Solution:

(a) Putting a charge of magnitude $+q$ in the centre of the shell causes a charge of magnitude $-q$ to be generated in the inner surface of the shell, and the reverse is true. Thus, the surface charge density at the inner surface of the shell can be calculated using the following equation:

$\sigma_{1}=$ Total charge $/$ Inner surface area $=-\mathrm{q} / 4 \pi \mathrm{r}_{1}^{2} \underline{-}(1)$

A charge $+q$ is induced on the shell’s outer surface, which is referred to as the shell’s surface charge. It is calculated that there is a total charge on the outer surface of the shell equal to $Q+q.$ The surface charge density of the shell’s outer surface is measured.

$\sigma_{2}=$ Total charge $/$ Outer surface area $=(\mathrm{Q}+\mathrm{q}) / 4 \pi \mathrm{r}_{1}^{2}=(2)$

(b) Yes. The electric field inside a cavity (with no charge) will always be zero, regardless of whether the shell is spherical or has any other irregular shape. Consider a closed-loop system in which part of the loop is inside the cavity along a field line and the remainder of the loop is inside the conductor. Due to the fact that the field within the conductor is zero, the net work done by the field in carrying a test charge over a closed loop is calculated. We are aware that this is not possible in an electrostatic field. As a result, there are no field lines inside the cavity (i.e., there is no field), and there is no charge on the inner surface of the conductor, regardless of the form of the conductor.