Potential Energy of Multi-Charge System

Total electrostatic PE of a system of n charges — sum over all unique pairs. For 3 charges: U = U₁₂ + U₁₃ + U₂₃. Counts each pair once (i < j condition).

Assembly argument

Bring charges one by one from infinity:

• $q_1$: $W_1 = 0$
• $q_2$: $W_2 = q_2 V_1(r_{12}) = \dfrac{q_1 q_2}{4\pi\varepsilon_0 r_{12}}$
• $q_3$: $W_3 = q_3[V_1(r_{13}) + V_2(r_{23})] = \dfrac{q_1 q_3}{4\pi\varepsilon_0 r_{13}} + \dfrac{q_2 q_3}{4\pi\varepsilon_0 r_{23}}$
• And so on...

Total PE is the sum of all pairwise contributions:

$$\boxed{U = \frac{1}{4\pi\varepsilon_0}\sum_{i < j} \frac{q_i q_j}{r_{ij}}}$$

The condition $i < j$ ensures each pair is counted exactly once.

Explicit form for 3 charges

$$U = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1 q_2}{r_{12}} + \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}}\right)$$

Alternative form

$$U = \frac{1}{2}\sum_{i=1}^{n} q_i V_i$$

where $V_i$ is the potential at the location of $q_i$ due to all other charges. The factor $\frac{1}{2}$ corrects for double-counting each pair.