Question

A solid sphere of radius $R$ has total charge $Q$, uniformly distributed. Find the minimum work done to decrease the radius of the sphere to $\frac{R}{2}$.

• A. $\frac{3}{5}\frac{K{Q}^2}{R}$
• B. $\frac{6}{5}\frac{K{Q}^2}{R}$
• C. $\frac{2}{5}\frac{K{Q}^2}{R}$
• D. $\frac{1}{5}\frac{K{Q}^2}{R}$

Answer 1

The correct option is A $\frac{3}{5}\frac{K{Q}^2}{R}$

We know that,
Self energy of a sphere of radius $R$ having uniformly distributed total charge $Q$ is,
$U_i=\frac{3}{5}\frac{K{Q}^2}{R}$

Now, energy of the sphere when it is compressed from radius $R$ to $R/2$,
$U_f=\frac{3}{5}\frac{K{Q}^2}{\left(R/2\right)}=\frac{6}{5}\frac{K{Q}^2}{R}$

Work done to compress the sphere will be equal to the change in self energy of the sphere.
i.e.$W_{comp}=U_f-U_i$

Substituting the values, we get

$W_{comp}=(\frac{6}{5}-\frac{3}{5})\frac{K{Q}^2}{R}$

$\therefore W_{comp}=\frac{3}{5}\frac{K{Q}^2}{R}$

Hence, option (a) is the correct answer.

Why this question? Self energy is just the work done in gathering the charges together from infinity. For a sphere, self energy $U=\frac{3}{5}\frac{K{Q}^2}{R}$