SATHEE: Electrostatics 4 Question 14

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 $B$

(d) total electric field flux through the surface of the cavity is $q / ε_{0}$ .

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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 $q / ε_{0}$ .

NOTE

Instead of an elliptical cavity, if it would had been a spherical cavity then options (a) and (b) were also correct.

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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

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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