Two thin conducting shells of radii R and 3R are shown in the figure. The outer shell carries a charge + Q and the inner shell is neutral. The inner shell is earthed with the help of a switch S.

$\begin{array}{r}{\mathrm{V}}_{\text{outer }}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{Q}}{3\mathrm{R}},\\ {\mathrm{V}}_{\text{inner }}=\frac{1}{4{\mathrm{\pi ϵ}}_0}\frac{\mathrm{Q}}{3\mathrm{R}}\end{array}$

When S is closed, let q is charge an inner sphere by the earth

${\mathrm{V}}_{\text{in }}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\mathrm{q}}{\mathrm{R}}+\frac{\mathrm{Q}}{3\mathrm{R}}\right]=0\Rightarrow \mathrm{q}=\frac{-\mathrm{Q}}{3}$

As the charge on the outer shell remains the same but the potential is decreased, so for having the same charge for lesser potential we can say that the capacitance of the system is increased.