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AQA A-Level Physics: Interpretation of the Area Under a Graph of Charge Against Potential Difference — mark scheme explained

Machine-verifiedchecked against the AQA A-Level Physics specificationlast verified 2 July 2026

The short answer

In AQA A-Level Physics, understanding the interpretation of the area under a graph of charge against potential difference (pd) is crucial for several reasons. This topic involves key equations and concepts that are essential for solving problems related to capacitors and energy storage.

The question

A capacitor with a capacitance of 3 μF is charged to a potential difference of 6 V. Calculate the energy stored in the capacitor. [Paraphrased for study — not reproduced from any exam paper.]

Mark scheme, decoded

What each mark is really for — in plain English — and the wording trap that loses it.

  • S1

    Identify the given values: C = 3 μF, V = 6 V

  • S2

    Use the equation E = 1/2 CV 2

  • S3

    Substitute the values into the equation: E = 1/2 × 3 × 10 -6 F × (6 V) 2

  • S4

    Calculate the energy: E = 1/2 × 3 × 10 -6 × 36

  • S5

    E = 54 × 10 -6 J = 54 μJ

Model answer

Worked through, with each step tagged to the mark it earns.

  1. S1

    Identify the given values: C = 3 μF, V = 6 V

  2. S2

    Use the equation E = 1/2 CV 2

  3. S3

    Substitute the values into the equation: E = 1/2 × 3 × 10 -6 F × (6 V) 2

  4. S4

    Calculate the energy: E = 1/2 × 3 × 10 -6 × 36

  5. S5

    E = 54 × 10 -6 J = 54 μJ

  6. Final answer: 54 μJ

Common mistakes

  • Forgetting to use the factor of 1/2 in the energy equations. — Always include the factor of 1/2 when using the energy equations E = 1/2 QV, E = 1/2 CV 2 , or E = Q 2 /2C.
  • Confusing charge (Q) with capacitance (C). — Double-check that you are using the correct symbol for each quantity. Charge is Q, and capacitance is C.
  • Using the wrong units when substituting values into equations. — Ensure that all units are consistent before substituting values into equations. Convert all units to the same base unit if necessary.
  • Misinterpreting the area under a graph of charge against potential difference. — Remember that the area under a linear graph of charge against potential difference is given by 1/2 × base × height, which corresponds to E = 1/2 QV.
  • Failing to rearrange equations correctly when solving for unknowns. — Practice rearranging equations step-by-step. For example, to solve for C in E = 1/2 CV 2 , first isolate CV 2 and then divide by V 2 .
  • Using the wrong form of the energy equation for a given problem. — Identify which quantities are given in the problem and choose the appropriate form of the energy equation. For example, if you know C and V, use E = 1/2 CV 2 .

Where the marks go

  • Full worked solution (all marking points)4 marks

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