There are two concentric spherical shells of radii r and 2r. Initially a charge Q is given to the inner shell. Now switch S₁ is closed and S₂ is opened then S₂ is closed and S₁ is opened and the process is repeated n times for both the keys alternatively. Find the final potential difference between the shells.

Now ‘S₁’ is closed.

Let q charge is on outer shell 

So, potential on outer surface should be zero.

i.e. $\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_{0}2\mathrm{r}}+\frac{{\mathrm{q}}^{\mathrm{\prime }}}{4{\mathrm{\pi \epsilon }}_{0}2\mathrm{r}}=0$

$\Rightarrow$q' = -Q . 

Now S₁ is opened S₂ is closed and charge on outer surface is -’Q’. Now V on inner surface should be zero.

So, $\frac{\mathrm{Q}+{\mathrm{q}}^{\mathrm{\prime }}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}-\frac{Q}{4{\mathrm{\pi \epsilon }}_0\left(2\mathrm{r}\right)}=0$

$\Rightarrow {\mathrm{q}}^{\mathrm{\prime }}=\frac{-\mathrm{Q}}{2}$

Potential difference $=\frac{1}{4}\left(\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}}\right)$

Similarly for n times potential difference 

$=\frac{1}{2^{\mathrm{n}+1}}\left(\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}\right)$