Question:

Shown in the figure are two point charges \(+Q\) and \(-Q\) inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If \(\sigma_1\) is the surface charge on the inner surface and \(Q_1\) net charge on it and \(\sigma_2\) the surface charge on the outer surface and \(Q_2\) net charge on it then:

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In a conducting shell the inner surface carries induced charge totalling minus the enclosed charge, and the outer-surface charge of a spherical shell is uniform regardless of the cavity contents.
Updated On: Nov 11, 2026
  • \(\sigma_1=0,\ Q_1=0,\ \sigma_2=0,\ Q_2=0\)
  • \(\sigma_1\neq0,\ Q_1=0,\ \sigma_2\neq0,\ Q_2=0\)
  • \(\sigma_1\neq0,\ Q_1\neq0,\ \sigma_2\neq0,\ Q_2\neq0\)
  • \(\sigma_1\neq0,\ Q_1=0,\ \sigma_2=0,\ Q_2=0\)
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The Correct Option is D

Solution and Explanation

The shell is a conductor, so the field in its material is zero. By Gauss's law the induced charge on the inner surface equals minus the enclosed charge: \(Q_1=-(+Q-Q)=0\). But +Q and -Q sit near the cavity wall at different places, so the induced charge density on the inner surface is non-uniform and non-zero: \(\sigma_1\neq0\) (negative near +Q, positive near -Q).
The shell is neutral, so \(Q_2=-Q_1=0\). The outer surface charge of a spherical conductor spreads uniformly (the interior pattern is screened), so \(\sigma_2=\frac{Q_2}{4\pi R^2}=0\).
Answer: (4) \(\sigma_1\neq0,\ Q_1=0,\ \sigma_2=0,\ Q_2=0\)
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