A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Dielectric Action in Capacitors and Relative Permittivity — mark scheme explained
The short answer
In this section, we will explore the dielectric action in capacitors and how it affects capacitance. We will also delve into the concept of relative permittivity and investigate the relationship between capacitance and the dimensions of a parallel-plate capacitor.
The question
A parallel-plate capacitor has plates with an area of 0.02 m 2 and a separation distance of 1 mm. The relative permittivity of the dielectric is 3. Calculate the capacitance.
[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 formulas, showing working steps, and providing the final answer with appropriate units. For conceptual questions, marks are given for clear and accurate explanations.
What the command words demand
- Calculate
- Perform a numerical calculation using given data and appropriate formulas.
- Determine
- Find the value of a quantity by applying a formula or method.
- Explain
- Provide a clear and detailed account of how something works or why it happens.
- Describe
- Give a detailed account of the characteristics or features of something.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5-7 minutes per question to ensure you have enough time to read, understand, and accurately answer each part.
- Identify the given values: A = 0.02 m 2 , d = 1 mm = 0.001 m, ε r = 3, and ε 0 = 8.85 × 10 -12 F/m.0 marks
- Use the formula for capacitance: C = ε 0 × ε r × (A / d).1 mark
- Substitute the values into the formula: C = 8.85 × 10 -12 × 3 × (0.02 / 0.001).1 mark
- Calculate the result: C = 8.85 × 10 -12 × 3 × 20 = 531 × 10 -12 F = 5.31 × 10 -10 F.1 mark
- The capacitance is 531 pF (5.31 × 10 -10 F).1 mark
Final answer: C = 531 pF (or 5.31 × 10 -10 F)
Work through every step correctly and you earn all 4 marks.
Another worked example
A parallel-plate capacitor has a capacitance of 20 pF without a dielectric and 60 pF with a dielectric. Calculate the relative permittivity (ε r ).
- Identify the given values: C = 60 pF, C 0 = 20 pF.0 marks
- Use the formula for relative permittivity: ε r = C / (C 0 ).1 mark
- Substitute the values into the formula: ε r = 60 pF / 20 pF.1 mark
- Calculate the result: ε r = 3.1 mark
- The relative permittivity is 3.0 marks
Final answer: 3
Work through every step correctly and you earn all 3 marks.
Common mistakes
Forgetting to convert units (e.g., mm to m) before using the formula for capacitance.
Why it happens: Students often forget to ensure all units are consistent, leading to incorrect calculations.
Fix: Always check and convert units to the same system (SI units) before substituting values into formulas.
Using the wrong formula for relative permittivity.
Why it happens: Students might confuse the formula for capacitance with the formula for relative permittivity.
Fix: Memorize and understand both formulas: C = ε 0 × ε r × (A / d) and ε r = C / (C 0 ).
Forgetting to include the permittivity of free space (ε 0 ) in calculations.
Why it happens: Students might overlook this constant, leading to incorrect results.
Fix: Always include ε 0 = 8.85 × 10 -12 F/m in the formula for capacitance.
Misinterpreting the effect of dielectric on electric field and capacitance.
Why it happens: Students might think that a dielectric increases the electric field, which is incorrect.
Fix: Understand that a dielectric reduces the electric field within the capacitor, thereby increasing its capacitance.
Confusing relative permittivity (ε r ) with absolute permittivity (ε).
Why it happens: Students might not clearly distinguish between these two concepts.
Fix: Remember that ε r is a dimensionless quantity, while ε = ε 0 × ε r has units of F/m.
Forgetting to align polar molecules in the presence of an electric field.
Why it happens: Students might not fully understand the mechanism by which dielectrics increase capacitance.
Fix: Understand that polar molecules align along the direction of the electric field, reducing the overall electric field within the dielectric material.
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 Capacitance | Use the parallel-plate formula with a dielectric to find the capacitance in farads. | 4 |
| Calculate Relative Permittivity | Find the relative permittivity from the capacitances with and without the dielectric. | 3 |
| Calculate Relative Permittivity | Rearrange the parallel-plate capacitor formula to find the relative permittivity from given values. | 4 |
| Calculate Capacitance | Use the parallel-plate formula with permittivity, area, and separation to find the capacitance. | 4 |
| Calculate Relative Permittivity | Use the ratio of capacitances with and without a dielectric to find relative permittivity. | 3 |
| Total across these question types | 18 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.