Question

A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radius 2RR that also carries charge QQ. The charge QQ is distributed uniformly over the insulating shell.
A. Find the magnitude of the electric field in the region 02R. Express your answer in terms of the variables R, r, Q, and constants
π
and
ε
0.

Answer 1

Answer:

$E=0$ at r < R;

$E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$ at 2R > r > R;

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$ at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

$\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon}$ (integral over a closed surface)

where,

$E$ = Electric field

$Q_{enclosed}$ = charged enclosed within the closed surface

$\epsilon$ = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

$Q_{enclosed}$ = 0 and hence $E$ = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

$Q_{enclosed}$ = Q,

therefore,

$E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}$      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to $4\pi r^{2}$)

or, $E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}$

at r >= 2R

$Q_{enclosed}$ = 2Q

Hence, by similar calculations, we get,

$E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}$