Potential Energy of a System of Two Charges in an External Field

Concept First

When a system of charges is placed in an external electric field, the total potential energy has two parts:

1. The energy needed to bring each charge from infinity to its position in the external field (against the external field alone).
2. The mutual interaction energy between the charges themselves (the work done to bring them close to each other against each other's fields).

The external field is assumed to be independent of the charges we bring in — it is produced by other sources (e.g., distant charges or a capacitor).

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Derivation: Two charges $q_1$ and $q_2$ in an external field $\mathbf{E}$

Let the external electric potential at a point $\mathbf{r}$ be $V(\mathbf{r})$, such that $\mathbf{E} = -\nabla V$.

Step 1: Bring $q_1$ from infinity to $\mathbf{r}_1$

Since there is no other charge yet, the only work is against the external field:

Step 2: Bring $q_2$ from infinity to $\mathbf{r}_2$

Now $q_1$ is already present. The work done on $q_2$ has two contributions:

• Against the external field: $q_2 V(\mathbf{r}_2)$
• Against the electric field of $q_1$: The potential due to $q_1$ at the location of $q_2$ is $\frac{1}{4\pi\varepsilon_0} \frac{q_1}{r_{12}}$, where $r_{12} = |\mathbf{r}_2 - \mathbf{r}_1|$. So the work done against $q_1$'s field is:

By the superposition principle, the total work in this step is:

$W_2 = q_2 V(\mathbf{r}_2) + \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}}$ …