Gujarati
1. Electric Charges and Fields
easy

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

Total electric field flux through the surface of the cavity is $q/{\varepsilon _0}$

C

Potential at $A = $ Potential at $B$

D

Both $(b)$ and $(c)$

(IIT-1999)

Solution

(d) Under electrostatic condition, all points lying on the conductor are in same potential. Therefore, potential at $A$ = potential at $B$.
From Gauss's theorem, total flux through the surface of the cavity will be $q/\varepsilon_0.$
Note :  Instead of an elliptical cavity, if it would had been a spherical cavity then options $(a)$ and $(b)$ were also correct.

Standard 12
Physics

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