

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