Question

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 surfaces as shown in the figure. Then

• A. Charge density at A = charge density at B
• B. Electric field near A in the cavity = Electric field near B in the cavity
• C. Total electric flux through the surface of the cavity is $q/{ϵ}_0$
• D. Potential at A = potential at B

Answer 1

Answer: (C), (D)

Electric field at A is different from field at B because, $E=\frac{K}{r^2}$
we know that conductor is equipotential surface, so potential will be same at A and B.
As charge density ,$\sigma \propto \frac{1}{r},$ then charge density is different at A and B.
Using Gauss's law, flux through the surface of the cavity, $\varphi =\frac{q_{enclosed}}{{ϵ}_0}=\frac{q}{{ϵ}_0}$
Ans: $\left(C\right),\left(D\right)$