Energy of a System of Charges

Two charges store potential energy $U = kq_1q_2/r$: positive if they repel, negative if they attract. What about three charges, or a dozen? The energy of the whole group is the work it takes to assemble it, bringing the charges in from far away one at a time.

Key idea

$$U = \sum_{\text{pairs } i < j}\frac{kq_iq_j}{r_{ij}}$$ One term for each pair, counted once. $N$ charges make $N(N-1)/2$ pairs.

Why it works

Assemble the charges one at a time. The first costs nothing: no other charges are there yet. The second costs $kq_1q_2/r_{12}$. The third moves through the potential of the first two, costing $q_3(V_1 + V_2) = kq_1q_3/r_{13} + kq_2q_3/r_{23}$. Each new charge adds one term for each charge already present, so in the end every pair appears exactly once. The order of assembly does not matter; the total is the same.

Three charges, three pairs. Drag them: pulling the attracting pair apart takes work, pushing the repelling pairs together takes work too.

This energy belongs to the arrangement as a whole, not to any one charge. Its sign tells a story. Positive $U$ means repulsion won while you built the group: let go, and the charges fly apart, turning $U$ into kinetic energy. Negative $U$ means attraction won: the group is bound, and you must supply $|U|$ to take it apart. The same bookkeeping explains why atoms and ionic crystals hold together.

Worked examples

Worked example: Three in a line

Charges of $+3.0$, $-2.0$ and $+1.0\ \mu\text{C}$ sit in a line, $0.10\ \text{m}$ apart in that order. Find $U$.

Three pairs: $(1,2)$ at $0.10\ \text{m}$, $(2,3)$ at $0.10\ \text{m}$, $(1,3)$ at $0.20\ \text{m}$.
$$U = k(10^{-12})\left(\frac{(3)(-2)}{0.10} + \frac{(-2)(1)}{0.10} + \frac{(3)(1)}{0.20}\right) = (8.99 \times 10^{-3})(-60 - 20 + 15)$$
$U = -0.584\ \text{J}$. Negative: assembling them released energy, and pulling them apart would take that much work.

Answer: $U \approx -0.584\ \text{J}$

Worked example: Four at the corners of a square

Four $2.0\ \mu\text{C}$ charges sit at the corners of a square with $0.10\ \text{m}$ sides. Find $U$.

$4 \cdot 3/2 = 6$ pairs: four along the sides ($r = a$) and two across the diagonals ($r = a\sqrt{2}$).
$$U = \frac{kq^2}{a}\left(4 + \frac{2}{\sqrt{2}}\right) = \frac{(8.99 \times 10^{9})(2.0 \times 10^{-6})^2}{0.10}(4 + \sqrt{2})$$
$U = 1.95\ \text{J}$.

Answer: $U \approx 1.95\ \text{J}$

Worked example: Adding one more

How much work does it take to bring a fifth charge, $+1.0\ \mu\text{C}$, from far away to the center of that square?

The work for the newcomer is its charge times the potential the others make there: $W = qV$. The energy among the first four does not change.
Each corner is $a/\sqrt{2} = 0.0707\ \text{m}$ from the center: $V = 4\cdot\dfrac{kq}{a/\sqrt{2}} = 1.02 \times 10^{6}\ \text{V}$.
$W = (1.0 \times 10^{-6})(1.02 \times 10^{6}) = 1.02\ \text{J}$.

Answer: $W \approx 1.02\ \text{J}$

Common mistake

Counting each pair twice, or skipping some. Summing $kq_iq_j/r_{ij}$ over every $i$ and every $j \ne i$ counts each pair twice and doubles $U$. Forgetting the diagonals of a square leaves pairs out. Check the number of terms: it must be $N(N-1)/2$, so 3 for three charges and 6 for four.

Energy of a System of Charges

Worked examples

Practice

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