Energy Stored in Capacitors

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. In this lesson we derive the capacitor energy formulae and show how to use them in exam-style problems.

This lesson continues OCR H556 Module 6.1.

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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.

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Deriving the Energy Formula

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

dW = v dq = (q/C) dq

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

W = ∫₀^Q (q/C) dq
  = (1/C) × [q²/2]₀^Q
  = Q² / (2C)

Using Q = CV we get three equivalent forms:

E = ½ QV = ½ CV² = Q² / (2C)

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…
E = ½ QV	charge and voltage
E = ½ CV²	capacitance and voltage
E = Q²/(2C)	charge and capacitance

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The Area Under a Q–V Graph

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

graph LR
    O((0,0)) -. "Q = CV is a straight line" .- P((V, Q))

Because Q = CV, the graph is a straight line through the origin with gradient C. The work done is the total of v dq, which is the area under the graph of V against Q (or equivalently above the graph of Q against V — the two are the same area).

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

Area = ½ × base × height = ½ QV

This is the same E = ½ QV we obtained from calculus. The factor of ½ 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-against-Q plot, and the area underneath is the energy. Forgetting the factor of ½ is the single most common error in this topic.

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Where Does the Other Half Go?

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

W_battery = QV = CV²

The energy stored in the capacitor at the end is

E_capacitor = ½ CV²