Unit 10 · Topic 10.5

10.5 Electric Potential

Electric potential is electric potential energy per unit charge, measured in volts. Unlike the field, it's a scalar, so the potential from several charges just adds as numbers. Equipotential lines connect points at the same potential, cross field lines at right angles, and their spacing tells you how strong the field is.

Key terms

electric potential
volt
potential difference
equipotential line (isoline)
scalar superposition

Potential and potential difference

The electric potential V at a point is the potential energy a charge would have there, per coulomb: V = $U_E/q$ . Its unit is the volt, 1 V = 1 J/C. Potential belongs to a location, whether or not a charge is sitting there.

What usually matters is the potential difference between two points, ΔV. Move a charge q across a potential difference ΔV and its potential energy changes by $\Delta U_E$ = qΔV. A 9 V battery keeps its terminals 9 V apart, so each coulomb passing through it gains 9 J. Batteries create this difference through chemical reactions that separate positive and negative charge.

Potential near point charges

The potential at distance r from a point charge q, taking V = 0 infinitely far away, is:

$V = \frac{kq}{r}$

Include the sign of q: potential is positive near positive charges and negative near negative charges. For several charges, add each charge's potential as plain numbers, with signs. This scalar superposition is much easier than adding field vectors.

A point can have zero potential but a nonzero field, or zero field but a nonzero potential. Midway between equal and opposite charges, V = 0 but E is large. Midway between two equal positive charges, E = 0 but V is positive.

Equipotential lines

An equipotential line (also called an isoline) connects points with the same potential, like contour lines on a hiking map. Key rules:

Equipotential lines are always perpendicular to field lines.
The field points from higher potential toward lower potential, “downhill.”
The field has no component along an equipotential, so moving a charge along one takes no work.
Where equipotentials (drawn at equal steps of V) are closer together, the field is stronger.
Around a point charge they're circles centered on the charge, crowded close to it. In a uniform field they're evenly spaced parallel lines.

Getting the field from the potential

The average field between two points is the potential difference divided by the distance between them:

$E = \left\lvert \frac{\Delta V}{\Delta r} \right\rvert$

That's why V/m and N/C are the same unit. On an equipotential map, measure the spacing between neighboring lines along a direction perpendicular to them, divide the step in V by that distance, and point the field toward the lower-potential line.

When conductors touch, electrons flow until every part of the combined conductor is at the same potential. The surface of any conductor at equilibrium is an equipotential.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Adding potentials
A +2.0 nC charge is 0.20 m from point P, and a +3.0 nC charge is 0.30 m from P. Find the electric potential at P.
Show the solutionHide the solution
1. Step 1: From the first charge: $V_1 = \dfrac{(9.0 \times 10^9)(2.0 \times 10^{-9})}{0.20} = 90$ V.
2. Step 2: From the second: $V_2 = \dfrac{(9.0 \times 10^9)(3.0 \times 10^{-9})}{0.30} = 90$ V.
3. Step 3: Potential is a scalar, so the directions to the charges don't matter: V = 90 + 90 = 180 V.
Answer: 180 V
Example 2Calculator allowed
Zero potential, nonzero field (classic trap)
A +4.0 nC charge and a −4.0 nC charge are 0.20 m apart. Find the electric potential and the electric field at the midpoint.
Show the solutionHide the solution
1. Step 1: Potential: $V = \dfrac{k(+4.0 \times 10^{-9})}{0.10} + \dfrac{k(-4.0 \times 10^{-9})}{0.10} = 360 - 360 = 0$ V.
2. Step 2: Field: each charge makes 3600 N/C at the midpoint, and both point toward the negative charge, so E = 7200 N/C (see 10.3).
3. Step 3: Zero potential does not mean zero field. The field depends on how fast V changes with position, not on V itself.
Answer: V = 0, but E = 7200 N/C toward the negative charge
Example 3Calculator allowed
Field from an equipotential map
On a map, the 30 V and 20 V equipotential lines are parallel and 2.0 cm apart. Find the size and direction of the electric field between them, and the work done by the field on a +5.0 μC charge that moves from the 30 V line to the 20 V line.
Show the solutionHide the solution
1. Step 1: $E = \left\lvert \dfrac{\Delta V}{\Delta r} \right\rvert = \dfrac{10 \text{ V}}{0.020 \text{ m}} = 500$ V/m, perpendicular to the lines and pointing from the 30 V line toward the 20 V line.
2. Step 2: Change in potential energy: $\Delta U_E$ = qΔV = (5.0 × 10⁻⁶)(20 − 30) = −5.0 × 10⁻⁵ J.
3. Step 3: The field does positive work equal to the drop in potential energy: +5.0 × 10⁻⁵ J.
Answer: 500 V/m from 30 V toward 20 V; the field does +5.0 × 10⁻⁵ J of work

Common mistakes

Adding potentials as vectors, or using magnitudes. Potential is a scalar with a sign; just add the signed values.
Assuming V = 0 means E = 0, or the reverse. They answer different questions.
Drawing the field along an equipotential line. Field lines cross equipotentials at right angles.
Using kq/r² for potential. Potential goes as 1/r.

On the exam

Expect to draw equipotential lines from a field map, or field vectors from an equipotential map, and to rank the field strength at points by line spacing.
Translation questions may pair an equipotential sketch with energy bar charts for a charge moving through the region.

Connected topics

Videos

AP Physics 2 - Unit 10 - Lesson 6 - Electric Potential
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Understanding Electric Potential
Flipping PhysicsWatch on YouTube (opens in a new tab)
Voltage (electric potential difference) | Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Equipotential Lines
Bozeman ScienceWatch on YouTube (opens in a new tab)
AP Physics 2 - Unit 10 - Lesson 7 - Reading Equipotential Lines
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
Voltage, Electric Energy, and Capacitors: Crash Course Physics #27
CrashCourseWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 10.5 Electric Potential. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

What is the electric potential 0.15 m from a +3.0 nC point charge? (Take the potential to be zero infinitely far away.)

Question 2 of 4Calculator allowed

A +4.0 nC charge is at x = 0 and a −2.0 nC charge is at x = 0.30 m. What is the electric potential at x = 0.10 m?

Question 3 of 4Calculator allowed

A large metal sphere and a small metal sphere, far apart, are joined by a long thin wire and given some positive charge. After the charges stop moving, which is true?

Question 4 of 4Calculator allowed

At point P the electric potential is 50 V, and at nearby point R it is 20 V. A proton is released from rest at the midpoint between them. Which way does it start to move?

0 of 4 answered