Question

A sphere of radius $R$ is charged with charge $Q$. Find the self energy of sphere, if the sphere is made of conducting material.

• A. $\frac{Q^2}{8\pi {ϵ}_{0}R}$
• B. $\frac{Q^2}{4\pi {ϵ}_{0}R}$
• C. $\frac{3Q^2}{40\pi {ϵ}_{0}R}$
• D. $\frac{6Q^2}{40\pi {ϵ}_{0}R}$

Answer 1

The correct option is A $\frac{Q^2}{8\pi {ϵ}_{0}R}$

The sphere can be said to be a spherical capacitor of $R=0$ to $R=\infty$.
Since,
$C=\frac{4\pi {ϵ}_0}{\frac{1}{R_1}-\frac{1}{R_2}}=4\pi {ϵ}_{0}R$

Self energy:
$U=\frac{Q^2}{2C}=\frac{Q^2}{2\times 4\pi {ϵ}_{0}R}=\frac{Q^2}{8\pi {ϵ}_{0}R}$