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AP Physics 2 · Unit 10 · Topic 3 of 4 · 6 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 8 practice questions.

The big idea

Electric potential V is electric potential energy per coulomb. It’s a scalar, so potentials from several charges just add, signs included. A charge q moving through ΔV changes its energy by ΔU = qΔV. The field points from high V to low V.

1. Potential and the field

40 V 30 V 20 V 10 V E E 1.0 cm E = ΔV/d = 1000 V/m
Field lines cross equipotentials at right angles and point toward lower potential. Closer equipotentials mean a stronger field.
V = kq/rpoint charge; keep the sign of q
ΔU = qΔVchange in potential energy
Wfield = −qΔVwork done by the field
E = ΔV/duniform field (V/m = N/C)

2. Which way do charges go?

Released from rest, a positive charge moves toward lower potential, and a negative charge toward higher potential. Either way, its potential energy drops and its kinetic energy rises: |q|ΔV.

Moving along an equipotential takes no work: ΔV = 0.

Midway between +q and −q, the potentials cancel (V = 0), but the fields don’t (both point toward −q). V = 0 doesn’t mean E = 0.

An electronvolt (eV) is the energy an electron gains crossing 1 V: 1.6 × 10⁻¹⁹ J.

3. Example, then try it

Example

Point P is 0.30 m from a +6.0 nC charge and 0.20 m from a −2.0 nC charge. What is the potential at P?

From +6.0 nC: (9.0 × 10⁹)(6.0 × 10⁻⁹) ÷ 0.30 = +180 V. From −2.0 nC: (9.0 × 10⁹)(−2.0 × 10⁻⁹) ÷ 0.20 = −90 V.

Scalars add: V = 180 − 90 = +90 V. No directions needed.

Try it · work it on paper first

A proton starts from rest at the positive plate and accelerates to the negative plate, 2.0 mm away, through 500 V. Find the field between the plates, the kinetic energy it gains, and its final speed. (mp = 1.67 × 10⁻²⁷ kg)

1

Field: E = ΔV/d = 500 ÷ 0.0020 = 2.5 × 10⁵ V/m.

2

Energy: K = qΔV = (1.6 × 10⁻¹⁹)(500) = 8.0 × 10⁻¹⁷ J (500 eV).

3

Speed: v = √(2K/m) = √(1.6 × 10⁻¹⁶ ÷ 1.67 × 10⁻²⁷) ≈ 3.1 × 10⁵ m/s.

Answer: 2.5 × 10⁵ V/m; 8.0 × 10⁻¹⁷ J; about 3.1 × 10⁵ m/s.

4. Common mistakes

Adding potentials as vectors.V is a scalar. Add with signs, not directions.
Dropping the sign of q.An electron moving to higher V loses potential energy: ΔU = (−e)(+ΔV) < 0.
“V = 0 means E = 0.”Between +q and −q, V = 0 but E is not zero.
Check yourself

An electron is released from rest in an electric field. Which way does it move?

A negative charge feels a force opposite to E, and E points toward lower V, so the electron heads toward higher potential. Its potential energy still decreases, because q is negative.

Ready to practice?

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

Unit 10 topics

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