An elliptical cavity is carved with in a prefect conductor figure. A positive charge q is placed at the center of the cavity. The points A and B are on the cavity surface as shown in the figure. Then

Options

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 = potential at B
D. Total electric field flux throught the surface of the cavity is `q//epsi_(0)`

Answer

C, D

Explanation

Under electrostatic condition, all points lying on the conductor are at same potential. Therefore, potential at A = potential at B. Hence, option (C) is correct. From Gauss theorem, total flux through the surface of the cavity will be `q//epsi_(0)`.