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(D) Under electrostatic conditions,the entire volume of a conductor is an equipotential region. Since points $A$ and $B$ lie on the surface of the con

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Both $(b)$ and $(c)$

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(D) Under electrostatic conditions,the entire volume of a conductor is an equipotential region. Since points $A$ and $B$ lie on the surface of the conductor,they must be at the same potential. Thus,Potential at $A = $ Potential at $B$.According to Gauss's Law,the total electric flux $\phi$ through any closed surface enclosing a charge $q$ is given by $\phi = q/\varepsilon_0$. Since the cavity surface encloses the charge $q$,the total flux through it is $q/\varepsilon_0$.For an ellipsoidal cavity,the electric field near the surface depends on the local curvature,so the electric field at $A$ is not necessarily equal to the electric field at $B$. Therefore,options $(b)$ and $(c)$ are correct.

An ellipsoidal cavity is carved within a perfect conductor. $A$ positive charge $q$ is placed at the centre of the cavity. The points $A$ and $B$ are on the cavity surface as shown in the figure. Then

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