Electric Potential Energy

Hold one proton fixed and push a second proton toward it. The repulsion fights you the whole way, so you have to do work, and when you let go the second proton flies off and gets that work back as kinetic energy. The work went somewhere in between: it was stored in the arrangement of the two charges. That stored energy is the electric potential energy of the system, and because the electric force is conservative it depends only on where the charges are, not on how they got there.

Potential energy as the work to assemble the charges

Suppose a charge $q_1$ sits fixed and you bring a second charge $q_2$ in from very far away. Move it slowly, so that at every instant your push exactly cancels the electric force and $q_2$ gains no kinetic energy. Then all of your work goes into the system's potential energy:

The work done by the field is the negative of yours because the two forces are equal and opposite at every point. This is the same bookkeeping as lifting a book slowly in mechanics: the work you do equals the gain in gravitational potential energy.

To get a formula, carry $q_2$ straight in along a radial line from $r = \infty$ to a separation $r$. The Coulomb force on $q_2$ is $\frac{q_1q_2}{4\pi\varepsilon_0 r'^2}$, pointing outward when the product $q_1q_2$ is positive, so your force is the opposite of that. Integrating your force along the path from $\infty$ in to $r$ (the limits take care of the inward direction), the work you do is

The path didn't matter. On any curved path, break the motion into small radial steps and small steps along circles centered on $q_1$. The force is radial, so the circular steps take no work, and the radial steps add up to the same integral. A force whose work depends only on the endpoints is conservative, and only conservative forces have a potential energy.

Choosing $U = 0$ when the charges are infinitely far apart (where they don't interact at all) gives the formula.

Notice the power of $r$: $\frac{1}{r}$, not the $\frac{1}{r^2}$ of the force. Doubling the separation halves the energy but quarters the force.

What the sign of the energy tells you

The sign of $U_E$ is the sign of $q_1q_2$, and each sign has a physical meaning.

• Like charges: $U_E > 0$. You had to push to bring them together. Released, they fly apart and convert the stored energy to kinetic energy. The closer they are, the more energy is stored, and $U_E$ grows without limit as $r \to 0$.
• Unlike charges: $U_E < 0$. They attract, so bringing them in from infinity, you hold them back and do negative work. Pulling them apart again takes positive work, exactly $|U_E|$, to get back to zero. A negative potential energy marks a bound system: the hydrogen atom, or an ion pair in a salt crystal.

For example, $+2.0\ \mu\text{C}$ and $-3.0\ \mu\text{C}$ held $0.15\ \text{m}$ apart have

so separating them completely, slowly, takes $+0.36\ \text{J}$ of external work.

The graph of $U_E$ against $r$ is a $\frac{1}{r}$ curve: above the axis for like charges and below it for unlike charges, approaching zero far away in both cases. As in mechanics, the force along the line joining the charges is the negative slope, $F_r = -\frac{dU_E}{dr}$. For like charges the slope is negative, so $F_r > 0$: the force pushes the charges apart, toward lower energy. For unlike charges the slope is positive, so $F_r < 0$: attraction. Either way, a system left alone moves toward lower potential energy. Differentiating checks the formula: $-\frac{d}{dr}\left(\frac{q_1q_2}{4\pi\varepsilon_0 r}\right) = \frac{q_1q_2}{4\pi\varepsilon_0 r^2}$, Coulomb's law.

Systems of three or more charges

Build the system one charge at a time. Bringing in the first charge costs nothing, since nothing else is there yet. The second costs the energy of the pair it forms with the first. The third interacts with both, and since forces add, the work to bring it in is the sum of the energies of its pairs with charges 1 and 2. Continuing, the total is a sum over every pair, each counted once:

For three equal charges $q$ at the corners of an equilateral triangle of side $a$, all three pairs are identical, so $U_E = \frac{3q^2}{4\pi\varepsilon_0 a}$.

Two consequences come up constantly.

• Adding or removing one charge changes the energy only by that charge's pairs. Pairs between the other charges are untouched, so they cancel out of $\Delta U$.
• Moving one charge while the others stay fixed changes only the pairs that involve the moving charge. If a question asks for the work to move it, the fixed charges' energy with each other is a constant you can leave out. (Topic 9.2 packages this as the charge times the electric potential at its location.)

Worked examples

Worked exampleExample 3: a charge between two fixed charges (free-response style)

Two particles, each with charge $+Q$, are fixed on the $x$-axis at $x = -a$ and $x = +a$. A particle with charge $-q$ can move along the $y$-axis. Express algebraic answers in terms of $Q$, $q$, $a$, $y$ and physical constants.

(a) Derive an expression for the electric potential energy of the three-particle system when the $-q$ particle is at position $y$.

(b) Sketch $U_E$ as a function of $y$ for $y$ from $-3a$ to $3a$.

(c) Use your answer to (a) to find the $y$-component of the electric force on the $-q$ particle. Verify it against Coulomb's law.

(d) Calculate the work an external agent must do to move the $-q$ particle slowly from the origin to a point very far away, for $Q = 4.0\ \text{nC}$, $q = 1.0\ \text{nC}$ and $a = 3.0\ \text{cm}$.

Show solution

(a) Sum over the three pairs. The two $+Q$ charges are $2a$ apart. The $-q$ particle is $\sqrt{a^2 + y^2}$ from each of them:

$$U_E(y) = \frac{1}{4\pi\varepsilon_0}\left[\frac{Q^2}{2a} - \frac{2Qq}{\sqrt{a^2 + y^2}}\right].$$

The first term is the fixed pair's energy; it is constant, but it is part of the system.

(b) The graph is symmetric about $y = 0$ because $y$ appears only as $y^2$. Its minimum is at $y = 0$, where the subtracted term is largest: $U_E(0) = \frac{1}{4\pi\varepsilon_0}\left(\frac{Q^2}{2a} - \frac{2Qq}{a}\right)$. As $|y|$ grows, $U_E$ rises smoothly and levels off toward $\frac{Q^2}{8\pi\varepsilon_0 a}$, the fixed pair's energy alone. The curve is flat (zero slope) at $y = 0$.

(c) The force component is the negative slope of the energy with respect to the moving particle's coordinate:

$$F_y = -\frac{dU_E}{dy} = -\frac{1}{4\pi\varepsilon_0}\frac{2Qqy}{(a^2 + y^2)^{3/2}}.$$

For $y > 0$ the force is negative and for $y < 0$ it is positive, so it always points back toward the origin: $y = 0$ is a stable equilibrium, as the valley in the graph shows. From Coulomb's law, each $+Q$ attracts $-q$ with magnitude $\frac{Qq}{4\pi\varepsilon_0(a^2 + y^2)}$. The $x$-components cancel by symmetry, and each $y$-component is that magnitude times $\frac{y}{\sqrt{a^2 + y^2}}$, toward the $x$-axis. Adding the two gives exactly the $F_y$ above.

(d) Slow motion means $W = \Delta U_E = U_E(\infty) - U_E(0)$. Write both energies in full, with $k = \frac{1}{4\pi\varepsilon_0}$. Far away only the fixed pair's term survives, so $U_E(\infty) = \frac{kQ^2}{2a}$, not zero:

$$W = \frac{kQ^2}{2a} - \left(\frac{kQ^2}{2a} - \frac{2kQq}{a}\right) = \frac{2kQq}{a} = \frac{2(9.0\times10^{9})(4.0\times10^{-9})(1.0\times10^{-9})}{0.030} = 2.4\times10^{-6}\ \text{J}.$$

The fixed pair's term cancels because that pair never moves.

Positive work is needed because the particle sits at the bottom of the energy valley.

Where students lose points

On the exam

Questions about this topic ask you to compute a system's energy from charges and distances, to say how it changes when a charge or distance changes by a factor, to sketch $U_E$ against separation, or to find the work done by an external agent or by the field. When you justify a claim, name the principle: the work done slowly by an external agent equals the change in the system's potential energy, and the field's work is its negative. Topic 9.3 adds kinetic energy to the bookkeeping for charges that are free to speed up.