Question

Consider an assembly of three conducting concentric spherical shell of radii a, b and c as shown in figure (31-E28). Find the capacitance of the assembly between the points A and B.

Answer 1

The spherical shells form two spherical capacitors: one made by A and B and the other made by B and C.



The capacitance of the spherical capacitor made by the shells of radii r₁ and r₂ is given by
$C=\frac{4\pi {\in }_0}{\left[{\displaystyle \frac{1}{r_1}}-{\displaystyle \frac{1}{r_2}}\right]}=\frac{4\pi {\in }_0{r}_1{r}_2}{r_2-r_1}$
The capacitance of the capacitor made by A and B is given by

$C_{\mathrm{AB}}=\frac{4\pi {\in }_{0}ab}{b-a}$

The capacitance of the capacitor made by B and C is given by

$C_{\mathrm{BC}}=\frac{4\pi {\in }_{0}bc}{c-b}$

As the capacitors are in series, the net capacitance is given by

$$\begin{gathered} \frac{1}{C}=\frac{1}{C_{\mathrm{AB}}}+\frac{1}{C_{\mathrm{BC}}} \\ \Rightarrow C=\frac{C_{\mathrm{AB}}{C}_{\mathrm{BC}}}{C_{\mathrm{AB}}+C_{\mathrm{BC}}}=\frac{{\displaystyle \frac{(4\pi {\in }_0{)}^{2}a{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \frac{4\pi {\in }_{0}ab}{(b-a)}}+{\displaystyle \frac{4\pi {\in }_{0}bc}{(c-b)}}} \\ \Rightarrow C=\frac{{\displaystyle \frac{4\pi {\in }_{0}a{b}^{2}c}{(b-a)(c-b)}}}{{\displaystyle \left(\frac{ab(c-b)+bc(b-a)}{(b-a)(c-b)}\right)}} \\ \Rightarrow C=\frac{4\pi {\in }_{0}a{b}^{2}c}{\left[ab\right(c-b)+bc(b-a\left)\right]} \\ \Rightarrow C=\frac{4\pi {\in }_{0}a{b}^{2}c}{b^2(c-a)}=\frac{4\pi {\in }_{0}ac}{(c-a)} \end{gathered}$$