Capacity of spherical capacitor with shells of radii ‘a’ and ‘b’ and outer shell earthed is given by

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It consists of two concentric spherical conducing shells of radii a and b, say b>a (see fig.). The outer shell is earthed. Place a charge +Q on the inner shell. It will reside on the outer surface of the shell. A charge −Q will be induced on inner surface of outer shell. A charge −Q will be induced on inner surface of outer shell. A charge +Q will flow from outer shell to the earth.

Consider a Gaussian spherical surface of radius r such that a<r<b.
From Gauss's law, electric field at distance r>a is

$$\begin{gathered} E=\frac{Q}{4\pi {\epsilon }_0{r}^2} \\ {\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}}=-{\int }_{\mathrm{a}}^{\mathrm{b}} \overrightarrow{\mathrm{E}}\cdot \mathrm{d}\overrightarrow{\mathrm{r}}=-{\int }_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{q}}{4\pi {\epsilon }_0{\mathrm{r}}^2}\mathrm{dr} \\ \text{ Since }{\mathrm{V}}_{\mathrm{b}}=0\text{, we have } \\ {\mathrm{V}}_{\mathrm{a}}={\int }_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\mathrm{dr} \\ =\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right)=\frac{\mathrm{Q}(\mathrm{b}-\mathrm{a})}{4{\mathrm{\pi \epsilon }}_0\mathrm{ab}} \\ \text{ Therefore, capacitance, }\mathrm{C}=\frac{\mathrm{Q}}{{\mathrm{V}}_{\mathrm{a}}-{\mathrm{V}}_{\mathrm{b}}}=\frac{\mathrm{Q}}{{\mathrm{V}}_{\mathrm{a}}}=\frac{4{\mathrm{\pi \epsilon }}_0\mathrm{ab}}{\mathrm{b}-\mathrm{a}} \end{gathered}$$