A Level Physics: Capacitance — Revision Notes

Condensed for the final weeks. For the full explanation, use the Capacitance study guide.

Assessed on Paper 4 (A Level structured, 2 hours, 100 marks, 38.5% of the A Level).

Definitions and equations

capacitance        C = Q / V          farads (F)

energy stored      W = 1/2 Q V
                     = 1/2 C V^2
                     = 1/2 Q^2 / C

Capacitance is the charge stored per unit potential difference. A capacitor stores charge separation and energy in the electric field between its plates. One farad is a large unit for practical circuits, so capacitance values are usually quoted in microfarads (μF, ×10⁻⁶), nanofarads (nF, ×10⁻⁹) or picofarads (pF, ×10⁻¹²) — always convert to farads before substituting into an equation.

Combining capacitors — the reverse of resistors

PARALLEL:   C_total = C1 + C2 + C3          (capacitances ADD)
SERIES:     1/C_total = 1/C1 + 1/C2 + ...   (reciprocals add)

This is the opposite of the resistor rules, and swapping them is the most common error in the topic. In parallel the effective plate area increases, so capacitance increases.

Worked example. Two capacitors, 4 μF and 12 μF, are connected first in series and then in parallel. Find the combined capacitance each way.

SERIES:    1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3
           C = 3 uF   (always LESS than the smallest individual value)

PARALLEL:  C = 4 + 12 = 16 uF   (always MORE than the largest individual value)

A quick sanity check for any series/parallel calculation: the series total must always come out smaller than the smallest capacitor in the combination, and the parallel total must always come out larger than the largest — if your answer doesn’t satisfy that, the formulas have likely been swapped.

Why energy is ½QV, not QV

As charge accumulates, the p.d. rises from zero. The work done to move each successive charge increases linearly, so the total is the area under the Q–V graph — a triangle, hence ½QV.

That “area under the graph” reasoning is worth quoting when asked to explain the factor of ½.

Discharge through a resistor

Q = Q0 e^(-t / RC)
V = V0 e^(-t / RC)
I = I0 e^(-t / RC)

All three decay exponentially with the same time constant.

Time constant τ = RC (seconds). After time τ the quantity falls to 1/e ≈ 37% of its initial value.

after 1 tau   -> 37%
after 2 tau   -> 13.5%
after 3 tau   -> 5%
after 5 tau   -> essentially fully discharged

Half-life: t½ = RC ln 2 ≈ 0.69 RC.

To find RC graphically, plot ln Q against t: since ln Q = ln Q₀ − t/RC, this gives a straight line with gradient −1/RC, letting you find the time constant from experimental discharge data without needing to identify a specific point where the charge has fallen to exactly 37%.

Worked example. A 100 μF capacitor is charged to 20 V. Find the energy stored.

W = 1/2 C V^2 = 0.5 x 100x10^-6 x 20^2 = 0.5 x 100x10^-6 x 400 = 0.02 J

Charging

Charge and p.d. rise as Q = Q₀(1 − e^(−t/RC)), while current decays exponentially from its initial maximum V₀/R.

Exam traps

• Series and parallel formulae are the reverse of the resistor ones.
• Energy is ½QV — not QV.
• τ = RC has units of seconds; check R in ohms and C in farads (not μF).
• Convert μF to F: 1 μF = 10⁻⁶ F.
• During charging, current decays while charge grows.
• The time constant is independent of the initial charge.
• Forgetting that a series combination must come out smaller than the smallest individual capacitor, and a parallel combination larger than the largest — a quick check that catches a swapped formula immediately.
• Leaving capacitance in μF when substituting into τ = RC or W = ½CV² — both equations require farads, so convert first.

Self-test

1. Define capacitance and give its unit.
2. Two 6 μF capacitors in series — find the total capacitance. And in parallel?
3. Why is the energy stored ½QV rather than QV?
4. What fraction remains after one time constant?
5. A 100 μF capacitor discharges through 10 kΩ. Find τ.
6. Two capacitors, 4 μF and 12 μF, are connected in series. Find the combined capacitance, and explain why your answer must be less than 4 μF.
7. A 100 μF capacitor is charged to 20 V. Find the energy stored.

Answers: 1. The charge stored per unit potential difference, C = Q/V, measured in farads. 2. Series: 1/C = 1/6 + 1/6 → 3 μF. Parallel: 12 μF. 3. The p.d. rises from zero as charge accumulates, so the work done per unit charge increases linearly; the total energy is the area under the Q–V graph, which is a triangle. 4. 1/e, about 37%. 5. τ = RC = 10 000 × 100 × 10⁻⁶ = 1.0 s. 6. 1/C = 1/4 + 1/12 = 1/3, so C = 3 μF; a series combination must be smaller than the smallest individual capacitance, because the same charge sits on each capacitor but the potential differences add, making the overall C = Q/V smaller. 7. W = ½CV² = 0.5 × 100×10⁻⁶ × 20² = 0.02 J.

For the full derivation of each formula and further worked examples, see the Capacitance study guide; for exam-style practice questions with complete worked answers, see the Capacitance practice questions.