Question
A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (2.36). Show
that the capacitance of a spherical capacitor is given by $\text{C}=\frac{{4}\pi\in_{0}\text{r}_1\text{r}_2}{\text{r}_1-\text{r}_2}$ where $r_1$ and $r_2$ are the radii of outer and inner spheres, respectively.

Answer

Radius of the outer shell $= r_1$
Radius of the inner shell $= r_2$​​​​​​​
The inner surface of the outer shell has charge +Q.
$\text{V}=\frac{\text{Q}}{{4}\pi\in_0\text{r}_2}-\frac{Q}{{4}\pi\in_0\text{r}_1}$
The outer surface of the inner shell has induced charge −Q. Potential difference between the two shells is given by,
Where,
$∈_0 =$ Permittivity of free space
$\text{V}=\frac{\text{Q}}{{4}\pi\in_0}\bigg[\frac{1}{\text{r}_2}-\frac{1}{\text{r}_1}\bigg]$
$\text{V}=\frac{\text{Q}(\text{r}_1-\text{r}_2)}{{4}\pi\in_0\text{r}_1\text{r}_2}$
Capacitance of the given system is given by,
$\text{C}\frac{\text{charge}(\text{Q})}{\text{potential difference}(\text{V})}$
$=\frac{{4}\pi\in_0\text{r}_1\text{r}_2}{\text{r}_1-\text{r}_2}$
Hence, proved.

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