A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Definition of Capacitance: C = Q / V — mark scheme explained
The short answer
Capacitance is a fundamental concept in the study of electrical circuits and fields. It describes how much electric charge a device can store for a given voltage. The definition of capacitance, as specified by AQA A-Level Physics, is: C = Q / V Where: C is the capacitance measured in farads (F).
The question
A capacitor with a capacitance of 20 μF is charged to a voltage of 15 V. Calculate the charge stored on 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 identifying the correct formula (1 mark), substituting the given values correctly (1 mark), performing the calculation accurately (1 mark), and providing the final answer with the correct units (1 mark). For explanation questions, marks are awarded for clarity, accuracy, and completeness of the response.
What the command words demand
- Calculate
- Perform a numerical calculation using the given formula.
- Determine
- Find the value of a variable using the provided information.
- Explain
- Provide a clear and concise explanation of a concept or relationship.
- Derive
- Show the steps to derive a formula from basic principles.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: For a 4-mark question, allocate approximately 3-4 minutes to read the problem, perform calculations, and check your work. Ensure you have enough time to double-check your units and final answer.
- Identify the given values: C = 20 μF, V = 15 V0 marks
- Convert the capacitance to farads: 20 μF = 20 × 10 -6 F1 mark
- Use the formula Q = C × V1 mark
- Substitute the values: Q = 20 × 10 -6 F × 15 V1 mark
- Calculate the charge: Q = 300 × 10 -6 C = 300 μC1 mark
Final answer: 300 μC
Work through every step correctly and you earn all 4 marks.
Another worked example
A capacitor stores a charge of 500 μC when the voltage across it is 25 V. Calculate the capacitance of the capacitor.
- Identify the given values: Q = 500 μC, V = 25 V0 marks
- Convert the charge to coulombs: 500 μC = 500 × 10 -6 C1 mark
- Use the formula C = Q / V1 mark
- Substitute the values: C = (500 × 10 -6 C) / 25 V1 mark
- Calculate the capacitance: C = 20 × 10 -6 F = 20 μF1 mark
Final answer: 20 μF
Work through every step correctly and you earn all 4 marks.
Common mistakes
Confusing the units of capacitance, charge, and voltage.
Why it happens: Students often mix up the units, especially when converting between microfarads (μF) and farads (F), or between microcoulombs (μC) and coulombs (C).
Fix: Always double-check the units and convert them to standard units (farads and coulombs) before performing calculations.
Using the wrong formula for capacitance, charge, or voltage.
Why it happens: Students may confuse the formulas C = Q / V, Q = C × V, and V = Q / C. This can lead to incorrect calculations.
Fix: Memorize the main formula C = Q / V and derive the other forms as needed. Practice using each form in different contexts.
Forgetting to convert units before performing calculations.
Why it happens: Students often forget to convert capacitance from microfarads (μF) to farads (F) or charge from microcoulombs (μC) to coulombs (C), leading to incorrect results.
Fix: Always check and convert units before substituting values into the formula. Use consistent units throughout the calculation.
Misinterpreting the relationship between charge, voltage, and capacitance.
Why it happens: Students may not fully understand that increasing the voltage across a capacitor increases the charge stored, assuming a constant capacitance. This can lead to incorrect problem-solving approaches.
Fix: Understand that the relationship is linear: Q = C × V. Practice problems that involve changing one variable while keeping another constant.
Using the wrong value for capacitance or charge in calculations.
Why it happens: Students may use the wrong value from a given problem, such as using the total charge instead of the charge on one plate of a capacitor.
Fix: Carefully read the problem statement and identify the correct values to use. Double-check your work to ensure you are using the right values in the formula.
Failing to check the final answer for reasonableness.
Why it happens: Students may not verify if their calculated value makes sense in the context of the problem. For example, a capacitance of 10 6 F is unrealistic.
Fix: Always check your final answer to ensure it is reasonable and consistent with the given values and units. Use estimation techniques to verify the order of magnitude.
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 Charge | Find the charge stored on a capacitor from its capacitance and voltage. | 4 |
| Calculate Capacitance | Find a capacitor's capacitance from its stored charge and voltage. | 4 |
| Calculate Capacitor Voltage | Rearrange the capacitance equation to find the voltage from charge and capacitance. | 4 |
| Calculate Capacitor Charge | Use the capacitance equation to find the charge stored on a capacitor from its voltage. | 4 |
| Total across these question types | 16 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.