An ellipsoidal cavity is carved within a perfect conductor. A positive charge q is placed at the center of the cavity. The points A and B are on the cavity surface as shown in the figure Then:
Text Solution
Verified by ExpertsThe correct answer is:
C
Since distance between q and A is less then distance between q and B.
σ A > σ B . & E A > E B .
but, V A = V B because surface of conductor is equipotential.
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