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 surface as shown in the figure. Then

• A. Electric field near A in the cavity = Electric field near B in the cavity
• B. Charge density at A = Charge density at B
• C. Potential at A is not equal to Potential at B
• D. Total electric field flux through the surface of the cavity is $\frac{q}{{ϵ}_0}$

Answer 1

The correct option is D Total electric field flux through the surface of the cavity is $\frac{q}{{ϵ}_0}$

Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at A = potential at B.
From Gauss's theorem, total flux through the surface of the cavity will be $\frac{q}{{ϵ}_0}$.