Skip to main content
United StatesPhysics 2Syllabus dot point

How does energy conservation link the voltage a charge crosses to the kinetic energy it gains?

Topic 10.7 Conservation of Electric Energy: apply conservation of energy to charges moving through potential differences, relating qV to kinetic energy.

A focused answer to AP Physics 2 Topic 10.7, covering conservation of energy for charges moving through electric potential differences, the relation between qV and kinetic energy, the electron-volt, and energy bookkeeping for charges accelerated by fields, with full worked examples.

Generated by Claude Opus 4.810 min answer

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

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. Energy conservation for charges
  3. Work, voltage and kinetic energy
  4. The electron-volt and the strategic payoff
  5. Try this

What this topic is asking

The College Board (Topic 10.7) wants you to apply conservation of energy to charges moving through potential differences, relating the work q ΔVq\,\Delta V to the kinetic energy gained, and to use this energy bookkeeping for charges accelerated by electric fields.

Energy conservation for charges

This is the same conservation of energy from mechanics, with electric potential energy playing the role gravitational potential energy played for a falling mass. A charge "falls" through the electric field, trading potential energy for kinetic energy. Because U=qVU = qV, the change in potential energy as a charge moves between two points is ΔU=q ΔV\Delta U = q\,\Delta V, so the kinetic energy gained is ΔKE=−q ΔV\Delta KE = -q\,\Delta V (or, in magnitude, qq times the voltage crossed).

Work, voltage and kinetic energy

The relation ΔKE=q ΔV\Delta KE = q\,\Delta V is the practical core of the topic and the standard recipe: the kinetic energy gained equals the charge times the voltage it crosses, then set that equal to 12mv2\tfrac{1}{2}m v^2 to find the speed. There is no factor of one half in q ΔVq\,\Delta V itself (a frequent trap); the one half appears only when you convert the kinetic energy to a speed. The direction rule is worth holding clearly: a charge always speeds up when it moves to lower potential energy, which for a positive charge is toward lower potential and for a negative charge (like an electron) is toward higher potential.

The electron-volt and the strategic payoff

Because the energies of single charges are tiny in joules, the electron-volt is the natural unit: 11 eV is the kinetic energy one elementary charge gains crossing 11 V, equal to 1.6×10−191.6 \times 10^{-19} J. A proton accelerated through 10001000 V gains 10001000 eV (or 1.6×10−161.6 \times 10^{-16} J). The strategic value of this topic is that it unifies the unit's energy ideas with the rest of physics: it is the electric version of "potential energy converts to kinetic energy," it explains how every particle accelerator and electron gun works, and it sets up the energy accounting of circuits. In Unit 11, the source's voltage delivers energy qVqV to each charge that flows, which is then dissipated in resistors or stored in capacitors, exactly the energy-per-charge bookkeeping introduced here.

Try this

Q1. Calculate the kinetic energy gained by a +3.0×10−6+3.0 \times 10^{-6} C charge accelerated through 250250 V. [2 points]

  • Cue. ΔKE=q ΔV=(3.0×10−6)(250)=7.5×10−4\Delta KE = q\,\Delta V = (3.0 \times 10^{-6})(250) = 7.5 \times 10^{-4} J.

Q2. State the direction (toward higher or lower potential) in which an electron gains kinetic energy. [1 point]

  • Cue. Toward higher potential (it lowers its potential energy because its charge is negative).

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)6 marksSection II (short FRQ). An electron (charge magnitude 1.6×10−191.6 \times 10^{-19} C, mass 9.11×10−319.11 \times 10^{-31} kg) is released from rest and accelerated through a potential difference of 200200 V. (a) Write the energy-conservation statement linking the potential energy lost and the kinetic energy gained. (b) Calculate the kinetic energy gained. (c) Calculate the final speed of the electron.
Show worked answer →

A 6-point FRQ on conservation of electric energy.

(a) Statement (2 points): the loss in electric potential energy equals the gain in kinetic energy, ΔKE=−ΔU=q ΔV\Delta KE = -\Delta U = q\,\Delta V (in magnitude), since the electric force is conservative.
(b) Kinetic energy (2 points): ΔKE=q ΔV=(1.6×10−19)(200)=3.2×10−17\Delta KE = q\,\Delta V = (1.6 \times 10^{-19})(200) = 3.2 \times 10^{-17} J.
(c) Speed (2 points): 12mv2=3.2×10−17\tfrac{1}{2}m v^2 = 3.2 \times 10^{-17}, so v2=2(3.2×10−17)9.11×10−31=7.03×1013v^2 = \dfrac{2(3.2 \times 10^{-17})}{9.11 \times 10^{-31}} = 7.03 \times 10^{13}, giving v=8.4×106v = 8.4 \times 10^6 m/s.

Markers reward equating lost potential energy to gained kinetic energy, computing qΔVq\Delta V, and solving for the speed.

AP 2023 (style)1 marksSection I (multiple choice). A charge qq is moved through a potential difference ΔV\Delta V by the electric force alone. The kinetic energy it gains is equal to (A) q/ΔVq/\Delta V (B) qΔVq\Delta V (C) ΔV/q\Delta V / q (D) 12qΔV\tfrac{1}{2}q\Delta V. Justify your reasoning.
Show worked answer →

A 1-point MCQ on the work-energy link. The answer is (B).

The work done by the electric force on a charge crossing a potential difference is W=q ΔVW = q\,\Delta V, and by the work-energy theorem this equals the kinetic energy gained (if only the electric force acts). The trap is (D): there is no factor of one half in qΔVq\Delta V for a single charge crossing a fixed potential difference.

Related dot points

Sources & how we know this