Question

A solid non-conducting charged sphere has total charge $Q$ and radius $R$. If energy stored outside the sphere is $V_0$ joules, then find the self energy of the sphere in terms of $V_0$.

• A. $\frac{5}{6}{V}_0$
• B. $\frac{6}{5}{V}_0$
• C. $3V_0$
• D. $\frac{V_0}{5}$

Answer 1

The correct option is B $\frac{6}{5}{V}_0$

Let us consider a non-conducting sphere charged uniformly with a charge $Q$ on it.

We know that, in the outer region of the sphere, electric field is exactly the same as that of a charged conducting sphere.

Consider an elemental shell of radius $r(>R)$ and thickness $dr$ as shown in the figure.

Field energy stored in the volume of this shell is given as
$d{E}_{out}=\frac{1}{2}{\epsilon }_0{E}^{2}dV$
$\Rightarrow d{E}_{out}=\frac{1}{2}{\epsilon }_0{E}^2\times 4\pi r^{2}dr$

So, field energy stored in the surroundings of this sphere from its surface to infinity can be given as
$E_{out}={\int }_R^{\infty }\frac{1}{2}{\epsilon }_0{E}^2\times 4\pi r^{2}dr$
$\Rightarrow E_{out}=\frac{Q^2}{8\pi {\epsilon }_{0}R}$

For a non-conducting sphere, $E\ne 0$ at interior points. So, field energy exits in the interior region also.

Field energy stored in a elemental shell of radius $r(<R)$
$d{E}_{in}=\frac{1}{2}{\epsilon }_0{\left(\frac{Qr}{4\pi {\epsilon }_0{R}^3}\right)}^{2}4\pi r^{2}dr[\because E=\frac{Qr}{4\pi {\epsilon }_0{R}^3}]$

Integrating on both sides, we get

$E_{in}=\underset{0}{\overset{R}{\int }}\left(\frac{Q^2{r}^3}{8{\epsilon }_0{R}^6}\right)dr$

$\Rightarrow E_{in}=\frac{Q^2}{40\pi {\epsilon }_{0}R}$

Self energy of solid non-conducting sphere is equal to the total field energy given by
$E_{self}=E_{in}+E_{out}=\frac{3}{5}\frac{K{Q}^2}{R}$
where, $K=\frac{1}{4\pi {\epsilon }_0}$

Given, $E_{out}=V_0=\frac{K{Q}^2}{2R}$

$\therefore \frac{E_{self}}{E_{out}}=\frac{6}{5}$

$\Rightarrow E_{self}=\frac{6V_0}{5}$

Hence, option (b) is the correct answer.