3.7.3.1FieldsCore

Capacitors and Energy Storage

A capacitor is one of the simplest components in electronics — just two conducting plates separated by an insulator — yet it can store and release a burst of electrical energy far faster than a chemical battery ever could. From camera flashes to smoothing out a DC power supply, understanding capacitance means understanding how much charge (and energy) a capacitor can hold for a given voltage.

40 min Video by Science Shorts 3.7.3 Capacitance

All of CAPACITORS in 5 mins — A-level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Define capacitance and use $C = \frac{Q}{V}$
Explain qualitatively how plate area, separation and the dielectric affect capacitance
Explain why energy stored is $\frac{1}{2} Q V$ , not $Q V$
Use the equivalent energy formulae $W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{1}{2} \frac{Q ^{2}}{C}$
Interpret a graph of charge against potential difference for a capacitor

Capacitance

A capacitor stores charge on two conducting plates, separated by an insulating material (a $dielectric$ ) that prevents charge from flowing directly between them. $Capacitance$ , $C$ , is defined as the charge stored per unit potential difference across the capacitor, measured in farads (F) — a genuinely huge unit in practice, so most real capacitors are rated in microfarads (µF), nanofarads (nF) or picofarads (pF).

C = \frac{Q}{V}

Capacitance: charge stored per unit p.d.

1Use

C = \frac{Q}{V}

C = \frac{6.0 \times 1 0 ^{- 4}}{12}

Answer

C = 5.0 \times 1 0^{- 5}

F (50 µF)

Tip — C = Q/V is a genuine constant of the capacitor (set by its physical construction), not a proportionality you can rearrange freely to imply capacitance "changes" if you charge it to a different voltage — Q simply scales with V, keeping C fixed.

What determines capacitance

For a simple parallel-plate capacitor, capacitance increases with a larger plate $area$ (more surface to store charge on) and decreases with a larger $separation$ between the plates (the electric field, and hence the ability to hold charge at a given voltage, weakens as plates move apart). Filling the gap with a dielectric material of higher permittivity than air also increases capacitance, since the dielectric reduces the effective field for a given charge, allowing more charge to be stored for the same p.d.

Tip — A quick way to remember the trend: bigger plates, closer together, better insulator between them — all three increase capacitance.

Energy stored in a capacitor

Unlike a battery (which does a constant amount of work per unit charge, since its e.m.f. doesn’t change as it discharges), a capacitor’s p.d. rises from zero as it charges — so the very first bit of charge is moved at almost no p.d., while the last bit is moved at the full final p.d. Because charge and p.d. rise together (in direct proportion, since $Q = C V$ ), a graph of $Q$ against $V$ is a straight line through the origin, and the energy stored equals the area under that line — a triangle, giving exactly half of $Q V$ , not the full $Q V$ you might first expect.

W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{1}{2} \frac{Q ^{2}}{C}

Three equivalent forms for the energy stored in a charged capacitor.

1Use

W = \frac{1}{2} C V^{2}

W = 0.5 \times 200 \times 1 0^{- 6} \times 9. 0^{2}

W = 0.5 \times 200 \times 1 0^{- 6} \times 81

Answer

W \approx 8.1 \times 1 0^{- 3}

J (8.1 mJ)

Tip — The most common exam trap: writing W = QV instead of ½QV. Only use QV for the work a BATTERY does moving charge (constant e.m.f.) — a capacitor’s own stored energy is always half that.

Equation recap

C = \frac{Q}{V}

Capacitance.

W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{1}{2} \frac{Q ^{2}}{C}

Energy stored in a charged capacitor.

Common mistakes to avoid

Using W = QV for the energy stored in a capacitor.

A capacitor’s p.d. rises as it charges, so the correct energy is W = ½QV — the factor of ½ comes from the linear Q–V relationship, exactly like the area of a triangle rather than a rectangle.

Thinking capacitance depends on how much charge or voltage is currently stored.

Capacitance is a fixed property of the capacitor’s construction (area, separation, dielectric) — Q and V both change together, but their ratio C stays constant.

Assuming a larger plate separation increases capacitance.

A LARGER separation DECREASES capacitance — closer plates give a stronger field and store more charge for the same p.d.

Key takeaways

Capacitance C = Q/V, measured in farads.
Capacitance increases with larger plate area, smaller separation, and a higher-permittivity dielectric.
Energy stored W = ½QV = ½CV² = ½Q²/C, because p.d. rises linearly with charge during charging.
A Q–V graph for a capacitor is a straight line through the origin; energy stored is the area under it.

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Capacitors and Energy Storage

40 min Video by Science Shorts 3.7.3 Capacitance