[NOTES/EM-03006]-Electrostatic Energy of a Uniformly Charged Solid Sphere

1. Uniformly charged solid sphere

We will compute the electrostatic energy of a uniformly charged sphere using \(W\)
\begin{eqnarray} W&=&\frac{\epsilon_0}{2}\int\int\int(\bar{E}\cdot\bar{E})\,d^3r \label{eq30}\\
d^3r&=&r^2\sin\theta dr\,d\theta d\phi\label{eq31}\\
\nonumber W&=&\frac{\epsilon_0}{2}\int_0^{\infty}dr\int_0^{\pi}d\theta \int_0^{2\pi}d\phi\, (\bar{E}\cdot\bar{E})\,r^2. \end{eqnarray}
The magnitude of the electric field of a uniformly charged sphere is
\begin{equation} \mid\bar{E}\mid =
\begin{cases} \dfrac{Q r}{(4\pi\epsilon_0) R^3}, & \text{for }   r<R,\\
\dfrac{q}{(4\pi\epsilon_0)r^2}, & \text{for }   r >R . \end{cases} \end{equation}
Thus the electrostatic energy is given by
\begin{eqnarray} W&=&\frac{\epsilon_0}{2}(4\pi)\int_0^{\infty}r^2\mid\bar{E}\mid^2dr\\
\label{eq34} \nonumber &=&(2\pi\epsilon_0) \int_0^R \Big(\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3} \Big)^2r^2\, dr+(2\pi\epsilon_0) \int_R^{\infty} \Big(\frac{1}{4\pi\epsilon_0}\frac{Q}{ r^2}\Big)^2\,dr\\
\nonumber &=&(2\pi\epsilon_0)\Big(\frac{Q}{4\pi\epsilon_0\,R^6}\Big)^2\Big\{ \int_0^Rr^4dr+\int_R^{\infty}\frac{1}{r^2}dr\Big\}\\
\nonumber &=&\frac{Q^2}{8\pi\epsilon_0}\Big(\frac{1}{5R}+\frac{1}{R}\Big)\\
\nonumber &=&\frac{Q^2}{8\pi\epsilon_0}\frac{6}{5R} \\
&=&\frac{3}{5}\Big(\frac{Q^2}{4\pi\epsilon_0R}\Big) \label{eq35}
\end{eqnarray}