A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Interpretation of the Area Under a Graph of Charge Against Potential Difference — mark scheme explained
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
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, marks are typically awarded for correct substitution of values into the appropriate equation, showing working, and providing the final answer with units. For explanation questions, marks are given for clear and accurate descriptions of concepts and equations.
What the command words demand
- Calculate
- Perform a numerical calculation to find the answer.
- Determine
- Find the value of a quantity using given data and equations.
- Explain
- Provide a clear and concise explanation, including relevant concepts and equations.
- Derive
- Show step-by-step how an equation or result is obtained from known principles.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 2-3 minutes per mark. For a 5-mark question, spend approximately 10-15 minutes.
- Identify the given values: C = 3 μF, V = 6 V0 marks
- Use the equation E = 1/2 CV 21 mark
- Substitute the values into the equation: E = 1/2 × 3 × 10 -6 F × (6 V) 21 mark
- Calculate the energy: E = 1/2 × 3 × 10 -6 × 361 mark
- E = 54 × 10 -6 J = 54 μJ1 mark
Final answer: 54 μJ
Work through every step correctly and you earn all 4 marks.
Another worked example
A capacitor stores 20 J of energy and has a charge of 8 C. Calculate the capacitance of the capacitor.
- Identify the given values: E = 20 J, Q = 8 C0 marks
- Use the equation E = Q 2 /2C1 mark
- Rearrange the equation to solve for C: C = Q 2 /2E1 mark
- Substitute the values into the equation: C = (8 C) 2 /2 × 20 J1 mark
- Calculate the capacitance: C = 64/40 F = 1.6 F2 marks
Final answer: 1.6 F
Work through every step correctly and you earn all 5 marks.
Common mistakes
Forgetting to use the factor of 1/2 in the energy equations.
Why it happens: Students often forget that the energy stored in a capacitor is half the product of charge and potential difference, leading to incorrect calculations.
Fix: 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).
Why it happens: Students sometimes mix up the symbols for charge and capacitance, leading to incorrect substitutions in equations.
Fix: 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.
Why it happens: Students often use inconsistent units, such as mixing microfarads (μF) with farads (F), leading to incorrect results.
Fix: 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.
Why it happens: Students may incorrectly interpret the area as simply Q × V, forgetting the factor of 1/2.
Fix: 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.
Why it happens: Students sometimes struggle with algebraic manipulation, leading to incorrect solutions when solving for variables like capacitance or potential difference.
Fix: 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.
Why it happens: Students may use an inappropriate form of the energy equation, such as E = 1/2 QV when they should be using E = 1/2 CV 2 or E = Q 2 /2C.
Fix: 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
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Calculate Capacitor Energy | Find the energy stored using the capacitance and potential difference values given. | 4 |
| Calculate Capacitance | Find the capacitance of a capacitor from its stored energy and charge. | 5 |
| Capacitor Energy Stored | Use charge and voltage to calculate the energy stored in a capacitor. | 3 |
| Calculate Potential Difference | Rearrange the capacitor energy equation to find the voltage from energy and capacitance. | 5 |
| Calculate Capacitor Charge | Rearrange the energy equation to find the charge stored on a capacitor. | 5 |
| Total across these question types | 22 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.