Question

A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge ‘Q’. A charge ‘q’ is placed at the centre of the shell. 

(a) What is the surface charge density on the (i) inner surface, (ii) outer surface of the shell? 

(b) Write the expression for the electric field at a point x >r₂ from the centre of the shell.

Answer 1

We have,

radii = r₁​, r_2​

charge = Q, q

So,

a) Surface charge density on the inner surface is \(\frac q{4\pi r_1^2}\)

Surface charge density on the outer surface is \(\frac {q+ Q}{4\pi r_2^2}\)

b) For finding the electric field at the distance x from the center is:

For this Apply  the Gauss law,

E × 4πx² = 4π(Q + q)

\(E = \frac{Q + q}{x^2}\)

Answer 2

(a) Charge Q resides on outer surface of spherical conducting shell. Due to charge q placed at centre, charge induced on inner surface is –q and on outer surface it is +q. So, total charge on inner surface -q and on outer surface it is Q + q.

Surface charge density on inner surface  = - q/4πr₁²

Surface charge density on outer  surface = (Q + q)/4πr₂²

(b) For external points, whole charge acts at centre, so electric field at distance x >r₂,​