Derive the formula for the electric potential energy of a system of two point charges.

(N/A) Consider two point charges $q_{1}$ and $q_{2}$ initially at infinity.
First,bring the charge $q_{1}$ from infinity to its position at point $P_{1}$. Since there is no external electric field,the work done $W_{1}$ is zero.
$W_{1} = 0$ ... $(1)$
This charge $q_{1}$ creates an electric potential $V_{1}$ at any point $P$ in space,given by $V_{1} = \frac{k q_{1}}{r_{1p}}$,where $r_{1p}$ is the distance from $q_{1}$ to point $P$.
Now,bring the charge $q_{2}$ from infinity to its position at point $P_{2}$,which is at a distance $r_{12}$ from $q_{1}$. The work done $W_{2}$ against the electric field of $q_{1}$ is:
$W_{2} = q_{2} V_{1} = q_{2} \left( \frac{k q_{1}}{r_{12}} \right) = \frac{k q_{1} q_{2}}{r_{12}}$ ... $(2)$
Since the electrostatic force is conservative,the total work done is stored as the electric potential energy $U$ of the system:
$U = W_{1} + W_{2} = 0 + \frac{k q_{1} q_{2}}{r_{12}}$
$\therefore U = \frac{k q_{1} q_{2}}{r_{12}}$
This expression is independent of the order in which the charges are brought to their positions. For a system of three charges,the potential energy is the sum of the interaction energies of all pairs:
$U = k \left[ \frac{q_{1} q_{2}}{r_{12}} + \frac{q_{1} q_{3}}{r_{13}} + \frac{q_{2} q_{3}}{r_{23}} \right]$
If $q_{1} q_{2} > 0$,the potential energy is positive (repulsive). If $q_{1} q_{2} < 0$,the potential energy is negative (attractive).