Electrostatics 4 Question 14
14. An elliptical 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
(1999, 3M)
(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 $q / \varepsilon_{0}$.
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Answer:
Correct Answer: 14. (c, d)
Solution:
- 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 $q / \varepsilon_{0}$.
NOTE
Instead of an elliptical cavity, if it would had been a spherical cavity then options (a) and (b) were also correct.
