A solid, conducting sphere having a charge ‘Q’ is surrounded by an uncharged, concentric conducting hollow spherical shell. Let the P.D between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of -3Q, the new PD between the same two surfaces is

In case of charged conducting sphere, 
$\begin{array}{l}{\mathrm{V}}_{\mathrm{inside}}={\mathrm{V}}_{\mathrm{center}}={\mathrm{V}}_{\mathrm{surface}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{R}}\\ {\mathrm{V}}_{\mathrm{outside}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{\mathrm{r}}\end{array}$
If a,b are the radii of sphere and of outer surface of spherical shell respectively, then potential at their surfaces will be
$$\begin{gathered} {\mathrm{V}}_{\mathrm{sphere}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Q}}{\mathrm{a}};{\mathrm{V}}_{\mathrm{shell}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{Q}}{\mathrm{b}} \\ \mathrm{V}={\mathrm{V}}_{\mathrm{sphere}}-{\mathrm{V}}_{\mathrm{shell}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{\mathrm{a}}-\frac{\mathrm{Q}}{\mathrm{b}}\right] \end{gathered}$$
Now when the shell is given charge -3Q  the potential will be
$$\begin{gathered} {\mathrm{V}}_{\mathrm{sphere}}^1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{\mathrm{a}}+\frac{(-3\mathrm{Q})}{\mathrm{b}}\right] \\ {\mathrm{V}}_{\mathrm{shell}}^1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{\mathrm{b}}+\frac{(-3\mathrm{Q})}{\mathrm{b}}\right] \\ {\mathrm{V}}_{{\mathrm{sphere}}_{ }}^1-{\mathrm{V}}_{{\mathrm{shell}}_{ }}^1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{Q}}{\mathrm{a}}-\frac{\mathrm{Q}}{\mathrm{b}}\right]=\mathrm{V} \end{gathered}$$