Question

A total charge $Q$ is distributed uniformly into a spherical volume of radius $R$. Find the electrostatic energy of this configuration.

• A. $\frac{Q^2}{4\pi {ϵ}_{0}R}$
• B. $\frac{3Q^2}{5\pi {ϵ}_{0}R}$
• C. $\frac{3Q^2}{20\pi {ϵ}_{0}R}$
• D. $\frac{3Q^2}{10\pi {ϵ}_{0}R}$

Answer 1

Answer: (C)

$\frac{3Q^2}{20\pi {ϵ}_{0}R}$

Given a total charge $Q$ is distributed uniformly into a spherical volume of radius $R$. We have to find the electrostatic energy of this configuration.

For this we use the formula, the electrostatic energy $V_o=\int \frac{{\epsilon }_0}{2}{E}^{2}dl$

where, ${\epsilon }_0=$ permeability of free space

$E=$ electric field

$dl=$ Volume element

For this the electric field can be easily found from Gauss' Theorem to be,

$E=\frac{Q}{4\pi {\epsilon }_0}\times \left\{\begin{array}{cc}\frac{r}{R^3}& (r<R)\\ \frac{1}{r^2}& (r>R)\end{array}\right\}$

The two intergrals needed are:

${\int }_0^R{\left(\frac{r}{R^3}\right)}^{2}4\pi r^{2}dr=\frac{4\pi }{5R}$

${\int }_R^{\infty }{\left(\frac{1}{r^2}\right)}^{2}4\pi r^{2}dr=\frac{4\pi }{R}$

So, we obtain $V_o={\int }_0^{\infty }\frac{{\epsilon }_0}{2}{E}^{2}4\pi r^{2}dr$

$[d\tau =4\pi r^{2}dr]$

So, $V_o=\frac{{\epsilon }_0}{2}{\left(\frac{Q}{4\pi {\epsilon }_0}\right)}^2(\frac{4\pi }{5R}+\frac{4\pi }{R})$

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

$=\frac{3}{20}\frac{Q^2}{\pi {\epsilon }_{0}R}$