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Learn7408Paper 2 · Paper 2Fields and their consequencesEnergy stored in a capacitor
Fields and their consequences · Subtopic 7 of 13

Energy Stored in a Capacitor

As a capacitor charges, each new bit of charge is pushed onto the plates against a rising p.d., so the p.d.–charge graph is a straight line through the origin. The energy stored is the area under it: W=12QV=12CV2=Q22CW=\tfrac12QV=\tfrac12CV^2=\dfrac{Q^2}{2C}. A battery charging a capacitor supplies QVQV, but only half is stored; the other half is dissipated in the circuit resistance.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 30 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Explain why the energy stored is the area under a VV–QQ graph.
  • Use W=12QV=12CV2=Q22CW=\tfrac12QV=\tfrac12CV^2=\dfrac{Q^2}{2C}.
  • Explain why only half the supplied energy is stored.
  • Calculate energy changes when capacitors charge or discharge.
Pause and think

A camera flash capacitor is charged to twice the voltage. How much brighter (more energy) is the flash?

Section 2 of 14

Core idea

Start hereHalf the energy goes missing

Charge a capacitor from a battery and exactly half the energy the battery supplies ends up stored; the other half heats the wires, whatever their resistance. The reason lies in the triangle under the QQ–VV graph.

The big idea

W=12QV=12CV2=Q22CW=\frac12QV=\frac12CV^2=\frac{Q^2}{2C} = area under QQ–VV.

1. Why ½

Energy stored: the triangle under Q–V
24681012246810120V (V)Q (mC)C = 1 mF½QV
At 10 V: 12(10×10−3)(10)=0.05\tfrac12(10\times10^{-3})(10)=0.05 J. Doubling VV quadruples the energy (∝V2\propto V^2).
W=12QVW=\frac12QV
  1. δW=V δQ\delta W=V\,\delta Q
    Adding a small charge at the current p.d.
  2. W=∫0QV dQ=∫0QQC dQW=\int_0^QV\,dQ=\int_0^Q\frac QC\,dQ
    VV rises from 0.
  3. =Q22C=12QV=12CV2=\frac{Q^2}{2C}=\tfrac12QV=\tfrac12CV^2
    Triangle area; substitute Q=CVQ=CV.
The battery supplies QVQV at constant voltage: half is stored, half dissipated.
Worked example1000 μF at 10 V

Energy; then at 20 V.

Goal
Two energies.
Strategy
12CV2\frac12CV^2.
  1. 12(10−3)(102)=0.050 J\tfrac12(10^{-3})(10^2)=0.050\text{ J}
    10 V.
  2. 12(10−3)(202)=0.20 J\tfrac12(10^{-3})(20^2)=0.20\text{ J}
    ×4.
Answer0.0500.050 J; 0.200.20 J.
✓ CheckV2V^2 dependence ✓.
Worked examplePartial discharge

2200 μF from 12 V to 6 V.

Goal
Energy released.
Strategy
Difference of stored energies.
  1. 12(2.2×10−3)(144)=0.158 J\tfrac12(2.2\times10^{-3})(144)=0.158\text{ J}
    Before.
  2. 12(2.2×10−3)(36)=0.040 J\tfrac12(2.2\times10^{-3})(36)=0.040\text{ J}
    After.
  3. 0.119 J0.119\text{ J}
    Released: 75%.
Answer0.1190.119 J.
✓ CheckHalf the voltage leaves a quarter of the energy ✓.
Common misconceptionQVQV without ½ The p.d. was not constant during charging.

Summary

  • W=12QV=12CV2W=\frac12QV=\frac12CV^2.
Key words
Energy stored
Area under Q–V.
Charge against p.d. for a 470 µF capacitor
246810121412345670p.d. V (V)charge Q (mC)Q = CVarea = ½QV(12 V, 5.64 mC)
The energy stored is the area under the QQ–VV graph: W=12QV=12×5.64×10−3×12=0.034W=\tfrac12QV=\tfrac12\times5.64\times10^{-3}\times12=0.034 J.
HistoryFranklin's battery. Benjamin Franklin wired several Leyden jars together in 17491749 to store more energy and called the arrangement a "battery", borrowing a military term for a group of cannons. The word stuck, though today it usually means chemical cells instead.
Fun factDefibrillators. A defibrillator charges a capacitor of about 100100 μF to around 20002000 V, storing 12×10−4×20002=200\tfrac12\times10^{-4}\times2000^2=200 J. Released in a few milliseconds through the chest, it can reset a heart's rhythm.
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