A signed scalar, counted pair by pair

For two point charges a distance $r$ apart, taking $U = 0$ at infinite separation,

$$U = \frac{kq_1q_2}{r}, \qquad \text{signs included}.$$

Absolute value bars belong to the force, where they keep a magnitude positive so that a drawn arrow can supply the direction. Energy has no direction to supply, so its sign is free to mean something, and it does:

1. Unlike pair: $U < 0$. The pair is bound, and energy has to be supplied to pull it apart.
2. Like pair: $U > 0$. The pair flies apart on its own, converting that $U$ into kinetic energy.

A proton and an electron therefore have $U = k(+e)(-e)/r$, which is negative. Separating them makes $r$ larger, which makes $U$ less negative, which is an increase. That is the check that catches the dropped sign: pulling a bound pair apart must raise $U$ toward zero.

The ladder from the first topic pays off here:

$$F \ \text{and} \ E \sim \frac{1}{r^2}, \qquad U \ \text{and} \ V \sim \frac{1}{r}.$$

So halving the separation of a pair doubles the magnitude of $U$ while it quadruples the force. Doubling the separation halves $|U|$.

The reason the exponents differ is worth one sentence, because it makes the pair memorable rather than arbitrary: energy is force accumulated over distance, and accumulating $1/r^2$ over a distance gives $1/r$. Any time you are unsure which rung you are on, ask whether the quantity has a direction. Vectors here carry $1/r^2$; scalars carry $1/r$.

The energy of a configuration is one $kq_iq_j/r_{ij}$ term per distinct pair. Three charges make three pairs, not three terms of some other kind:

$$U_{\text{total}} = \frac{kq_1q_2}{r_{12}} + \frac{kq_1q_3}{r_{13}} + \frac{kq_2q_3}{r_{23}}.$$

Four charges make six pairs. The count is $n(n-1)/2$, and listing the pairs before computing anything catches both failure modes at once.

The alternative route is walking through the charges and adding $q_iV_i$ for each, where $V_i$ is the potential at that charge's location due to all the others. That works, and it counts every interaction twice, once from each end, so it needs a factor of one half:

$$U_{\text{total}} = \tfrac{1}{2}\sum_i q_i V_i.$$

Use either method. Using the second without the one half is the common error, and so is bringing in a third charge and pairing it with only one of the two already placed.

There is a physical reading of that sum that makes it hard to miscount. Build the configuration by bringing charges in from infinity, one at a time, and add up the work you have to do.

1. The first charge costs nothing: there is nothing there yet to push against.
2. The second costs $kq_1q_2/r_{12}$: one new pair.
3. The third costs $kq_1q_3/r_{13} + kq_2q_3/r_{23}$: two new pairs, since it now has two partners.
4. The fourth costs three new pairs. And so on.

Add those up and you get $0 + 1 + 2 + 3 = 6$ pairs for four charges, which is the count again, arrived at by a route where forgetting a term feels like forgetting to push something. The total does not depend on the order you bring them in, which is what makes $U$ a property of the arrangement rather than of the history.