Capacitors
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This page is still being built and has known issues still to be fixed — some diagrams, examples and questions may be incomplete or change. Use it for an overview, but check your class notes and the SQA materials for anything you rely on.
Capacitance & charge
On the relationship sheet — given to you in the exam
Derived — 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.
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)
Energy stored in a capacitor
On the relationship sheet
Energy stored
E = ½QV = ½CV2 = ½
Q2C
pick the form from what you are given
Where the energy equations come from
The energy stored = the area under the Q–V graph.
Because Q ∝ V the graph is a triangle: area = ½ × base × height = ½VQ
So E = ½QV; then put in Q = CV to get E = ½CV2
All 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.
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
Answer
C = 1000 × 10⁻⁶ = 1.0 × 10⁻³ F (µF → F)
(a) Q = CV = 1.0 × 10⁻³ × 6.0 = 6.0 × 10⁻³ C
(b) E = ½CV² = 0.5 × 1.0 × 10⁻³ × 6.0² = 0.5 × 1.0 × 10⁻³ × 36 = 1.8 × 10⁻² J
Charging & discharging in RC circuits
Key idea
time to charge ∝ R × C
bigger R or C → slower charge / discharge
- 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. The product RC sets the timescale — the numerical value is not needed at Higher, only this effect.
- Fully charged: no current flows, the capacitor p.d. equals the supply p.d., and a capacitor blocks steady d.c. once charged.
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.
Beyond Higher — the time constant τ = RC (Advanced Higher)
This is Advanced Higher — you do not need it for Higher. It is just here as an interesting extra to explain why R 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.