United StatesPhysics 2Syllabus dot point

What is voltage, and how is it related to the electric field and to energy per charge?

Topic 10.5 Electric Potential and its Relation to the Electric Field: define electric potential, relate potential difference to field and to potential energy, and use equipotentials.

A focused answer to AP Physics 2 Topic 10.5, covering electric potential as energy per unit charge, the potential of a point charge, the relation between potential difference and the field, equipotential surfaces, and the work done moving a charge through a potential difference, with full worked examples.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Electric potential $V$ is the electric potential energy per unit charge at a point: $V = \dfrac{U}{q}$ , measured in volts (V $=$ J/C). A single point charge produces $V = \dfrac{kQ}{r}$ (a scalar, with sign). The potential difference (voltage) between two points is $\Delta V = \dfrac{\Delta U}{q}$ , and the work to move a charge between them is $W = q\,\Delta V$ . Potential is tied to the field: in a uniform field, $E = \dfrac{V}{d}$ , and the field always points from high to low potential. Equipotential surfaces connect points at the same potential; they are perpendicular to the field lines, and no work is done moving a charge along one. Released charges accelerate toward lower potential energy: positive charges move from high to low potential, negative charges the reverse. Potential is the energy-per-charge language that makes circuits (Unit 11) work, where voltage drives current.

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What this topic is asking
What electric potential is
Potential difference, field and work
Equipotential surfaces
Try this

What this topic is asking

The College Board (Topic 10.5) wants you to define electric potential as energy per unit charge, relate potential difference to the electric field and to potential energy, and use equipotential surfaces to reason about the work done moving a charge.

What electric potential is

Potential strips out the test charge from potential energy, leaving a property of the location alone: the energy a unit positive charge would have there. Because it is a scalar, adding the potentials of several charges is just arithmetic (add the signed values), far simpler than the vector addition needed for fields. The potential of a point charge falls as $1/r$ , like the potential energy it comes from.

Potential difference, field and work

The relation $W = q\,\Delta V$ is the workhorse of the topic and the bridge to circuits: it says voltage is energy per charge, so moving a charge across a voltage transfers energy $q\,\Delta V$ . The link $E = V/d$ lets you swap between field and potential in a uniform field, and tells you the field points "downhill" in potential. A released charge always moves to lower its potential energy, which for a positive charge means toward lower potential, and for a negative charge means toward higher potential.

Equipotential surfaces

Equipotential surfaces are surfaces on which the potential is constant. Three properties are tested: they are perpendicular to the field lines everywhere; no work is done moving a charge along an equipotential (because $\Delta V = 0$ ); and they are closely spaced where the field is strong. Around a point charge they are concentric spheres; between parallel plates they are flat planes parallel to the plates. The strategic value of potential is that it converts the vector problem of fields into a scalar one, simplifies energy calculations through $W = q\,\Delta V$ , and provides the concept of voltage that drives every circuit in Unit 11. Recognizing that the field points from high to low potential, that charges move to lower their potential energy, and that equipotentials are perpendicular to the field, ties the whole electrostatics unit together.

Accelerating a proton through a voltage

A proton (charge $+1.6 \times 10^{-19}$ C, mass $1.67 \times 10^{-27}$ kg) is accelerated from rest through a potential difference of $500$ V. Find the kinetic energy it gains and its final speed.

step 1 Kinetic energy from the voltage

The work done equals the charge times the potential difference: $\Delta KE = q\,\Delta V = (1.6 \times 10^{-19})(500) = 8.0 \times 10^{-17}$ J.

step 2 Set kinetic energy equal to one half m v squared

$\tfrac{1}{2}m v^2 = 8.0 \times 10^{-17}$ J, so $v^2 = \dfrac{2(8.0 \times 10^{-17})}{1.67 \times 10^{-27}}$ .

step 3 Solve for the speed

$v^2 = 9.58 \times 10^{10}$ , so $v = \sqrt{9.58 \times 10^{10}} = 3.1 \times 10^5$ m/s.

step 4 Interpret

The proton, being positive, accelerates from high to low potential, gaining $8.0 \times 10^{-17}$ J and reaching about $3 \times 10^5$ m/s. This is how particle accelerators work: voltage gives energy per charge.

Try this

Q1. State the work done by the electric force when a charge moves along an equipotential surface. [1 point]

Cue. Zero (the potential, and so the potential energy, does not change).

Q2. Parallel plates $0.010$ m apart have $50$ V across them. Calculate the field between them. [2 points]

Cue. $E = V/d = 50/0.010 = 5.0 \times 10^3$ V/m.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2024 (style)7 marksSection II (long FRQ). Two parallel plates are separated by

0.020

m with a potential difference of

120

V across them. (a) Calculate the magnitude of the uniform electric field between the plates. (b) An electron (charge magnitude

1.6 \times 10^{-19}

C) is released from rest at the negative plate. Calculate the kinetic energy it gains crossing to the positive plate. (c) State the shape of the equipotential surfaces between the plates and justify it.

Show worked answer →

A 7-point FRQ on potential, field and equipotentials.

(a) Field (2 points): for a uniform field, $E = \dfrac{V}{d} = \dfrac{120}{0.020} = 6.0 \times 10^3$ V/m.
(b) Kinetic energy (3 points): the work done equals the charge times the potential difference, $\Delta KE = q\,\Delta V = (1.6 \times 10^{-19})(120) = 1.92 \times 10^{-17}$ J. (All the potential energy lost becomes kinetic energy.)
(c) Equipotentials (2 points): flat planes parallel to the plates, because the field is uniform and perpendicular to the plates, so surfaces of equal potential are perpendicular to the field, parallel to the plates.

Markers reward $E = V/d$ , the work $q\Delta V$ for the kinetic energy, and flat equipotential planes perpendicular to the uniform field.

AP 2023 (style)1 marksSection I (multiple choice). A positive charge is moved along an equipotential surface. How much work is done by the electric force? (A) a positive amount (B) a negative amount (C) zero (D) it depends on the path length. Justify your reasoning.

Show worked answer →

A 1-point MCQ on equipotentials. The answer is (C).

Moving along an equipotential means the potential, and so the potential energy, does not change. The work done by the electric force equals minus the change in potential energy, which is zero. The trap is (D): no work is done regardless of how far along the equipotential the charge moves.

Related dot points

Sources & how we know this

AP Physics 2: Algebra-Based Course and Exam Description — College Board (2024)