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.

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Solution

(a) The charge of magnitude $+ q$ present inside the spherical shell induces a charge of magnitude $- q$ 0n the inner part of the shell. Therefore, for the charge on the inner part, the surface charge density will be due to $- q$ .
It is given by:
$σ_{1} = \frac{T o t a l c h a r g e}{s u r f a c e a r e a o f i n n e r p a r t}$
$\Rightarrow \frac{- q}{4 π r_{1}^{2}}$
The charge on the outer shell will be $+ q$ and the surface charge density at the outer part will be due to sum of $q$ and $Q$ charges present on the outer shell.
$σ_{2} = \frac{T o t a l c h a r g e}{O u t e r s u r f a c e a r e a} = \frac{Q + q}{4 π r_{2}^{2}}$ ...(ii)

(b) If we consider a loop inside an irregular conductor, then will be no work done on it as there is no field present inside the cavity. So, there will be no field inside the cavity even if it is irregular in shape.

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