United StatesPhysics C: Electricity and MagnetismSyllabus dot point

How does conservation of energy govern a charge moving through a potential difference?

Topic 9.3 Conservation of Electric Energy: apply conservation of energy to charges moving through potential differences, including charged particles accelerated by fields.

A calculus-based answer to AP Physics C E&M Topic 9.3, covering the work-energy theorem with electric forces, charges accelerated through a potential difference, the electronvolt, and energy conservation in combined fields.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

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

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Quick answer

A charge moving through a potential difference $\Delta V$ changes its potential energy by $\Delta U = q\,\Delta V$ . With only the electric force acting, energy is conserved: $\Delta K + \Delta U = 0$ , so the kinetic energy gained is $\Delta K = -q\,\Delta V$ . A positive charge gains kinetic energy moving toward lower potential; a negative charge gains kinetic energy moving toward higher potential. For a charge accelerated from rest, $\tfrac{1}{2}mv^2 = |q\,\Delta V|$ , giving $v = \sqrt{\dfrac{2|q\,\Delta V|}{m}}$ . The result is path-independent because the electric force is conservative. The electronvolt (eV), the energy gained by one elementary charge across $1$ V, is a convenient unit: $1$ eV $= 1.6\times10^{-19}$ J.

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What this topic is asking
Energy conservation with the electric force
Which way a charge speeds up
Acceleration from rest
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What this topic is asking

The College Board (Topic 9.3) wants you to apply conservation of energy to charges moving through potential differences: a charge speeds up or slows down depending on the sign of $q\Delta V$ , and the energy gained or lost shows up as kinetic energy. This is the work-energy theorem from mechanics applied to the electric force, and it underpins particle accelerators, oscilloscopes and cathode-ray tubes.

Energy conservation with the electric force

This is the work-energy theorem: the work done by the (conservative) electric force equals the gain in kinetic energy and the loss in potential energy. Because the force is conservative, the result depends only on the endpoints, not the path taken.

Which way a charge speeds up

The sign of $q\,\Delta V$ decides everything:

A positive charge released in a field drifts toward lower potential, converting $U$ to $K$ , so it speeds up going "downhill" in potential.
A negative charge does the opposite: it speeds up moving toward higher potential, because $\Delta U = q\,\Delta V < 0$ there.

In both cases the charge moves toward lower potential energy, just as a mass falls toward lower gravitational potential energy.

Acceleration from rest

A charge starting from rest and accelerated through a potential difference of magnitude $|\Delta V|$ reaches a speed set by

\tfrac{1}{2}mv^2 = |q\,\Delta V| \quad\Longrightarrow\quad v = \sqrt{\frac{2|q\,\Delta V|}{m}}

This single relation runs every electron gun and ion accelerator: choose the accelerating voltage, get the final speed.

Speed of an alpha particle across a potential difference

An alpha particle (charge $+2e$ , mass $6.64\times10^{-27}$ kg) is accelerated from rest through a potential difference of $2000$ V. Find its final speed.

step 1 Find the energy gained

$\Delta K = |q\,\Delta V| = (2e)(2000\,\text{V}) = (2)(1.6\times10^{-19})(2000) = 6.4\times10^{-16}$ J.

step 2 Set kinetic energy equal to the energy gained

Starting from rest, $\tfrac{1}{2}mv^2 = \Delta K = 6.4\times10^{-16}$ J.

step 3 Solve for the speed

$v = \sqrt{\dfrac{2\Delta K}{m}} = \sqrt{\dfrac{2(6.4\times10^{-16})}{6.64\times10^{-27}}} = \sqrt{1.93\times10^{11}} = 4.4\times10^5$ m/s.

step 4 Check the direction

The alpha particle is positive, so it accelerated toward the lower-potential plate, gaining $4000$ eV of kinetic energy.

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Q1. A charge of $+3.0\,\mu\text{C}$ moves from a point at $40$ V to a point at $10$ V. Find the kinetic energy it gains (electric force only). [2 points]

Cue. $\Delta K = -q\,\Delta V = -(3.0\times10^{-6})(10 - 40) = 9.0\times10^{-5}$ J.

Q2. State which way a negative charge speeds up: toward higher or lower potential. [1 point]

Cue. Toward higher potential (its potential energy $q\Delta V$ falls there).

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 2021 (style)1 marksSection I (multiple choice). A proton is accelerated from rest through a potential difference of

1000

V. Its final kinetic energy is (A)

1000

eV (B)

1000

J (C)

1.6\times10^{-16}

J only if the field is uniform (D) zero. Justify your reasoning.

Show worked answer →

A 1-point MCQ on energy gained across a potential difference. The answer is (A).

A charge $q$ moving through $\Delta V$ gains kinetic energy $\Delta K = |q\Delta V|$ . For a proton ( $q = e$ ) through $1000$ V, $\Delta K = e(1000\,\text{V}) = 1000$ eV (which equals $1.6\times10^{-16}$ J, regardless of whether the field is uniform). The answer is path-independent, so (C)'s condition is wrong, and (A) is exact.

AP 2024 (style)5 marksSection II (FRQ, quantitative). An electron (mass

9.11\times10^{-31}

kg, charge

-e

) starts from rest and is accelerated through a potential difference of

250

V. (a) Determine whether it moves toward higher or lower potential, and explain. (b) Calculate its final kinetic energy in joules. (c) Calculate its final speed.

Show worked answer →

A 5-point FRQ on accelerating a charge through a potential difference.

(a) Direction (2 points): a negative charge gains kinetic energy moving toward higher potential, because $\Delta U = q\Delta V$ is negative when $q < 0$ and $\Delta V > 0$ , so $U$ falls and $K$ rises.
(b) Energy (1 point): $\Delta K = |q\Delta V| = (1.6\times10^{-19})(250) = 4.0\times10^{-17}$ J.
(c) Speed (2 points): $\tfrac{1}{2}mv^2 = 4.0\times10^{-17}$ , so $v = \sqrt{\dfrac{2(4.0\times10^{-17})}{9.11\times10^{-31}}} = 9.4\times10^6$ m/s.

Markers reward the direction reasoning for a negative charge, the energy, and the speed.

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Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)