Learn9702Paper 4 · A Level Structured QuestionsCapacitanceCapacitors and capacitance

Capacitance · Subtopic 1 of 3

Capacitors and Capacitance

A capacitor stores separated charge: $+Q$ on one plate, $-Q$ on the other. Its capacitance is the charge stored per unit p.d.: $C=\dfrac QV$ , measured in farads (F). An isolated sphere also has capacitance, $C=4\pi\varepsilon_0r$ . Capacitors in parallel add: $C=C_1+C_2$ . In series, reciprocals add: $\dfrac1C=\dfrac1{C_1}+\dfrac1{C_2}$ , the opposite of resistors.

Cambridge 9702 Paper 4Edexcel YPH11 WPH14AQA 7408 Paper 2≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Define capacitance and the farad; use $C=\dfrac QV$ .
Find the capacitance of an isolated sphere.
Derive and use the series and parallel formulae.
Find charges and p.d.s in capacitor networks.

Pause and think

Connecting two identical capacitors in series halves the capacitance, while two resistors in series doubles the resistance. Why are they opposite?

Section 2 of 14

Core idea

Start hereA camera flash

A camera flash needs a sudden burst of energy that a small battery can't supply instantly. A capacitor charges slowly from the battery, then dumps its stored charge in a millisecond.

The big idea

$C=\frac QV$ (F). Parallel: $C=C_1+C_2$ . Series: $\frac1C=\frac1{C_1}+\frac1{C_2}$ .

1. Capacitance

Charging a parallel-plate capacitor

Electrons are pushed onto one plate and pulled off the other. No charge crosses the gap. The capacitor stores

+Q

and

-Q

; "charge stored" means

Q

Q against V is a straight line with gradient C

At 10 V,

Q=470

μC. The area under the line is the energy stored (next lesson).

C=\frac QV

$Q$: charge on one plate (C).
$V$: p.d. between plates.
$C$: farads.

Worked exampleFrom a Q–V graph

$Q=470$ μC at 10 V.

Goal: $C$ ; $Q$ at 6 V.
Strategy: Gradient of $Q$ – $V$ .

$C=\frac{470\times10^{-6}}{10}=47\ \mu\text{F}$
Definition.
$Q=47\times6=282\ \mu\text{C}$
$Q=CV$ .

Answer

47

μF;

282

μC.

✓ CheckProportional ✓.

2. Combinations

Series and parallel

$Q=Q_1+Q_2=C_1V+C_2V\Rightarrow C=C_1+C_2$
Parallel: same $V$ , charges add.
$V=V_1+V_2=\frac Q{C_1}+\frac Q{C_2}$
Series: same $Q$ on each.
$\frac1C=\frac1{C_1}+\frac1{C_2}$
Divide by $Q$ .

The opposite of resistor rules.

Worked example6 μF and 3 μF on 12 V

Series and parallel.

Goal: $C$ , charges, voltages.
Strategy: Rules above.

$\tfrac1C=\tfrac16+\tfrac13=\tfrac12\Rightarrow C=2\ \mu\text{F}$
Series.
$Q=24\ \mu\text{C};\ V=4\text{ V},\ 8\text{ V}$
Same $Q$ ; $V=Q/C$ .
$C=9\ \mu\text{F},\ Q=108\ \mu\text{C}$
Parallel.

AnswerSeries 2 μF (4 V, 8 V); parallel 9 μF, 108 μC.

✓ Check

4+8=12

V ✓; smaller capacitor takes the larger voltage ✓.

Common misconceptionResistor rules Capacitors in parallel add.

Summary

$C=Q/V$ .
Parallel adds; series reciprocals.

Farad: Coulomb per volt.

HistoryThe Leyden jar. In

1745

Pieter van Musschenbroek in Leiden charged water in a glass jar from a friction machine and received a shock so violent he said he wouldn't repeat it "for the whole kingdom of France". The Leyden jar was the first capacitor; Benjamin Franklin later used them to study lightning.

Fun factSupercapacitors. Ordinary capacitors are micro- or nanofarads, but supercapacitors reach thousands of farads by using porous carbon with huge surface areas. Some electric buses in China recharge supercapacitors at each stop in under a minute.

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