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Capacitance

Energy stored in a capacitor

  • 3 things to know
  • Component 2

Overview

Recall questions on E = ½QV and its equivalent forms, why the factor of a half appears, the energy as the area under a charge–voltage graph, the fourfold rise when the voltage doubles, and worked energy calculations.

What you need to know

Answer each one before you reveal it.

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Key term

State the equation for the energy stored in a capacitor in terms of charge and potential difference.

Key term

Why is the energy stored in a capacitor ½QV rather than QV?

Key term

How is the energy stored obtained from a graph of charge against potential difference?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the equation for the energy stored in a capacitor in terms of charge and potential difference.

    Model answer

    The energy stored is U = ½QV, where U is in joules, Q is the charge on one plate in coulombs and V is the potential difference in volts.

  2. 2

    Why is the energy stored in a capacitor ½QV rather than QV?

    Model answer

    The potential difference rises steadily from zero to V as the charge is transferred, so the average potential difference during charging is only ½V and the total work done is ½QV.

  3. 3

    Give two further forms of the energy equation for a capacitor.

    Model answer

    Substituting Q = CV gives U = ½CV², and substituting V = Q/C gives U = Q²/2C.

  4. 4

    How is the energy stored obtained from a graph of charge against potential difference?

    Model answer

    The energy is the area between the line and the potential difference axis. For the straight line through the origin that area is a triangle of value ½QV.

  5. 5

    Calculate the energy stored in a 470 μF capacitor charged to 12 V.

    Model answer

    The energy is U = ½CV² = 0.5 × 470 × 10⁻⁶ × 144 = 3.4 × 10⁻² J. The charge is 5.64 mC, and ½QV = 0.5 × 5.64 × 10⁻³ × 12 gives the same 0.0338 J.

  6. 6

    Calculate the energy stored in a 100 μF capacitor holding a charge of 2.0 mC.

    Model answer

    The energy is U = Q²/2C = (2.0 × 10⁻³)² ÷ (2 × 100 × 10⁻⁶) = 2.0 × 10⁻² J. The potential difference is 20 V, and ½QV = 0.5 × 2.0 × 10⁻³ × 20 gives the same 0.020 J.

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