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A2 · Practice questions · Energy stored in a capacitor

The area of a triangle.

Six original Cambridge-style questions on the energy stored in a capacitor, the three forms W = ½QV = ½CV² = ½Q²/C, and where the factor of one half comes from.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Half, because the p.d. climbs as the charge builds.

01
Calculation
[2 marks]

A 4700 μF capacitor is charged to a potential difference of 16 V. Calculate the charge stored and the energy stored.

  • Q = CV = 4700 × 10⁻⁶ × 16 = 7.5 × 10⁻² C. ✓
  • W = ½CV² = ½ × 4700 × 10⁻⁶ × 16² = 0.60 J (the same as ½QV = ½ × 0.0752 × 16). ✓
02
Analysis
[3 marks]

A capacitor holds charge Q at a potential difference V. A student argues that the energy stored must be QV, because that is the work done moving a charge Q through a p.d. V. Explain, with reference to a graph of p.d. against charge, why the energy stored is in fact ½QV.

  • The p.d. is not V throughout charging: it rises from zero to V as the charge builds up, so only the last small amount of charge is moved through the full p.d. V. ✓
  • The work done is the sum of (p.d. × small charge added), which is the area under the graph of p.d. against charge. ✓
  • That graph is a straight line through the origin, so the area is a triangle of base Q and height V, giving W = ½QV; equivalently the average p.d. during charging is ½V. ✓
03
Calculation
[3 marks]

A 150 μF capacitor in a camera flash unit must store 2.7 J of energy. Calculate the potential difference it must be charged to and the charge it then holds. The flash is then fired until the potential difference has fallen to half of this value; calculate the energy delivered to the flash tube.

  • V = √(2W / C) = √(2 × 2.7 / 150 × 10⁻⁶) = √36 000 = 190 V. ✓
  • Q = CV = 150 × 10⁻⁶ × 190 = 2.8 × 10⁻² C. ✓
  • Halving V quarters the energy, so 2.7 / 4 = 0.675 J remains and 2.7 − 0.675 = 2.0 J has been delivered. ✓
04
Calculation
[3 marks]

A capacitor stores 0.80 J of energy when it holds a charge of 4.0 mC. Calculate its capacitance and the potential difference across it, and check your answer using a second energy expression.

  • From W = ½Q²/C, C = Q² / 2W = (4.0 × 10⁻³)² / (2 × 0.80) = 1.0 × 10⁻⁵ F (10 μF). ✓
  • V = Q / C = 4.0 × 10⁻³ / 1.0 × 10⁻⁵ = 400 V. ✓
  • Check: ½CV² = ½ × 1.0 × 10⁻⁵ × 400² = 0.80 J, as given. ✓
05
Analysis
[2 marks]

Capacitor X has twice the capacitance of capacitor Y. Compare the energies stored in X and Y when (i) each is charged to the same potential difference, and (ii) each holds the same charge.

  • (i) At the same V, W = ½CV² is proportional to C, so X stores twice the energy of Y. ✓
  • (ii) At the same Q, W = ½Q²/C is inversely proportional to C, so X stores half the energy of Y (the same charge sits at half the p.d.). ✓
06
Calculation
[4 marks]

A 20 μF capacitor and a 30 μF capacitor are connected in series across a 50 V supply.

(a) Calculate the combined capacitance and the charge on each capacitor. (b) Calculate the potential difference across each capacitor. (c) Calculate the energy stored in each capacitor, and show that the two energies add to the energy stored by the combination.

  • (a) 1/C = 1/20 + 1/30 = 5/60, so C = 12 μF, and the series charge is Q = CV = 12 × 10⁻⁶ × 50 = 6.0 × 10⁻⁴ C on each. ✓
  • (b) V₂₀ = Q / C = 6.0 × 10⁻⁴ / 20 × 10⁻⁶ = 30 V and V₃₀ = 6.0 × 10⁻⁴ / 30 × 10⁻⁶ = 20 V (total 50 V). ✓
  • (c) W₂₀ = ½QV = ½ × 6.0 × 10⁻⁴ × 30 = 9.0 × 10⁻³ J and W₃₀ = ½ × 6.0 × 10⁻⁴ × 20 = 6.0 × 10⁻³ J. ✓
  • Sum = 1.5 × 10⁻² J, equal to ½CV² for the combination = ½ × 12 × 10⁻⁶ × 50² = 1.5 × 10⁻² J; the smaller capacitor stores more energy because it has the larger p.d. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.