Question 128 of 311

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:

A${\sigma }_{1}=0,{Q}_{1}=0$ , ${\sigma }_{2}=0,{Q}_{2}=0$B${\sigma }_{1}\neq 0,{Q}_{1}=0$ , ${\sigma }_{2}\neq 0,{Q}_{2}=0$C${\sigma }_{1}\neq 0,{Q}_{1}\neq 0$ , ${\sigma }_{2}\neq 0,{Q}_{2}\neq 0$D${\sigma }_{1}\neq 0,{Q}_{1}=0$ , ${\sigma }_{2}=0,{Q}_{2}=0$

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Answer: (D) ${\sigma }_{1}\neq 0,{Q}_{1}=0$ , ${\sigma }_{2}=0,{Q}_{2}=0$

By the property of electrostatic shielding in the conductors $E=0$ in the conductor. So, electric flux,$=0$ through a dotted Gaussian surface as shown



The net enclosed charge through the Gaussian surface$=0$,$\Rightarrow$ The net charge ${Q}_{1}$ on the inner surface$=0$, but the equal and opposite induced charge on the surface will be distributed non uniformly on the inner surface

So, ${\sigma }_{1}=0$

$∵{Q}_{1}=0$ on the inner surface

So, net charge ${Q}_{2}=0$ on the outer surface as conductor is neutral but $∵$ outer surface is free from any electric field so no charge density exists on the outer surface. So, ${\sigma }_{2}=0$ .