A solid conducting sphere of radius $R$ is surrounded by a concentric metallic shell of inner radius $2 R$ and outer radius $3 R$ and the shell is earthed. The inner sphere is connected to a switch $S$ by a thin conducting wire through a small hole in the shell. By closing the switch $S$ , the inner sphere is connected to a distant conducting sphere of radius $R$ having charge $Q$ .

The charge that flows through wire

A B

has magnitude of

\frac{2 Q}{3}

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The charge on the inner surface of shell is

- \frac{2 Q}{3}

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The charge on the inner surface of shell is

+ \frac{2 Q}{3}

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The charge on the distant sphere's surface is

\frac{Q}{3}

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Solution

The correct options are
A The charge that flows through wire $A B$ has magnitude of $\frac{2 Q}{3}$
B The charge on the inner surface of shell is $- \frac{2 Q}{3}$
D The charge on the distant sphere's surface is $\frac{Q}{3}$

Let charge on inner sphere be $q$ .
A charge $- q$ is induced on the inner surface of the shell.

Two spheres of equal size will share charge till both of them acquire same potential.

Potential of the inner sphere= potential of distant sphere
$\Rightarrow \frac{K q}{R} + K (\frac{- q}{2 R}) = K (\frac{Q - q}{R}) \Rightarrow q = 2 Q - 2 q$
$\Rightarrow q = \frac{2 Q}{3}$
$∴$ Charge induced on the inner surface of the shell $= - q = - \frac{2 Q}{3}$
$∴$ Charge that flows via wire $= \frac{2 Q}{3}$

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