Energy stored in a capacitor
A charged capacitor stores energy in the field between its plates: U = Q²/(2C) = ½CV² = ½QV. The board asks for the formulae only, not the derivation.
Where is the energy stored?
Charging a capacitor means moving charge from one plate to the other against the field. The work done is stored as electrostatic potential energy. The energy sits in the electric field between the plates. Its energy density is u = ½ε₀E² in vacuum. The board asks for the formulae only, not the derivation. The factor ½ appears because the voltage rises from 0 to V during charging, so the average voltage is V/2.
Which formula to use
| You know | Use | Typical case |
|---|---|---|
| C and V | U = ½CV² | A capacitor joined to a battery |
| Q and C | U = Q²/(2C) | A charged capacitor cut off from the battery |
| Q and V | U = ½QV | Charge and voltage both given |
Putting in a dielectric of constant K
| Quantity | Battery stays connected (V fixed) | Battery removed first (Q fixed) |
|---|---|---|
| Capacitance | Becomes KC₀ | Becomes KC₀ |
| Charge | Becomes KQ₀ | Stays Q₀ |
| Potential difference | Stays V₀ | Becomes V₀/K |
| Field between the plates | Stays E₀ | Becomes E₀/K |
| Energy stored | Becomes KU₀ | Becomes U₀/K |
Worked example: energy in a capacitor
A 20 μF capacitor is charged to 100 V. Find (a) the charge and (b) the energy stored. (c) What is the energy if it is charged to 200 V instead?
- (a) Q = CV = 20 × 10⁻⁶ × 100 = 2 × 10⁻³ C.
- (b) U = ½CV² = ½ × 20 × 10⁻⁶ × (100)² = 10 × 10⁻⁶ × 10⁴ = 0.1 J.
- (c) U ∝ V². Doubling V makes U four times: 4 × 0.1 = 0.4 J.
(a) 2 × 10⁻³ C; (b) 0.1 J; (c) 0.4 J
Worked example: sharing charge
A 4 μF capacitor charged to 150 V is joined to an uncharged 2 μF capacitor. Find (a) the common potential and (b) the energy lost.
- (a) Charge is conserved: V = (C₁V₁ + C₂V₂)/(C₁ + C₂) = (4 × 150 + 2 × 0)/(4 + 2) = 600/6 = 100 V.
- (b) Energy before: ½ × 4 × 10⁻⁶ × (150)² = 2 × 10⁻⁶ × 22500 = 0.045 J.
- Energy after: ½ × (4 + 2) × 10⁻⁶ × (100)² = 3 × 10⁻⁶ × 10⁴ = 0.030 J.
- Energy lost = 0.045 − 0.030 = 0.015 J, given out as heat in the wires (and as radiation).
(a) 100 V; (b) 0.015 J
Uses of the stored energy
- Camera flash: a capacitor stores energy and dumps it into the flash tube in a split second.
- Defibrillator: a large capacitor gives a controlled shock to restart a heart.
- Ceiling fans and motors: a capacitor helps them start and run.
- Chargers and computers: capacitors smooth out changes in voltage.
A hospital defibrillator stores energy in a large capacitor and releases it through the chest in a fraction of a second to restart a heart. A camera flash works the same way on a much smaller scale.
2 marks: “A capacitor is charged by a battery. The battery is removed and a dielectric slab of constant K is slid in. How do the charge, potential difference, field and energy change?” Q stays the same; V becomes V/K; E becomes E/K; U = Q²/(2C) becomes U/K, since C becomes KC.
3 marks: “A capacitor C charged to V is joined to an identical uncharged capacitor. Find the common potential and the energy lost.” Common potential V′ = CV/(2C) = V/2. Before: ½CV². After: ½(2C)(V/2)² = ¼CV². Loss = ¼CV², half the original, lost as heat.
1 marks: “Write the expression for the energy density of an electric field in vacuum.” u = ½ε₀E².
Half C V squared. Battery on: V fixed, so C, Q and U grow K times. Battery off: Q fixed, so V, E and U shrink K times.