Module 2 : Electrostatics
Lecture 9 : Potential Energy of a System of Charges
The extra factor of $\displaystyle{\frac{1}{2}}$in the last expression is to ensure that each pair $(i,j)$is counted only once. The sum excludes the terms $i=j$. Since the potential at the $i$- th position due to all other charges is
 
\begin{displaymath}\phi(r_i) = \frac{1}{4\pi\epsilon_0}\sum_{j\ne i} \frac{q_i}{r_{ij}}\end{displaymath}
  we get
 
   
  Energy of a continuous charge distribution
  If is the density of charge distribution at $\vec{r}$, we can generalize the above result
 
   
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