Question

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

• A. $\frac{4\pi {\rho }^2{R}^5}{15{ϵ}_0}$
• B. $\frac{\pi {\rho }^2{R}^5}{15{ϵ}_0}$
• C. $\frac{8\pi {\rho }^2{R}^5}{15{ϵ}_0}$
• D. $\frac{11\pi {\rho }^2{R}^5}{15{ϵ}_0}$

Answer 1

The correct option is A $\frac{4\pi {\rho }^2{R}^5}{15{ϵ}_0}$

Charge density $=\rho$ radius $=R$

The work $dw$ done in bringing up each increment of charge $dq$ to the radius

$dw=\frac{1}{4\pi {ϵ}_o}q\frac{\left(r\right)dq}{r}$

Now, $q\left(r\right)$ Amount of charge already assembled up to radius $r$.

$q\left(r\right)=\frac{4}{3}\pi r^3\rho$

$dq=\frac{4}{3}\pi 3r^2\rho dr$

$4\pi r^2\rho dr$

$\therefore dw=\frac{1}{4\pi {ϵ}_o}\left(\frac{4}{3}\pi r^3\rho \right)\left(4\pi r^2\rho \right)\frac{dr}{r}$

$=\frac{4\pi }{3{ϵ}_o}{\rho }^2{r}^{4}dr$

The toatl work done in assembling the sphere of charge to radius $R$ is

$\omega ={\int }_0^R\frac{4\pi }{3{ϵ}_o}{\rho }^2{r}^{4}dr$

$=\frac{4\pi }{15{ϵ}_o}{\rho }^2{R}^5$