Energy Stored in Capacitors

Spec mapping: OCR H556 Module 6.1 — Capacitors (energy stored in a charged capacitor; the three equivalent forms $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = Q^2/(2C)$W=21​QV=21​CV2=Q2/(2C); the area under a $V$V-against-$Q$Q graph as the work done in charging; the famous half-of-the-battery-work-is-dissipated result). Refer to the official OCR H556 specification document for exact wording.

So far we have seen how much charge a capacitor holds at a given voltage. But the real reason capacitors are so useful is that they store energy. A single camera flash capacitor can hold enough energy to deliver a pulse of many thousands of watts to the flash tube. A bank of capacitors on a fusion research facility can release gigawatts in microseconds. A defibrillator capacitor delivers a couple of hundred joules in five milliseconds to restart a fibrillating heart. In this lesson we derive the capacitor energy formulae and show how to use them in exam-style problems.

This lesson continues Module 6.1, building on the definition $C = Q/V$C=Q/V from lesson 1 and the combination rules of lesson 2. The energy formulae we derive here will be re-used in every subsequent capacitor question, and the factor of one-half is the single most-tested point in the topic — drop it and you fail the question.

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Where Does the Energy Come From?

Charging a capacitor means transferring electrons from one plate to the other against the electric field that develops as the charge builds up. The battery does work on each electron; that work is stored in the electric field between the plates as electric potential energy.

Key idea: The energy stored in a capacitor is the work done against the field to charge it up.

Crucially, it is not the same as the energy delivered by the battery. Half of the work done by the battery during charging is dissipated as heat in the resistance of the connecting wires, no matter how small that resistance is. This is a famous result which we will demonstrate below, and which is one of the cleanest demonstrations of energy conservation applied to a non-conservative process anywhere in the A-Level course.

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Deriving the Energy Formula

Suppose at some instant during charging the charge on the capacitor is $q$q and the voltage is $v = q/C$v=q/C. To move a further small charge $\mathrm{d}q$dq from one plate to the other, the battery must do work

$\mathrm{d}W = v\,\mathrm{d}q = \frac{q}{C}\,\mathrm{d}q.$dW=vdq=Cq​dq.

The total work to charge the capacitor from 0 to $Q$Q is therefore

$$\begin{aligned} W &= \int_0^Q \frac{q}{C}\,\mathrm{d}q \\ &= \frac{1}{C}\left[\frac{q^2}{2}\right]_0^Q \\ &= \frac{Q^2}{2C}. \end{aligned}$$W​=∫0Q​Cq​dq=C1​[2q2​]0Q​=2CQ2​.​

Using $Q = CV$Q=CV we get three equivalent forms:

$\boxed{\,W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}\,}$W=21​QV=21​CV2=2CQ2​​

All three are worth memorising. In an exam you will use whichever pair of variables is most convenient for the numbers given.

Form	Best used when you know…
$W = \tfrac{1}{2}QV$W=21​QV	charge and voltage
$W = \tfrac{1}{2}CV^2$W=21​CV2	capacitance and voltage
$W = Q^2/(2C)$W=Q2/(2C)	charge and capacitance

The factor of one-half is the signature of this topic. Physically, it arises because the voltage across the capacitor rises linearly from 0 to its final value $V$V during charging, so the average voltage that each unit of charge "feels" as it is being deposited is $V/2$V/2, not $V$V. Total energy $=$= total charge $\times$× average voltage $= Q \times V/2 = \tfrac{1}{2}QV$=Q×V/2=21​QV.

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The Area Under a Q-V Graph

A non-calculus derivation comes from plotting charge $Q$Q against voltage $V$V:

Because $Q = CV$Q=CV, the graph is a straight line through the origin with gradient $C$C. The work done is the total of $v\,\mathrm{d}q$vdq, which is the area under the graph of $V$V against $Q$Q (or equivalently above the graph of $Q$Q against $V$V — the two are the same shaded area).

For a straight-line graph from $(0, 0)$(0,0) to $(V, Q)$(V,Q), the area is a triangle:

$\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2}QV.$Area=21​×base×height=21​QV.

This is the same $W = \tfrac{1}{2}QV$W=21​QV we obtained from calculus. The factor of $\tfrac{1}{2}$21​ is sometimes called the "triangle factor" and it is the reason a capacitor only ever stores half the energy supplied by the battery that charged it.

Exam tip. If OCR show you a graph, it is almost certainly a $V$V-against-$Q$Q plot, and the area underneath is the energy. Forgetting the factor of $\tfrac{1}{2}$21​ is the single most common error in this topic.

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Where Does the Other Half Go?

Consider charging a capacitor $C$C to voltage $V$V through a resistor $R$R. The battery e.m.f. is $V$V and the total charge delivered is $Q = CV$Q=CV. The work done by the battery is

$W_\text{battery} = QV = CV^2.$Wbattery​=QV=CV2.

The energy stored in the capacitor at the end is

$W_\text{capacitor} = \tfrac{1}{2}CV^2.$Wcapacitor​=21​CV2.

So exactly half the battery's work — $\tfrac{1}{2}CV^2$21​CV2 — has been dissipated as heat in the resistor. This result is independent of $R$R: even if the resistance is tiny, so that the capacitor charges almost instantaneously, exactly half the energy is still lost. The only way to avoid the 50% loss is to charge through an inductor, or to ramp the voltage smoothly from 0 to $V$V, neither of which are on the OCR spec.

We can verify this by direct integration. During charging the current is $i(t) = (V/R)e^{-t/RC}$i(t)=(V/R)e−t/RC (this is derived in lesson 5), so the instantaneous power dissipated in the resistor is

$P_R = i^2 R = \frac{V^2}{R}e^{-2t/RC},$PR​=i2R=RV2​e−2t/RC,

and the total dissipated energy is

$W_R = \int_0^\infty \frac{V^2}{R} e^{-2t/RC}\,\mathrm{d}t = \frac{V^2}{R} \cdot \frac{RC}{2} = \tfrac{1}{2}CV^2.$WR​=∫0∞​RV2​e−2t/RCdt=RV2​⋅2RC​=21​CV2.

The dissipated energy is exactly $\tfrac{1}{2}CV^2$21​CV2 — independent of $R$R, as claimed.

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Worked Example 1 — Camera Flash

A camera flash capacitor has capacitance $220\ \mu$220 μF and is charged to $330$330 V. Calculate (a) the charge stored, (b) the energy stored, (c) the mean power delivered if the flash lasts $2.0$2.0 ms.

(a) Charge.

$Q = CV = (220 \times 10^{-6})(330) = 7.26 \times 10^{-2}\ \text{C}.$Q=CV=(220×10−6)(330)=7.26×10−2 C.

(b) Energy.

$$\begin{aligned} W &= \tfrac{1}{2}CV^2 \\ &= 0.5 \times 220 \times 10^{-6} \times 330^2 \\ &= 0.5 \times 220 \times 10^{-6} \times 108\,900 \\ &= 11.98\ \text{J} \\ &\approx 12\ \text{J}. \end{aligned}$$W​=21​CV2=0.5×220×10−6×3302=0.5×220×10−6×108900=11.98 J≈12 J.​

(c) Mean power.

$P = \frac{W}{t} = \frac{11.98}{2.0 \times 10^{-3}} \approx 5990\ \text{W} \approx 6.0\ \text{kW}.$P=tW​=2.0×10−311.98​≈5990 W≈6.0 kW.

A small photographic flash delivers kilowatts of power for a few milliseconds. The total energy is modest ($12$12 J, about the same as dropping a 1 kg book by $1.2$1.2 m), but the power is enormous. The trick is that the capacitor concentrates a slow trickle of energy from a small battery into a very short, very intense pulse.

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Worked Example 2 — Charge Redistribution

A $10\ \mu$10 μF capacitor is charged to $100$100 V. It is then disconnected from the battery and connected in parallel with an uncharged $40\ \mu$40 μF capacitor. Find (a) the final voltage across each, (b) the energy stored before and after, and (c) explain any discrepancy.

(a) Final voltage. Total charge is conserved on the isolated combination:

$Q_\text{before} = (10 \times 10^{-6})(100) = 1.0 \times 10^{-3}\ \text{C}.$Qbefore​=(10×10−6)(100)=1.0×10−3 C.

After connection the combined capacitance is $10 + 40 = 50\ \mu$10+40=50 μF and the charge is unchanged:

$V_\text{final} = \frac{Q}{C_\text{total}} = \frac{1.0 \times 10^{-3}}{50 \times 10^{-6}} = 20\ \text{V}.$Vfinal​=Ctotal​Q​=50×10−61.0×10−3​=20 V.

(b) Energies.

$$\begin{aligned} W_\text{before} &= \tfrac{1}{2}\,(10 \times 10^{-6})(100)^2 = 0.050\ \text{J} = 50\ \text{mJ}, \\ W_\text{after} &= \tfrac{1}{2}\,(50 \times 10^{-6})(20)^2 = 0.010\ \text{J} = 10\ \text{mJ}. \end{aligned}$$Wbefore​Wafter​​=21​(10×10−6)(100)2=0.050 J=50 mJ,=21​(50×10−6)(20)2=0.010 J=10 mJ.​

(c) Discrepancy. Energy has dropped from $50$50 mJ to $10$10 mJ — 80% has "disappeared". Where has it gone? The answer is that during the redistribution, a brief current flowed through the small but non-zero resistance of the connecting wires, and $40$40 mJ was dissipated as heat (and a little as electromagnetic radiation). This is another example of the "50% loss" result, just in a more extreme form: the larger the mismatch in $V$V before connection, the larger the fraction dissipated.

This is a beautiful OCR-style synoptic question because it combines charge conservation, energy storage, and the distinction between energy stored in a field and energy dissipated in resistance. It is also a clean example of a non-reversible process — the lost energy cannot be recovered by simply disconnecting the second capacitor again.

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Worked Example 3 — Defibrillator

A defibrillator stores energy in a capacitor and discharges it through the patient's chest in a few milliseconds. A typical clinical setting is $200$200 J delivered at $2500$2500 V. Calculate (a) the capacitance needed, (b) the charge stored, (c) the peak current if the pulse discharges through an effective chest resistance of $50\ \Omega$50 Ω.

(a) Capacitance. From $W = \tfrac{1}{2}CV^2$W=21​CV2:

$C = \frac{2W}{V^2} = \frac{2 \times 200}{(2500)^2} = \frac{400}{6.25 \times 10^{6}} = 6.4 \times 10^{-5}\ \text{F} = 64\ \mu\text{F}.$C=V22W​=(2500)22×200​=6.25×106400​=6.4×10−5 F=64 μF.

(b) Charge. $Q = CV = (6.4 \times 10^{-5})(2500) = 0.16\ \text{C}.$Q=CV=(6.4×10−5)(2500)=0.16 C.

(c) Peak current. At the start of the discharge ($t = 0$t=0),

$I_0 = \frac{V}{R} = \frac{2500}{50} = 50\ \text{A}.$I0​=RV​=502500​=50 A.

Fifty amps is a colossal current — about the same as a small arc-welder. The reason it does not kill the patient is that it lasts only a few milliseconds and is delivered into a single, controlled pathway through the chest. A continuous $50$50 A would, of course, be lethal.

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Energy Density — Beyond the Spec but Worth Knowing

The energy is really stored in the electric field between the plates, not on the plates themselves. For a parallel-plate capacitor, you can show that the energy per unit volume (the energy density) is

$u = \tfrac{1}{2}\varepsilon_0 E^2.$u=21​ε0​E2.

OCR does not examine this directly, but it is worth knowing because it explains why the energy stored grows with the square of the electric field — and why high-field dielectrics are prized by capacitor designers. It also previews the undergraduate idea that fields, not charges, are the "stuff" in which electromagnetic energy lives.

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Synoptic Links

• Energy conservation (Module 3.3): The 50%-of-battery-work-is-stored result is a direct application of conservation of energy applied to a charging RC circuit. The other 50% must show up somewhere; in a real wire it appears as heat, and in the idealised zero-resistance limit it appears as electromagnetic radiation from the transient current. Energy is conserved either way.
• Internal resistance and e.m.f. (Module 4.2): Just as a battery's e.m.f. and internal resistance govern how much of $\varepsilon$ε ends up in the load and how much in $r$r, the capacitor-and-resistor split governs how much of the battery's energy ends up in the capacitor and how much in the resistor. The 50/50 split here is the idealised limit of the more general internal-resistance result.
• Power and pulses (Module 4.1, Module 6.1, lesson 4): A defibrillator delivers $200$200 J in $5$5 ms — a peak power of $40$40 kW from a battery that supplies milliwatts. The capacitor is the energy-storage element that bridges the gap between a slow charging source and a fast discharging load. This is the conceptual key to every flash, pulse and timing application.

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Specimen question modelled on the OCR H556 paper format

Question (9 marks): A defibrillator uses a $32\ \mu$32 μF capacitor charged to $2500$2500 V.

(a) Calculate (i) the charge stored, (ii) the energy stored. [3]