Q.An electric dipole of dipole moment $6\times 10^{-9}\ \text{C}\,\text{m}$ is initially aligned with a uniform external field of magnitude $5\times 10^4\ \text{N/C}$. Find the work an external agent must do to rotate the dipole from this aligned orientation to an orientation perpendicular to the field.

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 $r_1$):

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