AQA A-Level Physics Paper 2, June 2025: Question 17

1 mark · Easy difficulty · Multiple Choice

Calculate the capacitance of a parallel-plate capacitor whose plate height is doubled and dielectric constant is 3.0 relative to an initial capacitor of 100 μF.

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Question

Question 17 shows diagrams of two parallel-plate capacitors, X and Y. Capacitor X has plates of a given height, width, and separation, with air in between, having a capacitance of 100 microfarads. Capacitor Y has plates double the height of X, the same width and separation, and the space between the plates is filled with a dielectric material of relative permittivity 3.0. Four options for the capacitance of Y are given: A (67 microfarads), B (150 microfarads), C (600 microfarads), and D (1200 microfarads).

Mark scheme

Show the mark scheme Mark scheme table for question 17 showing the correct answer as C (600 microfarads) with assessment objective AO2.

How to answer it

Parallel-Plate Capacitor Dimensions & Dielectrics

WHAT THIS QUESTION TESTS

This question assesses AO2 (application of knowledge) concerning the physical factors affecting the capacitance of a parallel-plate capacitor: plate surface area (via plate dimensions), plate separation, and the relative permittivity (dielectric constant) of the dielectric medium.

Question 17 (Multiple Choice)

Analysis of Capacitance Scaling

✅ Correct Answer

Option C: 600 μF

Doubling the height doubles the active plate area (×2), and introducing a material with dielectric constant 3.0 scales the capacitance by another factor of 3 (×3).
Total factor = 2 × 3 = 6.
100 μF × 6 = 600 μF.

💡 Key Knowledge

  • Capacitance formula for a parallel-plate capacitor:
    C = (ε₀ εᵣ A) / d
  • A (Plate Area): For rectangular plates, Area = height × width .
  • εᵣ (Dielectric Constant): For air/vacuum, εᵣ ≈ 1.0 . Inserting a dielectric multiplies capacitance by εᵣ .
  • d (Plate Separation): Inversely proportional to capacitance ( C ∝ 1/d ).

📐 Step-by-Step Solution

1 Identify constants and variables between X and Y:
  • Separation: dY = dX (constant)
  • Width: wY = wX (constant)
  • Height: hY = 2 × hX
2 Determine the change in plate area (A):
AY = hY × wY = (2 × hX) × wX = 2 × AX
3 Determine the effect of the dielectric material:
Capacitor X has air between plates: εᵣ = 1.0
Capacitor Y has dielectric material: εᵣ = 3.0
4 Calculate capacitance of Y relative to X:
CY = (εᵣ) × (AY / AX) × CX
CY = 3.0 × 2 × 100 μF = 600 μF

🧠 Exam Technique: Proportional Reasoning

  • Avoid full substitutions: You do not need to look up or calculate using the value of ε₀ = 8.85 × 10⁻¹² F m⁻¹ . Work entirely with scaling ratios.
  • Track independent factors: Write down each scaling factor clearly:
    • Area factor: ×2
    • Dielectric factor: ×3
    • Separation factor: ×1
    Multiply them together: 2 × 3 × 1 = 6.

❌ Common Errors & Distractor Analysis

  • Option A (67 μF): Dividing by 3 and doubling ( 100 × 2 / 3 ). Dielectrics increase capacitance, never decrease it.
  • Option B (150 μF): Applying the dielectric factor but dividing by the area factor ( 100 × 3 / 2 ), or forgetting to double the height entirely and confusing ratios.
  • Option D (1200 μF): Assuming that doubling height also doubles the width (scaling area by 4 rather than 2: 100 × 4 × 3 = 1200 μF ). Read carefully: the width is explicitly stated as the same.
Mark Scheme Reference: 1 mark (AO2). Requires selection of key C (600 μF).

Topics

Physics · 3.7 Fields and their consequences (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.