Unit 10 · Topic 10.4

10.4 Electric Potential Energy

Electric potential energy is energy stored in how charges are arranged. For two point charges it's $U_E$ = kq₁q₂/r, positive for like charges and negative for opposite ones, and zero when they're infinitely far apart. Because it's a scalar, the total for many charges is just the sum over every pair.

Key terms

electric potential energy
work by an external force
scalar
zero at infinite separation
pairs of charges

Energy stored in an arrangement

It takes work to push two positive charges closer together, because they repel. That work doesn't vanish; it's stored as electric potential energy of the system. Let go, and the charges fly apart, turning the stored energy into kinetic energy.

The electric potential energy of two point charges equals the work an external force must do to bring them from infinitely far apart to their current separation (without giving them any kinetic energy):

$U_E = \frac{kq_1q_2}{r}$

Plug in the signs of the charges here. Unlike Coulomb's law, the sign of $U_E$ carries meaning.

What the sign tells you

With like charges, $U_E$ is positive. You had to push them together, and the system will release energy if they separate. As r shrinks, $U_E$ grows.

With opposite charges, $U_E$ is negative. They attract, so the system lost energy as they came together, and you'd have to add energy to pull them apart to infinity. As r shrinks, $U_E$ becomes more negative.

Either way, $U_E$ approaches zero as r gets very large. A graph of $U_E$ against r for like charges is a curve in the positive region that drops toward zero; for opposite charges it's a mirror image below the axis, rising toward zero. Both follow 1/r, not 1/r².

Several charges

Potential energy belongs to pairs of charges. For three charges there are three pairs (1–2, 1–3, 2–3); for four charges there are six pairs. Find $U_E$ for each pair, with signs, and add them as ordinary numbers. No components are needed, because energy is a scalar.

The total is the work needed to assemble the whole arrangement from charges that started infinitely far apart. It's also the energy released if the arrangement flies apart.

Work and changes in energy

If an external force moves charges slowly (no change in kinetic energy), its work equals the change in potential energy: $W_{\text{ext}}$ = $\Delta U_E$ . If the charges are released and only electric forces act, energy is conserved, so any drop in $U_E$ shows up as kinetic energy: ΔK = $-\Delta U_E$ .

Compare gravity: $U_G = -\dfrac{Gm_1m_2}{r}$ is always negative because gravity always attracts. Electric potential energy has the same 1/r form but can take either sign.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Two charges
Find the electric potential energy of a +2.0 μC charge and a −3.0 μC charge 0.10 m apart. How much work must an external force do to pull them infinitely far apart?
Show the solutionHide the solution
1. Step 1: $U_E = \dfrac{kq_1q_2}{r} = \dfrac{(9.0 \times 10^9)(2.0 \times 10^{-6})(-3.0 \times 10^{-6})}{0.10} = -0.54$ J.
2. Step 2: At infinite separation $U_E$ = 0. The external work is the change: W = 0 − (−0.54 J) = +0.54 J.
Answer: $U_E$ = −0.54 J; it takes +0.54 J to separate them
Example 2Calculator allowed
Three charges in a triangle
Three charges, +1.0 μC, +1.0 μC and −1.0 μC, sit at the corners of an equilateral triangle with 0.10 m sides. Find the total electric potential energy.
Show the solutionHide the solution
1. Step 1: Every pair is 0.10 m apart, and every pair has $\lvert q_1q_2 \rvert = 1.0 \times 10^{-12}$ C², so each pair's energy has size $\dfrac{(9.0 \times 10^9)(1.0 \times 10^{-12})}{0.10} = 0.090$ J.
2. Step 2: The (+, +) pair: +0.090 J. The two (+, −) pairs: −0.090 J each.
3. Step 3: Total: 0.090 − 0.090 − 0.090 = −0.090 J.
Answer: −0.090 J
Example 3Calculator allowed
Pushing like charges closer
Two +1.0 μC charges are 0.30 m apart. How much work must you do to push them slowly to 0.10 m apart?
Show the solutionHide the solution
1. Step 1: Starting energy: (9.0 × 10⁹)(1.0 × 10⁻¹²)/0.30 = 0.030 J.
2. Step 2: Final energy: (9.0 × 10⁹)(1.0 × 10⁻¹²)/0.10 = 0.090 J.
3. Step 3: Slowly means no change in kinetic energy, so W = $\Delta U_E$ = 0.090 − 0.030 = 0.060 J.
Answer: 0.060 J

Common mistakes

Using 1/r² for potential energy. Energy uses 1/r; force and field use 1/r².
Dropping the signs. Opposite charges give negative $U_E$ , and that sign matters when you add pairs.
Missing pairs. Three charges have three pairs, four charges have six.
Splitting potential energy into components. It's a scalar, so just add the numbers.

On the exam

Expect energy bar charts showing kinetic and electric potential energy as charges move toward or away from each other.
Sketching $U_E$ against r for like and unlike charges is a common graphing task; get the sign and the 1/r shape right.

Connected topics

Videos

AP Physics 2 - Unit 10 - Lesson 5 - Potential Energy
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Electric Potential Energy Explained
Flipping PhysicsWatch on YouTube (opens in a new tab)
Electric potential energy of charges | Physics | Khan Academy
Khan Academy PhysicsWatch on YouTube (opens in a new tab)
Electric Potential Energy of 3 Point Charges
Flipping PhysicsWatch on YouTube (opens in a new tab)
Electric potential energy | Electrostatics | Electrical engineering | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.4 Electric Potential Energy. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the electric potential energy of a system of a +2.0 μC charge and a +3.0 μC charge that are 0.50 m apart?

Question 2 of 4Calculator allowed

Charges +q, +q and −q are placed at the corners of an equilateral triangle with side a. What is the total electric potential energy of the system?

Question 3 of 4Calculator allowed

Four charges, each +q, sit at the corners of a square with side a. What is the electric potential energy of the system?

Question 4 of 4Calculator allowed

A charge +q is brought at constant speed from very far away to a distance r from a fixed charge +Q. How much work is done by the external force that moves it?

0 of 4 answered