The questions are on the capacitor test. Take it under real conditions before reading the mark scheme.
Quick answer: A capacitor stores charge on two plates, and its capacitance is C = Q / V in farads. This mark scheme gives the model answers and the marks for the 8 questions of the Physics 11-12 test on capacitor, from defining capacitance to unit of the time constant, with the equivalent answers to accept.
Marks are awarded for a correct formula, correct conversion of prefixes, a clear substitution and an answer with its unit. Accept answers within rounding error and award method marks when an earlier slip is carried through consistently.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.7.4
Question 1 Defining capacitance (2 marks)
Capacitance is the charge stored on one plate per unit potential difference, C = Q / V. The unit is the farad, 1 F = 1 C/V.
Marking: 1 mark for the definition, 1 mark for the unit.
Question 2 Charge and energy (3 marks)
- Q = CV = 220 × 10⁻⁶ × 15 = 3.3 × 10⁻³ C.
- E = ½CV² = 0.5 × 220 × 10⁻⁶ × 15² = 0.025 J ≈ 2.5 × 10⁻² J.
Marking: (a) 1 mark. (b) 1 mark for the formula and substitution, 1 mark for the answer.
Question 3 Combining two capacitors (2 marks)
(a) 1/C = 1/8.0 + 1/8.0 = 0.25, so C = 4.0 µF. (b) C = 8.0 + 8.0 = 16 µF.
Marking: 1 mark each.
Question 4 Reading a discharge graph (4 marks)
- V₀ = 12 V.
- After one time constant V = 0.37 × 12 = 4.4 V. The graph reaches 4.4 V at about 10 s, so τ = 10 s.
- C = τ / R = 10 / (20 × 10³) = 5.0 × 10⁻⁴ F (500 µF).
Marking: (a) 1 mark. (b) 1 mark for 0.37V₀, 1 mark for τ. (c) 1 mark.
Question 5 Charging current (2 marks)
At the start the capacitor is uncharged, so all of the supply voltage is across the resistor and the current is V/R. As charge builds up, the voltage across the capacitor rises and the voltage across the resistor falls, so the current decreases. It reaches zero when the capacitor voltage equals the supply voltage.
Marking: 1 mark for all the voltage across R at the start, 1 mark for the capacitor voltage rising until it equals the supply.
Question 6 Voltage after a delay (3 marks)
τ = RC = 330 × 10³ × 4.7 × 10⁻⁶ = 1.55 s. V = V₀e−t/τ = 5.0 e−2.0/1.55 = 5.0 × 0.275 = 1.4 V.
Marking: 1 mark for τ, 1 mark for the substitution, 1 mark for the answer.
Question 7 Energy after one time constant (2 marks)
E ∝ V². After τ the voltage is V₀e⁻¹, so E = E₀(e⁻¹)² = E₀e⁻² = 0.135 E₀, which is 13.5 % ≈ 14 %.
Marking: 1 mark for E proportional to V², 1 mark for e⁻² = 0.135.
Question 8 Unit of the time constant (2 marks)
1 Ω = 1 V/A and 1 F = 1 C/V. So 1 Ω × 1 F = (V/A) × (C/V) = C/A = C/(C/s) = 1 s.
Marking: 1 mark for ohm and farad in base form, 1 mark for the cancellation to seconds.
