AQA A-Level Physics Paper 2, June 2023: Question 15

1 mark · Medium difficulty · Multiple Choice

Calculate the new force of attraction between two point charges when the separation is increased by 400 mm from 200 mm.

Practise this question

Question

Question 15 asks: 'Two point charges are separated by a distance of 200 mm. The force of attraction between them is 180 microNewtons. The distance between the point charges is increased by 400 mm. What is the new force of attraction?' Four options are given: A: 20 microNewtons, B: 45 microNewtons, C: 60 microNewtons, D: 90 microNewtons.
Question text

15 Two point charges are separated by a distance of 200 mm.

The force of attraction between them is 180 μN.

The distance between the point charges is increased by 400 mm.

What is the new force of attraction?

[1 mark]

A 20 μN

B 45 μN

C 60 μN

D 90 μN

Mark scheme

Show the mark scheme Mark scheme table row showing question 15, answer key 'A', and corresponding value '20 microNewtons'.

15 A 20 μN

How to answer it

Coulomb’s Law & Inverse-Square Relationships

📌 What this question tests

This question assesses your ability to apply Coulomb’s Law to calculate electrostatic force when separation changes, specifically testing:

  • Understanding the inverse-square law ( F ∝ 1 / r² ) for point charges.
  • Careful parsing of language: distinguishing between increasing to a value versus increasing by a value.
  • Rapid proportional reasoning under timed multiple-choice exam conditions without needing unnecessary full-equation substitutions.
Question 15 • Multiple Choice [1 Mark]

Full Solution & Analysis

AQA A-Level Physics • Electric Fields

✅ Correct Answer

A: 20 μN

The separation increases from 200 mm to 600 mm (a factor of 3). Because electrostatic force follows an inverse-square law, the new force is reduced by a factor of 3² = 9:

180 μN ÷ 9 = 20 μN

💡 Key Knowledge

  • Coulomb’s Law:
    F = (1 / 4πε₀) × (q₁q₂ / r²)
  • Since charge magnitudes q₁, q₂ and permittivity ε₀ remain constant, force is inversely proportional to distance squared:
    F ∝ 1 / r²  ⇒  F₂ / F₁ = (r₁ / r₂)²

📐 Step-by-Step Calculation

  1. Determine initial separation (r₁):
    r₁ = 200 mm
  2. Identify the new separation (r₂):
    The question states the distance is "increased by 400 mm".
    r₂ = r₁ + 400 mm = 200 mm + 400 mm = 600 mm
  3. Find the scaling factor for distance:
    r₂ / r₁ = 600 / 200 = 3  (the separation has tripled)
  4. Apply the inverse-square law:
    F₂ = F₁ / (r₂ / r₁)² = 180 μN / 3² = 180 μN / 9 = 20 μN

❌ Common Distractor Traps

  • Option B (45 μN) — The "Reading Trap":
    Assuming the new distance is 400 mm (doubled instead of tripled):
    180 / 2² = 45 μN . This is the most common student error.
  • Option C (60 μN) — The "Inverse Linear Trap":
    Correctly finding r₂ = 600 mm (tripled), but forgetting to square the factor:
    180 / 3 = 60 μN .
  • Option D (90 μN) — Double Failure:
    Assuming the distance doubled and using inverse linear proportionality:
    180 / 2 = 90 μN .

🧠 Exam Technique & Examiner Insight

  • Watch the Prepositions: "Increased by x" means r_new = r_old + x . "Increased to x" means r_new = x .
  • Avoid Full Formula Calculation: Do not waste time converting to standard SI units (metres, Coulombs) to calculate unknown charge values. Proportional scaling takes less than 20 seconds.
  • Keep the Units: Because the multiplier is dimensionless, you can keep the force in μN and distances in mm .
Mark Scheme Breakdown:
• Correct option: A (20 μN) [1 Mark]
• Incorrect choices receive 0 marks. No partial credit in Section A multiple choice.

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 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.