Question

Find the minimum work done to form a solid sphere of radius $R$ and total charge $Q$.
[Assume all the charges are at infinity initially]

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

Answer 1

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

Imagine that we assemble the sphere by building up a succession of thin spherical layers of infinitesimal thickness.
At each stage of the process, we gather a small amount of charge and put it in a thin shell from $r$ to $r+dr$.
We continue the process until we arrive at the final radius $R$.

If $Q_r$ is the charge of the sphere when it has been built up to the radius $r$, the work done in bringing a charge $dQ$ to it is
$dW=dQ\times V_r$
$\Rightarrow dW=\frac{Q_{r}dQ}{4\pi {\epsilon }_{0}r}...\left(1\right)$

If the density of charge in the sphere is $\rho$, the charge $Q_r$ is $Q_r=\rho \times \frac{4}{3}\pi r^3$.

and the charge $dQ$ is $dQ=\rho \times 4\pi r^{2}dr$

Substituting the value in $\left(1\right)$, we get
$dW=\frac{4\pi {\rho }^2{r}^{4}dr}{3{ϵ}_0}$.

The work done required to assemble the sphere is the integral of $dW$ from $r=0$ to $r=R$

$\int dW={\int }_0^R\frac{4\pi {\rho }^2{r}^{4}dr}{3{\epsilon }_0}$

$\Rightarrow W=\frac{4\pi {\rho }^2{R}^5}{15{\epsilon }_0}$

Work done in terms of the total charge $Q=\rho \frac{4}{3}\pi R^3$ of the sphere,

$W=\frac{4\pi {\left(\frac{3Q}{4\pi R^3}\right)}^2{R}^5}{15{\epsilon }_0}$

$\Rightarrow W=\frac{1}{4\pi {\epsilon }_0}\frac{3}{5}\frac{Q^2}{R}$

Putting $\frac{1}{4\pi {\epsilon }_0}=K$, we get $W=\frac{3}{5}\frac{K{Q}^2}{R}$

Hence, option (a) is the correct answer.

Why this question? Note : The work done in gathering the charges together from infinity is called the self energy of the body. It is also the work required to break the system apart [constant of binding energy]