A 470 μF capacitor is charged to a p.d. of 9.0 V. How much energy does it store?
- Change the unit: C = 470 × 10−6 F.
- W = ½CV2 = ½ × 470 × 10−6 × 9.02.
- W = ½ × 470 × 10−6 × 81 = 0.019 J.
Answer: 0.019 J (19 mJ)
Energy stored in a capacitor
Physics A2 · Capacitance. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
Work must be done to charge a capacitor, because each new bit of charge is pushed onto a plate that already repels it. This work is stored as electric potential energy.
The p.d. is proportional to the charge (V = Q/C), so a graph of V against Q is a straight line through the origin. The energy stored is the area under this graph, which is a triangle: half base times height.
The factor of a half is there because the p.d. rises from zero to V during charging. The average p.d. is ½V, so the work done is Q × ½V, not QV.
Rules to remember
- energy stored = ½ × charge × p.d. (W = ½QV)
- W = ½CV2 (use when C and V are known)
- W = Q2 ÷ (2C) (use when Q and C are known)
- Energy = area under the graph of p.d. against charge
- For a fixed capacitor, energy is proportional to V2: doubling V gives 4 times the energy
- Energy released when the p.d. falls from V1 to V2 = ½C(V12 − V22)
Common mistake
When the p.d. falls from V1 to V2, students use ½C(V1 − V2)2. Square each p.d. first and then subtract. Also change μF to F before calculating.
Worked example
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