AQA A-Level Physics Paper 2, June 2025: Question 5

✅ Correct Answer

Any one of the following clear statements:

• The capacitance (or charge stored for the same potential difference) with the dielectric is 7 times greater than with a vacuum (or air / free space).
• The permittivity of the material is 7 times the permittivity of free space: ε / ε₀ = 7.0

💡 Key Knowledge

The dielectric constant (also called relative permittivity, ε_r) is a dimensionless ratio:

ε_r = C / C₀ = ε / ε₀

It indicates how much more charge a capacitor stores per volt compared to having empty space (vacuum) between the plates.

❌ Common Errors

• Saying: "The relative permittivity is 7 times greater than free space" — this conflates ε_r with absolute permittivity ε. Relative permittivity is 7, not 7 times ε₀!
• Writing equations like ε_r / ε₀ = 7 (incorrectly dividing the ratio by ε₀ again).
• Forgetting to mention "for the same pd" or "same capacitor dimensions" when referencing charge.

🧠 Exam Technique

Always express ratio definitions in words clearly: "The capacitance with this dielectric is 7 times the capacitance with a vacuum between the plates." This is guaranteed to hit the mark scheme directly.

🔧 Step-by-Step Method

1 Identify initial current:
At t = 0, I₀ = 40 μA.

2 Calculate current at t = T:
I = I₀ / e = 40 / 2.718 ≈ 14.7 μA (read-off acceptable at 14 – 15 μA).

3 Read time from horizontal axis:
At I = 14.7 μA, follow horizontally to the curve and down to the time axis:
T = 73 ± 2 s (Acceptable range: 71 s – 75 s).

💡 Alternative Valid Methods

• Half-Life Method: Find the time taken to fall to half the initial value (from 40 μA to 20 μA):
  t_½ ≈ 51 s.
  Then use: T = t_½ / ln(2) = 51 / 0.693 ≈ 73.6 s.
• Point Substitution: Pick a well-defined point on the curve, e.g. (60 s, 17.5 μA), and solve:
  17.5 = 40 e^−60/T ⇒ T ≈ 72.6 s.

🧠 Exam Technique

Always show construction lines or written working on the graph! Mark 1 is an AO2 method mark awarded for demonstrating a valid method (e.g. calculating 40/e = 14.7 μA and drawing a line across). If you write down a final value without working and make a small reading error, you lose both marks.

❌ Common Traps

• Confusing time constant (time to fall to 37%) with half-life (time to fall to 50%). A reading of ~51 s is the half-life, not T!
• Misreading the minor grid divisions on the time axis: each small square represents 2 seconds.

🔧 Step-by-Step Calculation

1 Use exponential decay relation:
I = I₀ e^−t/T
At t = 5T:
I / I₀ = e^−(5T/T) = e⁻⁵

2 Evaluate e⁻⁵:
e⁻⁵ = 0.0067379...

3 Convert to percentage:
Percentage = 0.0067379 × 100% = 0.67% (or 0.674%)

✅ Correct Answer

0.67%

(Note: if an approximation of 0.37 for 1/e is used: 0.37⁵ = 0.00693 ⇒ 0.69% is also condoned by the mark scheme).

❌ Common Errors

• Forgetting to convert the decimal 0.0067 into a percentage (× 100%).
• Using values read from the graph at t = 5 × 73 = 365 s (the graph only goes to 120 s!). You must use the mathematical exponential relation.

🧠 Key Insight

Notice that the time constant T cancels out completely: t / T = 5T / T = 5 . You don't even need your answer from 05.2 to score full marks here!

✅ Key Graph Features Required

Your sketched line must have:

• Initial point: Starts at −20 μA at t = 0.
• Negative throughout: Stays in the negative region (below zero) for all times.
• Correct half-life coordinate: At t = 100 s, the current must pass through half its initial value: (−10 μA ± 1 division).
• Correct shape: A smooth curve with decreasing gradient tending asymptotically towards 0 μA (drawn out to at least 120 s).

📍 How to Plot the Graph Accurately

Plot these specific guide points before sketching the smooth curve:

• (0 s, −20 μA) — initial intercept
• (100 s, −10 μA) — one half-life
• (146 s, −7.4 μA) — one time constant (146 s)
• Ensure the line flattens out smoothly towards the horizontal time axis (0 μA) without crossing it.

❌ Common Pitfalls from Examiner Reports

• Missing the second resistor: Many students drew a curve starting at ±40 μA, failing to see the loop now has 2 resistors in series.
• Wrong sign: Drawing a positive curve starting at +20 μA because they forgot that charging and discharging currents flow through the ammeter in opposite directions.
• Wrong curvature: Drawing an S-shape or a curve that levels off at a non-zero current. Charging current decays to zero as the capacitor reaches full charge.