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AP Physics C: Electricity and Magnetism · Unit 9 · Topic 1 of 5 · 5 min lesson

Electric Potential Energy

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.

The big idea

Two charges have potential energy U = kq1q2/r (zero when they’re infinitely far apart). For a group of charges, add the energy of every pair. U is the work needed to assemble the charges; it turns into kinetic energy if they’re released.

1. Adding up pairs

+ + − +2 µC +2 µC −3 µC U = +0.12 J −0.18 J −0.18 J side 0.30 m Utotal = −0.24 J
Three charges make three pairs. Like charges add positive energy; unlike charges add negative energy.
U = kq1q2/rone pair, U = 0 at r = ∞
Utotal = Σ over pairsn charges: n(n − 1)/2 pairs
ΔU = qΔVcharge moving through a potential
Wext = ΔUmoving slowly
Ki + Ui = Kf + Ufreleased charges

2. What the sign means

U > 0: like charges. You had to push them together; released, they fly apart and gain kinetic energy.

U < 0: unlike charges. They attract; you’d have to do work to pull them apart.

A charge accelerated through ΔV gains K = |q|ΔV. An alpha particle (+2e) through 1000 V gains 2000 eV.

Count each pair once. Three charges have three pairs, four have six.

3. Example, then try it

Example

Find the total potential energy of the three charges in the figure below (triangle side 0.30 m).

+ + − +2 µC +2 µC −3 µC side 0.30 m

Positive pair: (9.0 × 10⁹)(2.0 × 10⁻⁶)(2.0 × 10⁻⁶) ÷ 0.30 = +0.12 J.

Each +/− pair: (9.0 × 10⁹)(2.0 × 10⁻⁶)(−3.0 × 10⁻⁶) ÷ 0.30 = −0.18 J. Total: 0.12 − 0.36 = −0.24 J.

Try it · work it on paper first

Two protons are released from rest 1.0 × 10⁻¹⁰ m apart. Find each proton’s kinetic energy and speed once they’re far apart. (e = 1.6 × 10⁻¹⁹ C, mp = 1.67 × 10⁻²⁷ kg)

1

Starting energy: U = ke²/r = (9.0 × 10⁹)(1.6 × 10⁻¹⁹)² ÷ (1.0 × 10⁻¹⁰) ≈ 2.3 × 10⁻¹⁸ J.

2

Share it: far apart, U → 0. By symmetry each proton gets half: K ≈ 1.15 × 10⁻¹⁸ J (about 7.2 eV).

3

Speed: v = √(2K/m) = √(2 × 1.15 × 10⁻¹⁸ ÷ 1.67 × 10⁻²⁷) ≈ 3.7 × 10⁴ m/s.

Answer: about 1.15 × 10⁻¹⁸ J (7.2 eV) and 3.7 × 10⁴ m/s each.

4. Common mistakes

Using 1/r² for energy.Force goes as 1/r²; energy and potential go as 1/r.
Counting pairs twice.Each pair once.
Dropping signs.Keep the sign of each charge; it tells you whether U is positive or negative.
Check yourself

A positive charge is moved closer to another fixed positive charge. What happens to the system’s potential energy?

U = kq1q2/r is positive for like charges, and it grows as r shrinks. You have to do work to push them together.

Ready to practice?

6 questions on Electric Potential Energy, each with a hint and an explanation for every wrong answer.

Unit 9 topics

Every topic has a short lesson or quick review and its own quiz.