Energy stored in a capacitor
| English | Chinese | Pinyin |
|---|---|---|
| energy stored | 储存的能量 | chǔ cún de néng liàng |
| defibrillator | 除颤器 | chú chàn qì |
Half the energy never reaches the capacitor
- Connect an ideal battery of e.m.f. $V$ to a capacitor through a wire. The battery pushes charge $Q = CV$ through a p.d. of $V$, so it gives out $QV$ of energy.
- The capacitor ends up holding $\tfrac12 QV$. Exactly half the energy has gone, as heat in the wire, and it makes no difference how good the wire is.
- The missing half is not an inefficiency to be engineered away. It is a consequence of the fact that the p.d. across the capacitor starts at zero and only reaches $V$ at the end.
- This lesson is the energy stored 储存的能量, where the factor of a half comes from, and what happens when charge is shared.
The three forms
- Charging a capacitor takes work, because each extra bit of charge must be pushed against the p.d. that is already there. Adding up all those bits gives:
- All three are the same expression with $Q = CV$ substituted differently. Choose the one that uses the two quantities the question gives you, and no rearranging is needed.
The energy stored in a capacitor is:
Equivalent forms: $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \dfrac{Q^{2}}{2C}$.
A $2.0\ \mu\text{F}$ capacitor is charged to $1000\ \text{V}$. How much energy does it store?
$W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 2.0 \times 10^{-6} \times (1000)^{2} = 1.0\ \text{J}$.
The area under the graph
- A graph of $V$ against $Q$ for a capacitor is a straight line through the origin, of gradient $1/C$.
- The energy stored is the area under that line up to the charge $Q$, which is a triangle of area $\tfrac12 QV$.
- That is where the half comes from: the p.d. grows from zero to $V$ as the capacitor charges, so the average p.d. during charging is $V/2$, not $V$.
A triangle, not a rectangle, and that is the whole reason for the half
Energy in a capacitor
E = ½C·V²
Stored energy grows with the square of the voltage.
Why is there a factor of ½ in the stored-energy formula?
The p.d. rises from 0 to V as it charges, so the average is V/2 — hence the triangle area $\tfrac{1}{2}QV$.
The energy stored equals the ____ under the V–Q graph.
The triangle under the line has area $\tfrac{1}{2}QV$ — the stored energy.
Worked example: energy from a graph
- A graph of $Q$ against $V$ is a straight line through the origin passing through $(10\ \text{V},\ 1.2 \times 10^{-3}\ \text{C})$. Find the capacitance, the energy at $10\ \text{V}$, and the extra energy needed to reach $12\ \text{V}$.
- The gradient of $Q$ against $V$ is the capacitance: $C = 1.2 \times 10^{-3}/10 = 1.2 \times 10^{-4}\ \text{F} = 120\ \mu\text{F}$.
- Energy at $10\ \text{V}$: the area, $\tfrac12 QV = \tfrac12 (1.2 \times 10^{-3})(10) = 6.0 \times 10^{-3}\ \text{J}$.
- At $12\ \text{V}$: $\tfrac12 CV^2 = \tfrac12 (1.2 \times 10^{-4})(144) = 8.6 \times 10^{-3}\ \text{J}$, so the extra energy is $2.6 \times 10^{-3}\ \text{J}$.
- Note that a 20% rise in voltage costs a 44% rise in energy, because $W \propto V^2$.
A graph of charge against p.d. for a capacitor is a straight line through the origin. What does its gradient give?
Q against V has gradient C, and the energy is the area under the V against Q line. Do not confuse the gradient with the area.
Energy goes as the square
- Because $W = \tfrac12 CV^2$, doubling the p.d. quadruples the energy for the same capacitor.
- That is why a camera flash or a defibrillator 除颤器 charges its capacitor to hundreds or thousands of volts rather than using a larger capacitor at a low voltage: voltage is the cheaper way to buy energy.
- It is also why the same capacitor at half the voltage stores only a quarter of the energy, which is worth checking your answers against.
Charging a capacitor through a wire wastes about half the supplied energy as heat.
The battery gives out $QV$ but the capacitor stores only $\tfrac{1}{2}QV$; the rest heats the wire, whatever its resistance.
A 120 uF capacitor is charged to 12 V. What is the energy stored, in mJ?
W = C V^2 / 2 = 1.2e-4 x 144 / 2 = 8.6e-3 J. At 10 V it would store only 6.0 mJ, because energy goes as the square of the voltage.
Doubling the p.d. across a capacitor doubles the energy it stores.
W is proportional to V squared, so doubling the p.d. gives four times the energy. That is why flash and defibrillator circuits use high voltages.
Worked example: sharing charge
- A charged capacitor $C$ at p.d. $V$ is disconnected and then connected across an uncharged capacitor of capacitance $3C$. Find the final p.d. and the energy lost.
- Charge is conserved. The total charge is still $Q = CV$, now on a parallel pair of total capacitance $4C$, so the common p.d. is $V' = Q/4C = V/4$.
- Energy is not conserved. Before: $\tfrac12 CV^2$. After: $\tfrac12 (4C)(V/4)^2 = \tfrac18 CV^2$. Three-quarters of the energy has gone.
- It is dissipated as heat in the connecting wires, and as a spark or radiation, while the charge moves, however small the resistance. Saying where it went is the mark.
A charged capacitor is connected across an uncharged one. Put the reasoning in order.
Charge is conserved, energy is not. Naming where the lost energy goes is what the final mark is for.
Energy during a discharge falls twice as fast
- During a discharge the charge and the p.d. both decay as $e^{-t/RC}$, but the energy goes as $V^2$, so it decays as $e^{-2t/RC}$.
- After one time constant the p.d. has fallen to $37\%$ of its start, but the energy has fallen to $e^{-2} = 14\%$.
- After two time constants the p.d. is $14\%$ and the energy only $1.8\%$.
- So the energy has an effective half-life of half that of the charge. A question that gives you a time constant and asks for the remaining energy is testing exactly this factor of two.
- The same square is why a capacitor charged to twice the p.d. stores four times the energy.
A capacitor discharges for exactly one time constant. What percentage of its stored energy remains?
The p.d. falls to 37%, but energy goes as V squared, so e^-2 = 13.5% remains. Quoting 37% for the energy is the trap this question is built on.
Marks that slip away
- The factor of $\tfrac12$ comes from the average p.d. being $V/2$ during charging, or equivalently from the area of a triangle. Say one of those, not "it is in the formula".
- $W \propto V^2$, so doubling the voltage gives four times the energy, not twice.
- In a charge-sharing question, charge is conserved but energy is not. Name where the lost energy goes.
- Half the battery's energy is lost as heat during charging whatever the wire's resistance. It is not a fault in the circuit.
You've got it
- energy stored $W = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}$; pick the form matching the quantities given
- it is the area under the $V$ against $Q$ line, a triangle, and the $\tfrac12$ is because the average p.d. while charging is $V/2$
- $W \propto V^2$: double the p.d., four times the energy
- when charge is shared, charge is conserved and energy is not: the difference is dissipated as heat in the wires, whatever their resistance