[NOTES/EM-03002]-Electrostatic Energy

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1. Electrostatic energy of system of point charges.

For a system of point charges the electrostatic energy is defined to be the work required to assemble the points charges by bringing them from infinity to their current position. Consider a system of n charges $q_1,q_2,q_3 \ldots$ at positions $r_1,r_2,r_3 \ldots$ respectively. Therefore, work done in bringing the first charge $q_1$ from $\infty$ to $r_1$ is $W_1
$ \begin{eqnarray} W_1=0 \label{eq1} \end{eqnarray}
Then the work done in bringing the second charge $q_2$ is $W_2$ and so on, therefore; 
\begin{eqnarray} W_2&=&\frac{q_1q_2}{4\pi\epsilon_0\mid\bar{r_1}-\bar{r_2}\mid} \label{eq2} \\
W_3&=&\frac{q_1q_3}{4\pi\epsilon_0\mid\bar{r_1}-\bar{r_3}\mid}+\frac{q_2q_3}{ 4\pi\epsilon_0\mid\bar{r_3}-\bar{r_2}\mid} \label{eq3} \\
W_4&=&\frac{q_1q_4}{4\pi\epsilon_0\mid\bar{r_1}-\bar{r_4}\mid}+\frac{q_2q_4}{ 4\pi\epsilon_0\mid\bar{r_2}-\bar{r_4}\mid}+\frac{q_3q_4}{4\pi\epsilon_0\mid\bar{r_3 }-\bar{r_4}\mid} \label{eq4}\end{eqnarray}
Therefore
\begin{eqnarray}
W_k&=&\frac{q_k}{4\pi\epsilon_0}\Sigma_{i=1}^{k-1}\frac{q_i}{\mid\bar{r_i}-\bar{ r_k}\mid} \label{eq5}\\
W&=&W_1+W_2+W_3+\ldots+W_n \label{eq6} &=&\frac{1}{4\pi\epsilon_0}\Sigma_{k=1}^n\Sigma_{i=1}^{k-1}\frac{q_iq_k}{ \mid\bar{r_i}-\bar{r_k}\mid} \label{eq7}. \end{eqnarray}

The above result can be cast in an alternate form;
\begin{eqnarray}
W&=&\frac{1}{4\pi\epsilon_0}\Sigma_{k=1}^n\Sigma_{i<k}\frac{q_iq_k}{\mid\bar{r_i
}-\bar{r_k}\mid} \label{eq8}\\
&=&\frac{1}{4\pi\epsilon_0}\Sigma_{i}\Sigma_{i\neq
k}\frac{q_iq_k}{\mid\bar{r_i}-\bar{r_k}\mid}\times \frac{1}{2} \label{eq9}\\
W&=&\frac{1}{2}\Sigma_i q_i \phi_k(\bar{r})\label{eq10}.
\end{eqnarray}
where $\phi_k(\bar{r})$ is the potential due to all charges other than the $i^{th}$ charge.