Electrostatic potential energy

Push two like charges toward each other and you feel them shove back — get close enough and let go, and they fly apart, converting stored energy into motion. That stored energy is electrostatic potential energy: the work an external agent must do, against the mutual Coulomb force, to assemble a set of charges starting from infinite separation. It is not a new force law but a bookkeeping tool built directly on the potential $V$ you already know — bring one charge in for free, then charge $q_2V_1$ to bring the next one in. This article builds that bookkeeping for two charges, extends it by superposition to three, and then applies the same energy idea to a dipole sitting in an external field, where the natural variable is not distance but orientation angle $\theta$ .

The core insight

The electrostatic potential energy of a charge configuration is the work needed to assemble it from infinite separation, computed for point charges as $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ summed over every pair, and for a dipole in a uniform field as $U=-\vec p\cdot\vec E=-pE\cos\theta$ .

The core idea

Two point charges. Imagine bringing $q_1$ and $q_2$ in from infinity to a final separation $r_{12}$ , with no leftover kinetic energy. Placing $q_1$ first costs nothing — there is no field yet to push against. Bringing $q_2$ in from infinity then costs $W=q_2V_1$ , where $V_1$ is the potential $q_1$ creates at the final spot. That work is stored as potential energy:

U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}

where: $q_1,q_2$ are the point charges (C, with sign), $r_{12}$ is their final separation (m), and $\dfrac{1}{4\pi\varepsilon_0}=k=9\times10^{9}\ \text{N m}^2\text{C}^{-2}$ . The sign of $U$ carries physical meaning: same-sign charges give $U>0$ (they repel, so an agent must push them together, storing positive energy), opposite-sign charges give $U<0$ (they attract, so assembling them releases energy — the agent actually restrains them).

Three (or more) point charges. By superposition, the total assembly energy is just the sum over every distinct pair, because the order you bring the charges in does not matter:

U=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_1q_2}{r_{12}}+\frac{q_2q_3}{r_{23}}+\frac{q_1q_3}{r_{13}}\right]

where: $r_{12},r_{23},r_{13}$ are the three pairwise separations. For $N$ charges this generalizes to a sum over all $\binom{N}{2}$ distinct pairs — never $N$ terms, and never double-count a pair.

A dipole in a uniform external field. Here the "distance" being varied is really an orientation angle $\theta$ between the dipole moment $\vec p$ and the field $\vec E$ . Taking the reference $U=0$ at $\theta=90°$ (dipole perpendicular to the field):

U=-\vec p\cdot\vec E=-pE\cos\theta

where: $p$ is the dipole moment magnitude (C·m), $E$ is the field strength (N/C or V/m), and $\theta$ is the angle between $\vec p$ and $\vec E$ . This is the negative of the work the field's torque does rotating the dipole from $90°$ to $\theta$ . At $\theta=0°$ , $U=-pE$ : the deepest, most stable configuration, matching the zero-torque stable equilibrium you already met when studying dipoles in a field. At $\theta=180°$ , $U=+pE$ : the unstable equilibrium. The system always tends toward lower $U$ , so a dipole released at some angle rotates toward $\theta=0°$ .

Figure

Potential energy of a dipole plotted against orientation angle showing a minimum at zero degrees and a maximum at one hundred eighty degrees

Derivation APotential energy of two point charges from the work of assembly

The goal is to derive $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ from first principles by computing the work an external agent does bringing two point charges together from infinite separation, using only the definition of electric potential and the fact that the first charge placed experiences no field at all.

Step 1. Place $q_1$ at its final position. Since space is initially empty of any field, this step requires zero work: $W_1=0$ .

Step 2. Bring $q_2$ from infinity to a distance $r_{12}$ from $q_1$ , moving it slowly (no kinetic energy gained or lost) against the field of $q_1$ . The work done by the external agent equals $q_2$ times the potential $V_1$ that $q_1$ produces at the final location: $W_2=q_2V_1$ , and $V_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{12}}$ .

Step 3. Combine the two steps. Total work done in assembling the pair is $W=W_1+W_2=0+q_2\times\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1}{r_{12}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ .

Step 4. Identify this work with stored potential energy. Because the charges gained no kinetic energy, all the external work went into the configuration itself, so $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$ . The result is symmetric in $q_1$ and $q_2$ (it does not matter which charge you place first), and it correctly reduces to $U=0$ at $r_{12}\to\infty$ , matching the reference level chosen at the start.

Worked example

Two point charges, $q_1=+3\ \mu\text{C}$ and $q_2=-2\ \mu\text{C}$ , start infinitely far apart. Find the work done by an external agent bringing them to a separation of $15\ \text{cm}$ , and state which direction the electrostatic force itself did work.

Step 1. Since the charges begin at infinite separation where $U=0$ , the work done by the external agent equals the final potential energy: $U=k\dfrac{q_1q_2}{r}$ .

Step 2. Substitute values, keeping careful track of the negative sign on $q_2$ : $U=(9\times10^{9})\times\dfrac{(3\times10^{-6})\times(-2\times10^{-6})}{0.15}=(9\times10^{9})\times\dfrac{-6\times10^{-12}}{0.15}$ .

Step 3. Finish the arithmetic: $U=(9\times10^{9})\times(-4\times10^{-11})=-0.36\ \text{J}$ .

Step 4. Interpret the sign. The charges are opposite in sign, so they attract; an agent bringing them together slowly must actually hold them back, doing negative work on the system. The external agent does $-0.36\ \text{J}$ of work; equivalently, the electrostatic force itself does $+0.36\ \text{J}$ of positive work as the charges come together.

Competitive stretch

Now place a third charge $q_3=-4\ \mu\text{C}$ so that all three sit at the corners of an equilateral triangle of side $r=10\ \text{cm}$ , with $q_1=+2\ \mu\text{C}$ , $q_2=+3\ \mu\text{C}$ as before (a fresh triangle, not the pair above). All three pairwise separations equal $r$ , so $U=\dfrac{k}{r}\left[q_1q_2+q_2q_3+q_1q_3\right]$ . In units of $10^{-12}\ \text{C}^2$ : $q_1q_2=6$ , $q_2q_3=-12$ , $q_1q_3=-8$ , summing to $-14$ . Then $U=\dfrac{9\times10^{9}}{0.10}\times(-14\times10^{-12})=-1.26\ \text{J}$ . The negative total shows the two attractive pairs ( $q_2q_3$ and $q_1q_3$ ) outweigh the one repulsive pair ( $q_1q_2$ ), so the electrostatic forces do net positive work bringing this particular arrangement together from infinity.

Common traps

Assuming two-charge potential energy is always positive — it is negative whenever the charges have opposite signs, since assembling attracting charges releases rather than costs net work.
Missing pairs in a multi-charge sum — an $N$ -charge system has $\binom{N}{2}$ distinct pairs, not $N$ terms; a three-charge system has exactly 3 pair-terms, not 2 or 4.
Mixing up $U=-pE\cos\theta$ with the torque formula $\tau=pE\sin\theta$ — one gives energy, the other gives rotational force, and they use different trig functions of the same angle.
Thinking dipole potential energy is zero at $\theta=0°$ — it is instead the minimum, $-pE$ ; zero is only the conventional reference at $\theta=90°$ .

Ask the tutor: want a walkthrough of a four-charge potential-energy sum, or a check on how the dipole energy formula connects to the torque formula through work-energy reasoning?

Why does placing the first charge cost zero work but the second one costs something?

Work is required only when a charge moves through a field created by other charges already present. When you place the very first charge, no field exists yet anywhere, so there is nothing to push against and the work is zero. Once that charge sits there, it produces a potential $V_1$ everywhere in space, so bringing in a second charge $q_2$ from infinity to a point where that potential is nonzero costs $W=q_2V_1$ . With three or more charges, each new charge you add sees the combined potential of all the ones already placed, which is exactly why the total energy ends up as a sum over pairs.

Does the order in which I bring the charges together change the final potential energy?

No. Potential energy is a property of the final configuration alone, not of the path taken to reach it — this is a consequence of the electrostatic force being conservative. You could bring $q_2$ in first and $q_1$ second, or bring all three in simultaneously in any order, and the total work done always sums to the same $U=\dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1q_2}{r_{12}}+\dfrac{q_2q_3}{r_{23}}+\dfrac{q_1q_3}{r_{13}}\right]$ . This is exactly why superposition works so cleanly here: each pair contributes independently of assembly order.

How is the dipole formula U equals negative p dot E related to the two-charge formula?

They describe different situations and are not derived from one another in this course: the two-charge formula gives the energy of assembling a pair of point charges from infinity, while the dipole formula gives the energy of a fixed dipole (already assembled) as its orientation changes inside someone else's uniform external field. Both, however, share the same underlying idea — potential energy as stored work — and both use the same reference-point convention, choosing $U=0$ at a natural "neutral" configuration ( $r\to\infty$ for two charges, $\theta=90°$ for the dipole).

Electrostatic potential energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads