Question

There are two concentric spherical shells of radii r and 2r. Initially a charge Q is given to the inner shell.Now, switch $S_1$ is closed and opened then $S_2$ is closed and opened and the process is repeated n times for both the keys alternatively. Find the final potential difference between the shells

Answer 1

When $S_1$ is closed
Since outer shell is grounded $V_{outer}=0$ and let $q_1$ be the charge on outer sphere after some time t.
$V_{outer}=0$
$\frac{kQ}{2r}+\frac{k{q}_1}{2r}=0$
$q_1=-Q$
When $S_2$ is closed
Let $q_1^{\prime }$ be charge on inner surface after some time t.
$-\frac{kQ}{2r}+\frac{k{q}_1^{\prime }}{r}=0$
$q_1^{\prime }=\frac{Q}{2}$

Now, again when $S_1$ is closed
Let $q_2$ be the charge on outer sphere
$\frac{kQ}{2\times 2r}+\frac{k{q}_2}{2r}=0$
$q_2=\frac{Q}{2}$

Now when $S_1$ is opened and $S_2$ is closed
$\frac{-kQ/2}{2r}+\frac{k{q}_2^{\prime }}{r}=0$

Similarly proceeding after n times,
$q_n=-\frac{Q}{2^{n-1}}$
$q_n^{\prime }=\frac{Q}{2^n}$

Therefore, potential difference between the two shells after n times
$V_{AB}=k{q}_n^{\prime }(\frac{1}{r}-\frac{1}{2r})=\frac{kQ}{2^{n+1}r}$