A spherical conducting shell of inner radius `r_(1)` and outer radius `r_(2)` has a charge Q.
ls the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape ? Explain.

Explanation

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 and + q charge induced on the outer surface of the shell.
`:. ` Surface charge density at the inner surface of the shell is ,
`sigma=(q)/(A)=(-q)/(4pir_(1)^(2))`
Surface charge density at the outer surface of the shell is
`sigma_(0)=(q)/(A)=((+q)(+Q))/(4pir_(2)^(2))`
Where `r_(1)` and `r_(2)` are radius of inner and outer of shell and Q is the magnitude of charge placed on the outer surface of the shell

(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 zeru because the field inside the conductor is zero. Hence, electric field is zero whatever is the shape.