Two charges +4.0\,μC and +4.0\,μC are 0.20 m apart. Find their potential energy (k = 8.9910^9). [2 points]

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How is electric potential energy defined through the work done by the electric force?

Topic 9.1 Electric Potential Energy: relate electric potential energy to the work done by the electric force and compute it for point-charge systems.

A calculus-based answer to AP Physics C E&M Topic 9.1, covering work done by the electric force, the path independence of a conservative force, the potential energy of point-charge pairs, and assembling charge configurations.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The electric force is conservative, so the work it does moving a charge between two points depends only on the endpoints, not the path. Electric potential energy $U$ is defined so that the work done by the electric force is $W_{elec} = -\Delta U$ . For two point charges, taking $U = 0$ at infinite separation, $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r} = k\dfrac{q_1 q_2}{r}$ , keeping the signs of the charges: positive for like charges (it takes work to push them together) and negative for unlike charges (a bound pair). For several charges, sum $U$ over every distinct pair. Energy conservation then links $U$ to kinetic energy: $\Delta K + \Delta U = 0$ for the electric force alone.

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What this topic is asking
Work and potential energy
Potential energy of point charges
Assembling several charges
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What this topic is asking

The College Board (Topic 9.1) wants you to define electric potential energy through the work done by the electric force, recognize that the electrostatic force is conservative (so the work is path-independent), and compute the potential energy of systems of point charges. This is the energy half of electrostatics, the mirror of the force-and-field half from Unit 8.

Work and potential energy

The work done by the field is $W = \displaystyle\int \vec{F}\cdot d\vec{r} = q\int \vec{E}\cdot d\vec{r}$ , and because the force is conservative this depends only on the endpoints. Positive work by the field lowers $U$ (the charge moves "downhill"); to raise $U$ , an external agent must do work against the field.

Potential energy of point charges

Taking $U = 0$ when the charges are infinitely far apart, the potential energy of a pair is

U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = k\frac{q_1 q_2}{r}

Note the single power of $r$ (the force goes as $1/r^2$ ; the energy, being the integral of force over distance, goes as $1/r$ ). The signs are kept: like charges give $U > 0$ (you must do work to assemble them), unlike charges give $U < 0$ (a bound system, energy released when assembled).

Assembling several charges

For a configuration of charges, the total potential energy is the sum over every distinct pair:

U_{total} = k\sum_{i<j}\frac{q_i q_j}{r_{ij}}

The condition $i < j$ counts each pair once. Physically, this is the total work needed to bring the charges in from infinity one at a time, each new charge interacting with all those already in place.

Energy to assemble three charges on a line

Charges $+q$ , $+q$ , $+q$ sit at $x = 0$ , $x = d$ , $x = 2d$ . Find the total potential energy.

step 1 List the distinct pairs

Three charges give $\binom{3}{2} = 3$ pairs: (1,2) separated by $d$ , (2,3) separated by $d$ , and (1,3) separated by $2d$ .

step 2 Write each pair's energy

$U_{12} = \dfrac{kq^2}{d}$ , $U_{23} = \dfrac{kq^2}{d}$ , $U_{13} = \dfrac{kq^2}{2d}$ .

step 3 Sum the pairs

$U_{total} = \dfrac{kq^2}{d} + \dfrac{kq^2}{d} + \dfrac{kq^2}{2d} = \dfrac{kq^2}{d}\left(1 + 1 + \tfrac{1}{2}\right) = \dfrac{5kq^2}{2d}$ .

step 4 Interpret

$U_{total} > 0$ : all charges repel, so positive external work was needed to assemble them, and that energy would be released if they were freed.

Try this

Q1. Two charges $+4.0\,\mu\text{C}$ and $+4.0\,\mu\text{C}$ are $0.20$ m apart. Find their potential energy ( $k = 8.99\times10^9$ ). [2 points]

Cue. $U = k\dfrac{q_1 q_2}{r} = (8.99\times10^9)\dfrac{(4.0\times10^{-6})^2}{0.20} = 0.72$ J.

Q2. State what a negative potential energy of a two-charge system tells you. [1 point]

Cue. The pair is bound (attractive); external work is needed to separate them to infinity.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)1 marksSection I (multiple choice). Two point charges,

+q

and

-q

, are separated by a distance

r

. Their electric potential energy is (A)

\dfrac{kq^2}{r}

(B)

-\dfrac{kq^2}{r}

\dfrac{kq^2}{r^2}

. Justify your reasoning.

Show worked answer →

A 1-point MCQ on point-charge potential energy. The answer is (B).

$U = \dfrac{kq_1 q_2}{r}$ keeps the signs of the charges. With $q_1 q_2 = (+q)(-q) = -q^2$ , $U = -\dfrac{kq^2}{r}$ . The negative energy means the pair is bound: work must be done to pull them apart. The trap is using $|q_1 q_2|$ and losing the sign, or using the $1/r^2$ force form.

AP 2024 (style)5 marksSection II (FRQ, quantitative). A charge

q_1 = +2.0\,\mu\text{C}

is fixed at the origin. A charge

q_2 = +3.0\,\mu\text{C}

of mass

5.0\times10^{-3}

kg is released from rest

0.10

m away. (a) Calculate the potential energy of the pair initially. (b) Determine the speed of

q_2

when it is very far away. (c) Explain the energy transformation. Use

k = 8.99\times10^9

Show worked answer →

A 5-point FRQ on energy conservation with point-charge $U$ .

(a) Initial $U$ (1 point): $U_i = \dfrac{kq_1 q_2}{r} = (8.99\times10^9)\dfrac{(2.0\times10^{-6})(3.0\times10^{-6})}{0.10} = 0.539$ J.
(b) Final speed (3 points): at infinity $U_f = 0$ . Energy conservation: $U_i = \tfrac{1}{2}mv^2$ , so $v = \sqrt{\dfrac{2U_i}{m}} = \sqrt{\dfrac{2(0.539)}{5.0\times10^{-3}}} = 14.7$ m/s.
(c) Transformation (1 point): the stored electric potential energy converts entirely to kinetic energy as the repulsive force does positive work pushing the charges apart.

Markers reward the signed $U$ , energy conservation to infinity, and naming the conversion.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)