The equation of potential energy of a system of two charges is,

$\mathrm{U}(\theta)=q_{1} \mathrm{~V}\left(r_{1}\right)+q_{2} \mathrm{~V}\left(r_{2}\right)+\frac{k q_{1} q_{2}}{r_{12}}$

For dipole $q_{1}=+q, q_{2}=-q$ and their position vectors are $\overrightarrow{r_{1}}$ and $\overrightarrow{r_{2}}$. $\therefore \quad U^{\prime}(\theta)=q\left[V\left(r_{1}\right)-V\left(r_{2}\right)\right]-\frac{k q_{1} q_{2}}{r_{12}}$

$\ldots$ (1)

The potential difference between positions $r_{1}$ and $r_{2}$ equals the work done in bringing a unit positive charge against field from $r_{2}$ to $r_{1}$. The displacement parallel to the force is $2 a \cos \theta$.

$\therefore \mathrm{V}\left(r_{1}\right)-\mathrm{V}\left(r_{2}\right) =-\mathrm{E} \times 2 a \cos \theta[\mathrm{W}=\mathrm{F} d]$

$=-p \mathrm{E} \cos \theta \quad[\because p=q(2 a)=2 a]$

$\therefore \mathrm{U}^{\prime}(\theta)=-p \mathrm{E} \cos \theta-\frac{k q^{2}}{2 a}$

$\therefore \mathrm{U}^{\prime}(\theta)=-(\vec{p} \cdot \overrightarrow{\mathrm{E}})-\frac{k q^{2}}{2 a}$

$\ldots$ $(2)$

For a given dipole $U(\theta)$ differs from $U\left(\theta^{\prime}\right)$ only by a constant.

If $\theta_{0}=\frac{\pi}{2}$ be taken, then $q\left[\mathrm{~V}\left(r_{1}\right)-\mathrm{V}\left(r_{2}\right)\right]=0$

If we drop the second term in equation (2) then,

$\mathrm{U}^{\prime}(\theta)=-(\vec{p} \cdot \overrightarrow{\mathrm{E}})$ is like $\mathrm{U}=-(\vec{p} \cdot \overrightarrow{\mathrm{E}})$