A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ 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.

$(a)$ Charge placed at the centre of a shell is $+q$. Hence, a charge of magnitude  $-q$ will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is $- q$.

Surface charge density at the inner surface of the shell is given by the relation,

$\sigma_{1}=\frac{\text { Total charge }}{\text { Inner surface area }}=\frac{-q}{4 \pi r_{1}^{2}}$

A charge of $+q$ is induced on the outer surface of the shell. A charge of magnitude $Q$ is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is $Q+q .$ Surface charge density at the outer surface of the shell,

$\sigma_{2}=\frac{\text { Toter surface of the shell, }}{\text { Outer surface area }}=\frac{Q+q}{4 \pi r_{2}^{2}}$

$(b)$ Yes

The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.