POTENTIAL ENERGY OF CHARGES IN AN EXTERNAL ELECTRIC FIELD
Derive an expression for the potential energy in the cases of (i) single point charge and (ii) system of two charges in an external electric field.
Potential Energy of a Single Charge in an External Electric Field :
Let an external field $\vec{E}$ have different values of electric potential at different points.
Consider a point P, distant $\vec{r}$ from the origin in this field having electric potential as $V(\vec{r})$.
Then, work done in bringing charge $q$ from infinity to point P is given by,
$W = qV(\vec{r})$.
This work done is stored as potential energy of the charge $q$.
Thus, potential energy of charge $q$ at position vector $\vec{r}$ in external field is given by,
$U = qV(\vec{r})$
Potential energy of a system of two charges in an external electric field.
Let $q_1$ and $q_2$ be two charges placed at points P and Q having position vectors $\vec{r}_1$ and $\vec{r}_2$ respectively in an external field $\vec{E}$ (Figure 25). In bringing $q_1$ from infinity to point P, work done, $W_1 = q_1V(\vec{r}_1)$, where $V(\vec{r}_1)$ is the potential at P due to external electric field.
In bringing charge $q_2$ from infinity to point Q, work done, $W_2 = q_2V(\vec{r}_2)$, where $V(\vec{r}_2)$ is the potential at Q due to external electric field.
If $r_{12}$ is the distance between point P and point Q, then,
Work done on charge $q_2$ against the electric field due to charge $q_1$,
$W_3 = \frac{q_1q_2}{4\pi\epsilon_0 r_{12}}$.
Using superposition principle, potential energy of the system of two charges in an external electric field is given by
$ U = W_1 + W_2 + W_3$
$U=q_1V(\vec{r}_1) + q_2V(\vec{r}_2) + \frac{q_1q_2}{4\pi\epsilon_0 r_{12}}$