On the relationship sheet — given to you in the examDerived — rearrange the given one, or just remember it
On the relationship sheet
Capacitance
C =QV
capacitance = charge ÷ p.d.
On the relationship sheet
Charge stored
Q = It
for a constant charging current
A capacitor stores charge (and energy) when a p.d. is applied across it. Its capacitance C is the charge stored per volt: C = Q ÷ V, measured in the farad (F).
Definition to learn: a capacitor of 1 farad stores 1 coulomb of charge when the p.d. across it is 1 volt. A farad is huge — real capacitors are µF (×10⁻⁶), nF (×10⁻⁹) or pF (×10⁻¹²).
As a capacitor charges, the p.d. across it rises until it equals the supply p.d.; then charging stops. The charge stored is Q = CV.
If a capacitor is charged with a constant current for a time t, the charge delivered is Q = It — the same charge relationship from Current, p.d., power & resistance.
The capacitor (two parallel plates) charges up through R; the voltmeter reads the p.d. across the capacitor.
Fill the gaps — say it like the SQA
Pick the precise word for each gap, then check. The marks here are for using the right terms.
A capacitor stores
when a potential difference is applied across it. Its capacitance is the charge stored per
,
and is measured in the
.
If a constant current charges it, the charge delivered is Q =
.
Match the term to its meaning
Tap a term on the left, then its meaning on the right. Correct pairs lock green.
Term
Meaning
Substitute & solve
A capacitor stores 6.0 × 10⁻⁴ C of charge when the p.d. across it is 12 V. Calculate its capacitance. 3 marks
C=Q ÷ Von the sheet
C=6.0 × 10⁻⁴ ÷ 12
C=5.0 × 10⁻⁵ F (50 µF)
Rearrange — numbers in first
A 470 µF capacitor has a p.d. of 9.0 V across it. Calculate the charge stored. 3 marks
Convert first
C = 470 × 10⁻⁶ = 4.70 × 10⁻⁴ F µF → F
C=Q ÷ V
4.70 × 10⁻⁴=Q ÷ 9.0numbers in first
Q=4.70 × 10⁻⁴ × 9.0now rearrange
Q=4.2 × 10⁻³ C
Practice 1
A constant current of 2.0 mA charges a capacitor for 5.0 s. The final p.d. across the capacitor is 8.0 V. Calculate (a) the charge stored and (b) the capacitance. 4 marks
Answer
I = 2.0 × 10⁻³ A (mA → A)
(a) Q = It = 2.0 × 10⁻³ × 5.0 = 1.0 × 10⁻² C
(b) C = Q ÷ V = 1.0 × 10⁻² ÷ 8.0 = 1.25 × 10⁻³ F (1.3 mF)
Concept 2
Energy stored in a capacitor
On the relationship sheet
Energy stored
E = ½QV = ½CV2 = ½Q2C
pick the form from what you are given
Where the energy equations come fromThe energy stored = the area under the Q–V graph.Because Q ∝ V the graph is a triangle: area = ½ × base × height = ½VQSo E = ½QV; then put in Q = CV to get E = ½CV2All three forms give the same energy — choose the one that uses the quantities the question gives you.
The energy stored in a charged capacitor equals the area under its charge–p.d. (Q–V) graph. The graph is a straight line through the origin, so the area is a triangle: ½ × V × Q.
The other two forms come from substituting Q = CV: E = ½CV2 and E = ½Q2/C.
Two classic slips: don't forget the ½, and square the V (or Q) where the formula tells you to.
Why the ½? Unlike a resistor, a capacitor only reaches full p.d. gradually — on average the charge moves through half the final p.d., so the energy is ½QV, not QV.
The shaded triangle under the Q–V line is the energy stored: ½ × V × Q.
Check your calculation
A 220 µF capacitor is charged to a p.d. of 12 V. Work out the energy stored (in J), type your answer, then check — you'll get a hint if it's a common slip.
Substitute & solve
A 220 µF capacitor is charged to a p.d. of 12 V. Calculate the energy stored. 3 marks
Convert first
C = 220 × 10⁻⁶ = 2.20 × 10⁻⁴ F µF → F
E=½CV2given C and V
E=0.5 × 2.20 × 10⁻⁴ × 12²square the p.d.
E=0.5 × 2.20 × 10⁻⁴ × 144
E=1.6 × 10⁻² J
Choose the right form — rearrange
A capacitor stores 3.6 × 10⁻³ J of energy when it holds a charge of 6.0 × 10⁻⁴ C. Calculate the p.d. across the capacitor. 3 marks
E=½QVgiven E and Q
3.6 × 10⁻³=0.5 × 6.0 × 10⁻⁴ × Vnumbers in first
3.6 × 10⁻³=3.0 × 10⁻⁴ × V
V=3.6 × 10⁻³ ÷ 3.0 × 10⁻⁴now rearrange
V=12 V
Practice 2
A 1000 µF capacitor is charged to 6.0 V. Calculate (a) the charge stored and (b) the energy stored. 4 marks
Charging (through a resistor R): the instant the switch closes the capacitor is uncharged, so the current is a maximum and the p.d. across the capacitor is zero. As charge builds up, the capacitor p.d. rises towards the supply voltage while the current falls towards zero. Both curves are exponential in shape.
Discharging: the capacitor p.d. falls from a maximum towards zero. The current now flows the opposite way, so it is negative — it starts at its largest reverse value and rises back up to zero.
Effect of R and C: increasing either R or C makes charging and discharging take longer (a more gradual curve, smaller initial current). Decreasing them makes it faster. A bigger R means a smaller current, so charge flows on or off more slowly; a bigger C needs more charge for the same p.d.
Fully charged: no current flows, the capacitor p.d. equals the supply p.d., and a capacitor blocks steady d.c. once charged.
ChargingDischarging — the current is negative (it flows the opposite way), starting below the axis and rising to zero.
capacitor p.d. current
Charging and discharging simulator
Switch between charge and discharge, then change R and C. The time axis stretches so the whole curve always fits: watch its numbers.
The capacitor is fully charged. The switch moves to discharge. Predict what the current does.
Switch at charge: the supply pushes charge on to the capacitor.
Current at the start—
p.d. across C at the end—
Charge when full, Q = CV—
Energy when full, ½QV—
Switch
Predict & justify
A capacitor is charged through a resistor from a d.c. supply. The resistor is replaced with one of larger resistance. The time taken to fully charge the capacitor will…
The ammeter (in series) reads the capacitor current; the voltmeter (across C) reads its p.d. Record both at regular time intervals (a data logger helps for fast changes), then plot against time.
Read the charging curves
Use the graphs and the simulator above, then check both answers.
1. At the instant the switch closes (capacitor uncharged), the charging current is…
2. As the capacitor charges, the p.d. across it…
Practice 3
A capacitor is charged through a resistor from a d.c. supply. (a) Describe and explain what happens to the current in the circuit from the instant the switch is closed. (b) The resistor is replaced with one of larger resistance. State the effect on the time taken to fully charge the capacitor. 3 marks
Answer
(a) The current is a maximum at the instant the switch closes (the
capacitor is uncharged, so the p.d. across it is zero), then decreases
towards zero as the capacitor charges and its p.d. rises to oppose
the supply.
(b) The capacitor takes longer to fully charge.
Go further (Advanced Higher): the time constant τ = RC
This is Advanced Higher — you do not need it for Higher. It is just here as an interesting extra to explain whyR and C change the curves.
The product RC is called the time constant τ (unit: seconds): τ = RC. It sets roughly how long the capacitor takes to charge or discharge — a larger τ means a slower, more gradual curve. At Higher you only need the effect of R and C, never a numerical time constant.
The simulator's graphs always show at least five time constants. That is why the curves fill the graph whatever R and C you choose: watch the numbers on the time axis change instead.
Check yourself
Recap — fill the gaps
Pull the whole topic together: choose the right word for each gap, then check. Counts towards your badges.
A capacitor stores charge: its capacitance C = Q ÷
.
The energy stored equals the area under the
graph, which works out as E = ½QV. While charging through a resistor, the current starts at a
and falls to zero, while the p.d. across the capacitor
to the supply voltage. Increasing R or C makes this take
.
Mixed multiple choice
Exam-style multiple choice, 1 mark each. Choose an answer for each, then mark them.
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