Question

An elliptical 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 = potential at B
• D. total electric field flux through the surface of the cavity is $q/{\epsilon }_0$

Answer 1

Answer: (C), (D)

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 $q/{\epsilon }_0.$
NOTE Instead of an elliptical cavity, if it would had been a spherical cavity then options (a) and (b) were also correct.