$(a)\;\large\frac{kQ}{2^{n} r}\qquad(b)\;\large\frac{kQ}{2^{n-1}r}\qquad(c)\;\large\frac{kQ}{r2^{n+1}}\qquad(d)\;\large\frac{kQ}{2r}$

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Answer : (c) $\;\large\frac{kQ}{r2^{n+1}}$

Explanation :

When $\;S_{1}\;$ is closed

Let charge $\;q_{1}\;$ comes from the earth on outer shell

$V_{outer}=0$

$k\;(\large\frac{Q}{2r}+\large\frac{q_{1}}{2r})=0\; $ $\quad \; [q_{i}\;$ is charge on outer sphere after i times ]

$q_{1}=-Q$

When $\;S_{2}\;$ is closed

$k\;(\large\frac{q_{i}^{|}}{r}-\large\frac{Q}{2r})=0$ $\quad\;[q_{i}^{|}\;$ is charge on inner sphere after i times]

$q_{i}^{|}=\large\frac{Q}{2}$

Now again when switch $\;S_{1}\;$ is closed

$k\;(\large\frac{Q}{2\times2r}+\large\frac{q_{2}}{2r})=0$

$q_{2}=\large\frac{Q}{2}$

Now when $\;S_{1}\;$ opened and $\;S_{2}\;$ closed

$k\;(\large\frac{q_{2}^{|}}{r}-\large\frac{\large\frac{Q}{2}}{2r})=0$

Proceeding similarly we get after n times

$q_{n}=-\large\frac{Q}{2^{n-1}} \quad $ & $\;q_{n}^{|}=\large\frac{Q}{2^{n}}$

Potential difference between shells after n times

$V_{AB}=k\;q_{n}^{|}\;[\large\frac{1}{r}-\large\frac{1}{2r}]$

$V_{AB}=\large\frac{kQ}{2^{n}\times2}=\large\frac{kQ}{2^{n+1} r}\;.$

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