What it costs to put charges where they are

Move a charge through an electrostatic field and the field does work on it. That work depends only on where the charge started and where it ended, never on the route: the electrostatic force is conservative, exactly like gravity. A force with that property earns a potential energy, defined so that

$$W_{\text{field}} = -\Delta U = -(U_f - U_i).$$

Read the minus sign carefully, because it is the single most-flipped sign in the unit. Positive work by the field DRAINS potential energy. The stored energy goes down and, if nothing else is acting, kinetic energy goes up by the same amount. A charge that speeds up is a charge whose $U$ fell.

An external agent that moves the charge slowly, so it starts and ends at rest, is doing the opposite bookkeeping:

$$W_{\text{ext}} = +\Delta U = q\,\Delta V.$$

Two agents, two ledgers, opposite signs, and they sum to the change in kinetic energy, which for a slow move is zero. The way to stay out of trouble is to name whose work you are computing before writing a single number, and never to reuse one $W$ as the other.

Because only differences in $U$ carry physics, the zero is a choice. The standard choice for point charges is $U \to 0$ as $r \to \infty$: infinitely far apart is defined as costing nothing. Every formula below assumes it.

For two point charges a distance $r$ apart, the potential energy of the pair is

$$U = \frac{kq_1q_2}{r}.$$

Compare it to the force magnitude, $F = k|q_1||q_2|/r^2$, and note the two deliberate differences.

1. One power of $r$, not two. $F$ and $E$ go as $1/r^2$; $U$ and $V$ go as $1/r$. Doubling the separation quarters the force and only halves the energy.
2. The signs stay in. There are no absolute-value bars here, and putting them in is not a simplification, it is a different quantity. $U$ is a scalar, so its sign is not a direction and does not need to be argued away.

What the sign of $U$ actually encodes is bound versus unbound. Unlike charges give $U$ negative: the pair sits in a well, and separating them to infinity requires that you supply energy, raising $U$ up toward zero. Like charges give $U$ positive: the pair is loaded like a compressed spring, and releasing it converts that positive $U$ into kinetic energy as they fly apart.

So a proton and an electron $0.053$ nm apart have $U = -4.3\times10^{-18}$ J, not $+4.3\times10^{-18}$ J. Report the positive number and you have claimed that a hydrogen atom will spontaneously fly apart, which is a strong claim to make by dropping a minus sign.

The next topic builds the whole unit out of one quotient, so it is worth naming the distinction now rather than untangling it later.

$$V = \frac{U}{q_0} \qquad \Longleftrightarrow \qquad U = qV.$$

$V$ is energy per unit charge: a property of a point in space, produced by whatever source charges are around, and it exists at that point whether or not anything is sitting there. $U$ is energy: it belongs to a specific charge placed at that point, together with the sources.

The units are the tell. A volt is a joule per coulomb. So "the potential at $P$ is $12$ V" and "the energy of the charge at $P$ is $12$ J" are different statements about different quantities, and only one of them mentions the charge. Put an electron at that $12$ V point and its potential energy is

$$U = qV = (-1.6\times10^{-19}\ \text{C})(12\ \text{V}) = -1.9\times10^{-18}\ \text{J},$$

which is neither $12$ of anything nor positive. Double the charge you place there and $U$ doubles while $V$ does not move at all, because $V$ never knew about your charge.

A useful convenience unit falls straight out of $U = qV$. One electron volt is the energy an elementary charge picks up crossing one volt: $1\ \text{eV} = 1.6\times10^{-19}$ J. It turns most atomic-scale answers into small whole numbers.

The energy stored in a group of charges is the total work an external agent had to do to bring them in from infinity, one at a time, to their final positions. For $n$ charges that total is

$$U = \sum_{\text{distinct pairs}} \frac{kq_iq_j}{r_{ij}}.$$

Every pair contributes once. Three charges have exactly three pairs, $12$, $13$ and $23$, so three terms; four charges have six. Listing the pairs by name before adding anything is the audit that catches the two standard failures.

1. Dropping a term. Bring in the third charge and it interacts with both of the first two, not just the nearest. Two terms arrive with the third charge, not one.
2. Counting every term twice. The tempting shortcut $\sum_i q_iV_i$, where $V_i$ is the potential at charge $i$ from all the others, sees pair $12$ once from charge 1 and again from charge 2. It gives exactly $2U$, so $$U = \tfrac{1}{2}\sum_i q_iV_i.$$

Neither the total nor any individual pair term depends on the order you assemble the charges in. The step-by-step works genuinely differ, since the first charge in costs nothing and the last one arrives into a crowded field, but they always sum to the same total, because each pair term depends only on the final separation.