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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 throught the surface of the cavity is `q//epsi_(0)`

Text Solution

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The correct Answer is:
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//epsi_(0)`.
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