[NOTES/EM-03014] Discussion of Electrostatic Energy

We have two expressions for the electrostatic energy \begin{eqnarray} W'&=&\frac{1}{8\pi\epsilon_0}\sum_{i\neq j}\frac{q_iq_j}{\mid\vec {r_i}-\vec {r_j}\mid}\label{eq22}\\ W''&=&\frac{\epsilon_0}{2}\int\vec {E}\cdot\vec {E}d^3r \label{eq23}. \end{eqnarray}\(W^\prime\) is for electrostatic energy of a system of point charges  and \(W^{\prime\prime}\) for continuous  charge distributions. Here $W'$ can be negative whereas $W''$ is always positive.
To understand the difference in the two expressions, we may regard a point charge as a limit of uniformly charged sphere as its radius goes to zero. $W''$ for a uniformly charged sphere is
\begin{eqnarray}
W''&=&\frac{1}{4\pi\epsilon_0}\frac{3Q^2}{5R}\label{eq24}
\end{eqnarray}
where $R$ is the radius of charge distribution. Note that in the limit  \(R\to 0\), the energy $W'' \rightarrow \infty $. Thus, the difference between \(W''\) and \(W'\) arises from the fact that \(W'\) does not include the energy required to assemble point charges. We will continue to use $W'$ for point charges and $W''$ for continuous charge distributions.

Superposition principle
Electrostatic energy does not obey superposition principle like electric fields
do. If \(\vec E=\vec E_1+\vec E_2\), where \(\vec E_{1,2}\) are electric fields of two charge distributions, we have
\begin{eqnarray}
W&=&\frac{\epsilon_0}{2}\int\mid\vec {E}\mid^2d^3r \label{eq25}\\
&=&\frac{\epsilon_0}{2}\int\mid\vec {E_1}\mid^2d^3r+\frac{\epsilon_0}{2}
\int\mid\vec {E_2}\mid^2d^3r +
\frac{\epsilon_0}{2}\int\mid\vec {E_1}\cdot\vec {E_2}\mid d^3r\label{eq26}\\
\nonumber&\neq&\text{energy due to charge distribution 1 + energy due to charge
distribution 2} \label{eq27}
\end{eqnarray}
The last  term in \eqref{eq26} is the {\tt interaction energy } of the two charge distributions.