# Spherical Capacitor

## Spherical Capacitor

A spherical capacitor consists of two concentric spherical conducting shells of radii $$a$$ and $$b$$, say $$b>a$$. 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 the inner surface of the outer shell. A charge $$+Q$$ will flow from the outer shell to earth.

Consider a Gaussian spherical surface of radius $$r$$ such that $$a<r<b$$. Form Gauss’s law, the electric field at distance $$r>a$$ is $$E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}$$.

The potential difference is: $${{V}_{b}}-{{V}_{a}}=-\int\limits_{a}^{b}{\overrightarrow{E}.\overrightarrow{dr}}=-\int\limits_{a}^{b}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}dr$$.

Since, $${{V}_{b}}=0$$ we have:

$${{V}_{a}}=\int\limits_{a}^{b}{\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}}dr=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{a}-\frac{1}{b} \right)=\frac{Q\left( b-a \right)}{4\pi {{\varepsilon }_{0}}ab}$$.

Now, the capacitance is: $$C=\frac{Q}{{{V}_{a}}-{{V}_{b}}}=\frac{Q}{{{V}_{a}}}=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}$$.

Therefore, the capacitance of a spherical capacitor is: $$C=\frac{4\pi {{\varepsilon }_{0}}ab}{\left( b-a \right)}$$.