JEE AdvancedElectrostaticsEasy
Question
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
Options
A.electric field near A in the cavity = electric field near B in the cavity
B.charge density at A = charged density at B
C.potential at A = potemtial at B
D.total electric field flux through the surface of the cavity is q/ε0
Solution
Under electrostatic condition, all points lying on the conductor are at same potential. Therefore at A = potential at B. Hence, option (c) is correct. From Gauss theorem, total flux through the surface to the cavity will be q/ε0
Note : Instead of an elliptical cavity, if it would had been a spherical cavity then option (a) and (b) were also correct.
Note : Instead of an elliptical cavity, if it would had been a spherical cavity then option (a) and (b) were also correct.
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