Specification: AQA A-level Physics (7408) 3.7.4.3 | Paper: 2 | Data and formulae booklet: given in the exam
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Energy stored by a capacitor: the work done to move charge onto the plates against the repulsion of the charge already there, stored as electric potential energy in the field between the plates. E = ½QV = ½CV² = ½Q² ÷ C.
Q–V graph: a graph of pd V against charge Q (or Q against V) for a capacitor is a straight line through the origin. The energy stored is the area under the graph, which is a triangle, so E = ½QV, not QV.
Charging from a battery: the battery transfers energy QV to the circuit, but the capacitor stores only ½QV. The other half is dissipated as heat in the resistance of the circuit, whatever the value of the resistance.
The equation
energy stored = ½ × charge × pd E = ½QV = ½CV² = ½Q² ÷ C
| Quantity | Symbol | Unit |
|---|---|---|
| energy stored | E | joules (J) |
| charge stored (on either plate) | Q | coulombs (C) |
| potential difference across the capacitor | V | volts (V) |
| capacitance | C | farads (F) |
Rearranged: V = √(2E ÷ C) | Q = √(2EC) | C = 2E ÷ V² | Pick the form that uses the quantities you are given: ½CV² when V is fixed (connected to a supply), ½Q² ÷ C when Q is fixed (isolated capacitor).
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A capacitor stores a charge of 3.0 mC when the pd across it is 12 V. Calculate the energy stored.
Show solution
Convert first: 3.0 mC = 3.0 × 10−3 C
F: E = ½QV
I: E = ½ × 3.0 × 10−3 × 12
F: E is already the subject
A: 0.018 J (18 mJ)
Q2 (Core) A 470 µF capacitor is charged to 9.0 V. Calculate the energy stored, to 2 significant figures.
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Convert first: 470 µF = 470 × 10−6 F
F: E = ½CV²
I: E = ½ × 470 × 10−6 × 9.0²
F: E is already the subject
A: 0.0190… J = 0.019 J (19 mJ, 2 s.f.)
Watch out: square the pd only, not the whole of ½CV.
Q3 (Core) A 10 µF capacitor stores a charge of 200 µC. Calculate the energy stored in mJ.
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Convert first: 10 µF = 1.0 × 10−5 F; 200 µC = 2.0 × 10−4 C
F: E = ½Q² ÷ C
I: E = ½ × (2.0 × 10−4)² ÷ 1.0 × 10−5
F: E is already the subject
A: 2.0 × 10−3 J = 2.0 mJ
Q4 (Core) A 2200 µF capacitor must store 0.50 J of energy. Calculate the pd it must be charged to, to 2 significant figures.
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Convert first: 2200 µF = 2.2 × 10−3 F
F: E = ½CV²
I: 0.50 = ½ × 2.2 × 10−3 × V²
F: V = √(2E ÷ C) = √(1.0 ÷ 2.2 × 10−3) = √454.5…
A: 21.3… V = 21 V (2 s.f.)
Watch out: forgetting the square root (giving 450 V) or the ½ (giving 15 V).
Q5 (Exam) A graph of charge Q against pd V for a capacitor is a straight line from the origin to Q = 6.0 mC at V = 15 V. (a) Calculate the capacitance. (b) Use the area under the graph to calculate the energy stored at 15 V. (c) Calculate the energy stored when the pd is 7.5 V, to 2 significant figures.
Show solution
Step 1 – (a) capacitance from the gradient
C = Q ÷ V = 6.0 × 10−3 ÷ 15
A: 4.0 × 10−4 F = 400 µF
Step 2 – (b) area under the graph
F: E = area of triangle = ½QV
I: E = ½ × 6.0 × 10−3 × 15
A: 0.045 J (45 mJ)
Step 3 – (c) energy at 7.5 V
E = ½CV² = ½ × 4.0 × 10−4 × 7.5² = 0.01125 J
A: 0.011 J (2 s.f.), a quarter of the energy at 15 V, because E ∝ V²
Watch out: halving the pd does not halve the energy. Both Q and V halve, so E falls to a quarter.
Q6 (Exam) The flash unit of a camera uses a 150 µF capacitor charged to 330 V. The capacitor discharges through the flash tube in 2.0 ms. Calculate (a) the energy stored and (b) the mean power delivered to the flash tube, each to 2 significant figures.
Show solution
Convert first: 150 µF = 150 × 10−6 F; 2.0 ms = 2.0 × 10−3 s
Step 1 – (a) energy stored
F: E = ½CV²
I: E = ½ × 150 × 10−6 × 330²
A: 8.1675 J = 8.2 J (2 s.f.)
Step 2 – (b) mean power
P = E ÷ t = 8.1675 ÷ 2.0 × 10−3 = 4083.75 W
A: 4.1 kW (2 s.f.)
a battery could not supply this power directly, but it can charge the capacitor slowly, and the capacitor then releases the energy very quickly
Q7 (Exam) A defibrillator uses a 150 µF capacitor to deliver 300 J of energy to a patient’s heart. (a) Calculate the pd the capacitor must be charged to. (b) Calculate the charge stored. (c) The capacitor discharges in 6.0 ms. Calculate the mean current through the patient. Assume all the stored energy is delivered.
Show solution
Convert first: 150 µF = 1.5 × 10−4 F; 6.0 ms = 6.0 × 10−3 s
Step 1 – (a) pd
F: E = ½CV²
I: 300 = ½ × 1.5 × 10−4 × V²
F: V = √(2 × 300 ÷ 1.5 × 10−4) = √(4.0 × 106)
A: 2000 V (2.0 kV)
Step 2 – (b) charge
Q = CV = 1.5 × 10−4 × 2000
A: 0.30 C (check: ½QV = ½ × 0.30 × 2000 = 300 J)
Step 3 – (c) mean current
I = ΔQ ÷ Δt = 0.30 ÷ 6.0 × 10−3
A: 50 A
Q8 (Exam) The pd across a 1000 µF capacitor is increased from 6.0 V to 12 V. Calculate the extra energy stored.
Show solution
Convert first: 1000 µF = 1.0 × 10−3 F
Step 1 – energy at each pd
F: E = ½CV²
I: at 6.0 V: E = ½ × 1.0 × 10−3 × 6.0² = 0.018 J; at 12 V: E = ½ × 1.0 × 10−3 × 12² = 0.072 J
Step 2 – difference
ΔE = 0.072 − 0.018
A: 0.054 J (54 mJ)
Watch out: ½C(ΔV)² = ½ × 1.0 × 10−3 × 6.0² = 0.018 J is wrong. Find the energy at each pd, then subtract.
Q9 (Stretch) An uncharged 470 µF capacitor is charged through a 2.2 kΩ resistor by a 12 V battery until it is fully charged. Calculate, to 2 significant figures, (a) the charge that flows, (b) the energy transferred by the battery, (c) the energy stored by the capacitor and (d) the energy dissipated in the resistor. (e) State how your answer to (d) would change if a 4.7 kΩ resistor were used instead.
Show solution
Convert first: 470 µF = 4.7 × 10−4 F
Step 1 – (a) charge
Q = CV = 4.7 × 10−4 × 12 = 5.64 × 10−3 C
A: 5.6 × 10−3 C (2 s.f.)
Step 2 – (b) energy from the battery
the whole charge Q passes through the battery’s constant emf of 12 V, so W = QV = 5.64 × 10−3 × 12 = 0.06768 J
A: 0.068 J (2 s.f.)
Step 3 – (c) energy stored
E = ½QV = ½ × 5.64 × 10−3 × 12 = 0.03384 J
A: 0.034 J (2 s.f.)
Step 4 – (d) energy dissipated
0.06768 − 0.03384 = 0.03384 J
A: 0.034 J (2 s.f.), exactly half the energy from the battery
Step 5 – (e) larger resistor
no change: the charging takes longer (larger time constant), but the battery still supplies QV and the capacitor still stores ½QV, so half is still dissipated
Watch out: the battery’s pd is constant at 12 V, so it supplies QV. The capacitor’s pd rises from 0 to 12 V, so it stores only ½QV.
Q10 (Stretch) An air-filled parallel-plate capacitor of capacitance 50 pF is charged to 200 V and then disconnected from the supply. The plates are pulled apart until their separation is three times its original value. Calculate (a) the charge stored, (b) the energy stored before and after the plates are moved, (c) the new pd and (d) the work done in pulling the plates apart.
Show solution
Convert first: 50 pF = 5.0 × 10−11 F
Step 1 – (a) charge
Q = CV = 5.0 × 10−11 × 200
A: 1.0 × 10−8 C, which stays constant because the capacitor is isolated
Step 2 – (b) energy before and after
Q is fixed, so use F: E = ½Q² ÷ C
before: E = ½ × (1.0 × 10−8)² ÷ 5.0 × 10−11 = 1.0 × 10−6 J
C ∝ 1 ÷ d, so the new C = 50 ÷ 3 = 16.66… pF = 1.666… × 10−11 F
after: E = ½ × (1.0 × 10−8)² ÷ 1.666… × 10−11 = 3.0 × 10−6 J
Step 3 – (c) new pd
V = Q ÷ C = 1.0 × 10−8 ÷ 1.666… × 10−11
A: 600 V (check: ½QV = ½ × 1.0 × 10−8 × 600 = 3.0 × 10−6 J)
Step 4 – (d) work done
the plates attract each other, so work must be done to separate them; it is stored as extra energy
W = 3.0 × 10−6 − 1.0 × 10−6
A: 2.0 × 10−6 J
Watch out: ½CV² would suggest the energy falls when C falls, but V is not constant here. With Q fixed, use ½Q² ÷ C.
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