A student charges a capacitor using a 6 V battery. The capacitor stores a charge of 12 μC. The student then replaces the battery with a 12 V battery and charges the same capacitor fully. Explain how the capacitance, charge stored, and energy stored change when the higher voltage battery is used.

AQA A-Level Physics (7408) — 3.7.4 Capacitance · Explain · 5 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Model answer (5 marks)

Capacitance does not change – it is a property of the capacitor.

The charge stored is proportional to the potential difference: Q = CV. With C constant, doubling V from 6 V to 12 V doubles Q from 12 µC to 24 µC.

The energy stored is U = ½CV². With C constant, increasing V from 6 V to 12 V multiplies U by (12/6)² = 4. Thus the energy stored quadruples.

Examiner tips

  • State that C remains constant. Explain Q = CV and show the doubling of Q. Use U = ½CV² to show the factor‑of‑four increase.
  • common_mistakes
  • :
  • Saying the capacitance changes. Confusing the factor of two for the energy increase. Not showing the equations or the ½CV² relationship.

Mark scheme (5 marks)

  1. Capacitance does not change / remains the same
  2. Charge stored doubles (to 24 μC) because charge is proportional to potential difference
  3. Correct reasoning linking charge increase to the definition Q = CV with C constant
  4. Energy stored increases
  5. Energy stored quadruples / increases by a factor of four because energy is proportional to V²

Key terms in this question

capacitance · charge · energy stored

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