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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
When two points are connected with a conducting path in electrostatic condition, then the potential of the two points is equal. Thus potential at A=potential at B
(c) is correct option.
`Option (d) is result fo Gauss's law
Total electric fulx through cavity `=q/epsilon_0`
Option (a) and (b) are dependent on the curvature which is different at points A and B.
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