A spherical capacitor has outer sphere of radius \(5 \mathrm{~cm}\) and inner sphere of radius \(2…
A spherical capacitor has outer sphere of radius \(5 \mathrm{~cm}\) and inner sphere of radius \(2 \mathrm{~cm}\). When the inner sphere is earthed, its capacity is \(C_1\) and when the outer sphere is earthed its capacity is \(C_2\). Then \(\frac{C_1}{C_2}\) is
\(\frac{5}{2}\)
\(\frac{2}{5}\)
\(\frac{7}{3}\)
\(\frac{3}{7}\)
Solution
Key Idea If a spherical capacitor has outer radius \(R_2\) and inner radius \(R_1\). Then the capacitance,
(a) when outer shell is earthed,
\(C=4 \pi \varepsilon_0 \frac{R_1 R_2}{R_2-R_1}\)
(b) When inner shell is earthed,
\(C^{\prime}=C+4 \pi \varepsilon_0 R_2\)
Given, radius of outer sphere of capacitor, \(R_1=2 \mathrm{~cm}\) and Radius of inner sphere of capacitor, \(R_2=5 \mathrm{~cm}\)
\(\begin{aligned}
\therefore \quad C_1 & =4 \pi \varepsilon_0\left[\frac{R_1 R_2}{R_2-R_1}+R_2\right]=4 \pi \varepsilon_0\left[\frac{2 \times 5}{5-2}+5\right] \\
C_1 & =4 \pi \varepsilon_0 \frac{25}{3} \text { Farad }
\end{aligned}\)
and \(C_2=4 \pi \varepsilon_0 \frac{R_1 R_2}{R_2-R_1}=4 \pi \varepsilon_0 \frac{2 \times 5}{5-2}=4 \pi \varepsilon_0 \frac{10}{3}\)
\(\frac{C_1}{C_2}=\frac{\frac{25}{3}}{\frac{10}{3}}=\frac{25}{10}=\frac{5}{2}\)
Hence, the correct option is (a).