Physics 9702 · syllabus 19
Cambridge A Level Physics 9702 — Capacitance topical past papers
Storing charge and energy: capacitance of plates and spheres, capacitors in series and parallel, stored energy and exponential discharge through a resistor.
- Questions
- 64
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 26
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What Capacitance covers
A capacitor stores equal and opposite charges on two conductors separated by an insulator, with a p.d. between them. Capacitance is the charge stored per unit p.d., C = Q/V, measured in farads. The definition applies both to a parallel-plate capacitor and to an isolated conducting sphere, whose capacitance follows from its potential: C = 4πε₀r. From C = Q/V you also derive the rules for combining capacitors, which are the reverse of those for resistors.
The energy stored is the area under a graph of p.d. against charge, which is a triangle, so W = ½QV = ½CV². Not all the energy a supply transfers is stored: charging a capacitor from a fixed supply through a resistor dissipates as much energy in the circuit as ends up in the capacitor, as comparing QV with ½QV shows.
When a capacitor discharges through a resistor, the current is proportional to the charge remaining, so charge, p.d. and current all fall exponentially: x = x₀e^(−t/RC). The time constant τ = RC is the time taken to fall to 1/e, about 37%, of the starting value. This is A Level content, examined on Paper 4; it uses electric fields and D.C. circuits, and capacitors return in smoothing rectified a.c.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 19 Capacitance
- 19.1 Capacitors and capacitance
- 19.2 Energy stored in a capacitor
- 19.3 Discharging a capacitor
What you should be able to do
- Define capacitance and the farad, and apply C = Q/V to parallel plates and isolated spheres
- Derive and use the formulae for capacitors in series and in parallel
- Find stored energy from the area under a p.d.–charge graph and use W = ½QV = ½CV²
- Use τ = RC and x = x₀e^(−t/RC) for charge, current or p.d. during discharge
- Find a time constant from a discharge graph, from the 37% point or from a graph of ln V against t
Key ideas
- C = Q/V (1 F = 1 C V⁻¹); isolated sphere of radius r: C = 4πε₀r
- Parallel: C = C₁ + C₂ + …; series: 1/C = 1/C₁ + 1/C₂ + …
- Energy stored W = ½QV = ½CV² = Q²/2C
- Discharge: x = x₀e^(−t/RC), with time constant τ = RC (s); after one time constant x = 0.37x₀
- ln V = ln V₀ − t/RC, so a graph of ln V against t is a straight line of gradient −1/RC
How Capacitance has been examined
From 2022 to 2026, Capacitance was on Paper 4 · A Level Structured Questions in 9 of 11 exam series, worth about 8% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · A Level Structured Questions | 26 | 8% | 9 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 10.
Filed in Momentum under: Capacitance.
A worked example
A 470 μF capacitor charged to 12.0 V discharges through a 22 kΩ resistor. Find the time constant, the time for the p.d. to fall to 3.0 V, and the energy transferred from the capacitor in that time.
- τ = RC = 22 × 10³ × 470 × 10⁻⁶ = 10.3 s
- V = V₀e^(−t/RC), so t = RC ln(V₀/V) = 10.34 × ln(12.0/3.0) = 10.34 × 1.386 = 14.3 s
- Energy stored at 12.0 V: ½CV² = ½ × 470 × 10⁻⁶ × 12.0² = 3.38 × 10⁻² J
- Energy stored at 3.0 V: ½ × 470 × 10⁻⁶ × 3.0² = 0.21 × 10⁻² J
- Energy transferred = 3.38 × 10⁻² − 0.21 × 10⁻² = 3.17 × 10⁻² J
Answer: Time constant 10 s; time 14 s; energy transferred 3.2 × 10⁻² J
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Combining capacitors with the resistor rules — capacitances in parallel add directly, and in series their reciprocals add
- Leaving out the half in W = ½QV — the energy is the area of a triangle under the V–Q graph, not of a rectangle
- Multiplying μF by kΩ and calling the result seconds — 470 μF × 22 kΩ is 10.3 s only after converting to farads and ohms
- Treating discharge as linear — a steady fall would reach zero after a fixed time, but exponential decay halves the charge every 0.69RC whatever its starting value, so falling to a quarter takes twice as long as falling to a half
Exam technique
- To find τ from a discharge curve, read the time at which the value has fallen to 37% of its starting value, or plot ln V against t and use the gradient
- Rearrange the exponential by taking natural logarithms, t = RC ln(x₀/x), and keep three figures in the working before rounding
- For energy questions, decide whether Q or V stays fixed (an isolated capacitor or one connected to a supply) before choosing between ½QV, ½CV² and Q²/2C
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.