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An ellipsoidal cavity is carved within a...

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 = potential at B

D

total electric field flux through the surface of the cavity is `q//epsilon_(0)`.

Text Solution

Verified by Experts

The correct Answer is:
(c,d)
Points A and B lie on the ellipsoidal cavity carved within a perfect conductor . Since the potential of all points which lie on the conductor , is same the potentials at A and B are same.
Potential at A = Potential at B . Option (c ) is correct.
According to Gauss theorem,
Total flux through cavity - surface `(q)/(epsilon_(0))`
Hence option (d) is correct.
Options (a) and (b) are incorrect as the cacity is ellipsoidal. It may , however , be noted that options (a) and (b) hold good in case of a spherical cavity.
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