A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q.
i) 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?
ii) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

A ‘+ q’ charge which is placed at the centre of the shell will induce a ‘–q’ charge on the inner surface of the shell.
As a result of induction of '-q' on the inner surface, the charge on the outer surface will increase by '+q'. 
Therefore, there will be total (Q + q) charge on the outer surface of the shell and, '–q' charge on the inner surface of the shell.

 

i) Now,
$S\mathrm{urface} \mathrm{are} \mathrm{of} \mathrm{inner} \mathrm{sphere} = 4{{\mathrm{\pi r}}_1}^2$$S\mathrm{urface} \mathrm{area} \mathrm{of} \mathrm{outer} \mathrm{sphere} = 4{{\mathrm{\pi r}}_2}^2$

Thus,
Charge density on the outer surface$= \frac{\mathrm{Q}+\mathrm{q}}{4{{\mathrm{\pi r}}_2}^2}$
and
Charge density on the inner surface $= \frac{-\mathrm{q}}{4{{\mathrm{\pi r}}_1}^2}.$ 

ii) As charge resides on the outer surface so, the net charge on the inner surface of the cavity is zero as per the Gaussian theorem. Although, the net charge is zero yet the electric field may not be zero if the cavity is not spherical because, the surface may not have equal number of positive and negative charges. 
We assume a loop for this reason, some portion of which is inside the cavity and rest of its part is inside the conductor. Now, consider that there is some electric field inside the cavity. Since inside the conductor total electric field is zero and net work done by the field in bringing a test charge over this loop will not be zero. But this is not possible for an electrostatic field. Therefore, we must conclude that there is no electric field inside the cavity irrespective of its shape.