Unit 9 · Topic 9.1

9.1 Electric Potential Energy

Electric potential energy is the energy stored in a system of charges because of their positions. For two point charges it's $U_E = \dfrac{kq_1q_2}{r}$ , with zero at infinite separation, and for a group of charges you add the energy of every pair.

Key terms

electric potential energy
work done by an external force
zero at infinity
system of charges
sum over pairs

Energy belongs to the system

Potential energy isn't owned by one charge. It belongs to the system of interacting charges, the same way gravitational potential energy belongs to an object-Earth system. For two point charges a distance r apart, $U_E = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}$

Here you keep the signs of the charges. The zero of potential energy is chosen at infinite separation, where the charges no longer interact. Like electric force, U depends on 1/r, but it's a scalar and falls off more slowly than the 1/r² force.

What the sign tells you

Like charges: U is positive. You had to push them together, doing positive work, and they'd fly apart if released, turning U into kinetic energy. U gets bigger as they get closer.
Opposite charges: U is negative. They pull together on their own. To separate them to infinity, you must add energy. U gets more negative as they get closer.
In both cases, released charges move in the direction that lowers U.

Work and potential energy

The electric force is conservative, so the work it does depends only on the start and end positions. The change in potential energy is the negative of the work done by the electric force: $\Delta U_E = -W_{\text{field}} = -\displaystyle\int\vec{F}_E\cdot d\vec{r}$ .

Bring $q_2$ in from infinity to distance r, with $F = kq_1q_2/r'^2$ pointing outward: $\Delta U = -\displaystyle\int_\infty^r\frac{kq_1q_2}{r'^2}\,dr' = \frac{kq_1q_2}{r}$ . That's where the formula comes from.

If an outside agent moves a charge slowly, so its kinetic energy doesn't change, the work done by that agent equals the change in potential energy: $W_{\text{ext}} = \Delta U_E$ .

Systems of several charges

The total potential energy of a group of point charges is the sum over every pair: $U = \displaystyle\sum_{\text{pairs}}\frac{kq_iq_j}{r_{ij}}$ . This equals the work needed to assemble the group, one charge at a time, from far apart. Each pair counts once.

Number of charges	Number of pairs
2	1
3	3
4	6
N	N(N − 1)/2

Graphs of U against r

For like charges, the graph of U against r is a curve that is high and positive at small r and drops toward zero as r grows. For opposite charges, it's the mirror image below the axis: very negative at small r and rising toward zero. The slope connects to force: the force along r is the negative slope, F = −dU/dr, so a steep curve means a strong force.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Work to push two charges closer
Charges of +2.0 μC and +3.0 μC are 0.30 m apart. How much work must an outside agent do to move them slowly to 0.10 m apart?
Show the solutionHide the solution
1. Step 1: $W_{\text{ext}} = \Delta U = kq_1q_2\left(\dfrac{1}{r_f} - \dfrac{1}{r_i}\right)$ .
2. Step 2: $kq_1q_2 = (9.0\times10^{9})(2.0\times10^{-6})(3.0\times10^{-6}) = 0.054$ N·m².
3. Step 3: $\dfrac{1}{0.10} - \dfrac{1}{0.30} = 10 - 3.33 = 6.67$ m⁻¹, so W ≈ (0.054)(6.67) = 0.36 J.
4. Step 4: The work is positive, as it should be: you're pushing like charges together.
Answer: About 0.36 J
Example 2Calculator allowed
Assembling a triangle of charges
Three +2.0 μC charges are placed at the corners of an equilateral triangle with sides of 0.10 m. How much energy is stored in the system?
Show the solutionHide the solution
1. Step 1: There are 3 pairs, each separated by 0.10 m.
2. Step 2: Each pair: $\dfrac{kq^2}{a} = \dfrac{(9.0\times10^{9})(2.0\times10^{-6})^2}{0.10} = 0.36$ J.
3. Step 3: Total: 3 × 0.36 = 1.08 J. That's also the work needed to bring the charges in from far apart.
Answer: About 1.1 J
Example 3Calculator allowed
Mixed signs (classic trap)
Now one of the three charges in the triangle is changed to −2.0 μC. Find the system's potential energy, and say what the sign means.
Show the solutionHide the solution
1. Step 1: Pairs: (+,+) gives $+kq^2/a$ ; each of the two (+,−) pairs gives $-kq^2/a$ .
2. Step 2: Total: $\dfrac{kq^2}{a}(1 - 1 - 1) = -0.36$ J.
3. Step 3: The negative total means the system is bound: you'd have to add 0.36 J to pull all three charges apart to infinity.
4. Step 4: The traps are dropping the signs (which would give +1.08 J again) and counting six pairs instead of three (which doubles the answer).
Answer: −0.36 J; the system is bound

Common mistakes

Using 1/r² for potential energy. Force goes as 1/r²; potential energy goes as 1/r.
Dropping the signs of the charges. In U, the signs matter: they decide whether the system is bound.
Counting each pair twice when summing over a group of charges.
Mixing up the work done by the field with the work done by an outside agent. They're equal in size and opposite in sign when kinetic energy doesn't change.

On the exam

Questions often ask how much work is needed to bring in one more charge. That's that charge times the potential at its final spot from the charges already there (see 9.2).
In free response, name the system before you talk about its potential energy, and state your zero (infinite separation).

Connected topics

Videos

AP Physics C E&M - Unit 9 - Lesson 1 - Electric Potential Energy
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Topic 9.1 - Electric Potential Energy
Lessons With LondotWatch 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 Explained
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C E&M - Unit 9 - Lesson 2 - Deriving Electric Potential Energy
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Physics 38 Electrical Potential (1 of 22) Electrical Potential Energy
Michel van BiezenWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.1 Electric Potential Energy. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A +2.0 μC charge and a −3.0 μC charge are 0.20 m apart. Taking the energy to be zero when they are infinitely far apart, what is the electric potential energy of the pair?

Question 2 of 4Calculator allowed

Three charges, each +q, are brought from very far away and placed at the corners of an equilateral triangle with side a. How much work must an outside agent do in total? ( $k = \frac{1}{4\pi\varepsilon_0}$ )

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? ( $k = \frac{1}{4\pi\varepsilon_0}$ )

Question 4 of 4Calculator allowed

A system of two point charges has negative electric potential energy. What does that tell you?

0 of 4 answered