Question

A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer 1

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. so, -q charge induced on inner surface and +q charge induced on outer surface.

now, charge on inner surface, $q_{in}$ = -q

charge on outer surface, $q_{out}$ = (Q + q)

(a) surface charge density on the inner surface of the shell is $\sigma_{in}=\frac{q_{in}}{4\pi r_1^2}$

= $\frac{-q}{4\pi r_1^2}$

and on the outer surface of the shell is $\sigma_{out}=\frac{q_{out}}{4\pi r_2^2}$

= $\frac{(Q+q)}{4\pi r_2^2}$

(b) electric flux linked with any closed surface inside the conductor is zero as electric field inside the conductor is zero.

so, from Gauss's theorem, net charge enclosed by closed surface S is also zero.i.e., Qnet = 0

so,if there is no charge inside the cavity , then there cannot be any charge on the inner surface of the sell and hence electric field inside the cavity will be zero, even though the shell may not be spherical.