Q.Deduce the expression for the potential energy of a system of two point charges q1 and q2 brought from infinity to the points with positions r1 and r2 respectively in presence of external electric field E. OR A parallel plate capacitor is charged to a potential difference V by a DC source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state the reason, how the following will change:

The total potential energy is the work done against the external field to place each charge, plus the mutual interaction energy of the two charges.

Let $V(r_1)$ and $V(r_2)$ be the potentials due to the external field $\vec E$ at positions $r_1$ and $r_2$.

Step 1 — bring $q_1$ from infinity to $r_1$: Work done against the external field $= q_1V(r_1)$. (No other charge is present yet, so this is the only contribution.)

Step 2 — bring $q_2$ from infinity to $r_2$: Work is done against two fields — the external field, and the field of $q_1$ (now fixed at $r_1$):

$W_2 = q_2V(r_2) + \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}$ …