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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

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When two points are connected with a conducting path in electrostatic condition, then the potential of the two points is equal, therefore option(c) is the correct option option (d) follows from gauss's law. Options (a) and (b) are dependence on the curvature which are different at point A and B
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