Capacitance

Press the shutter on a camera in a dark room and, a heartbeat later, the flash fires — a dazzling burst far brighter than the little battery inside could ever manage on its own. Listen closely beforehand and you may even hear a rising whine as the camera gets ready. What is happening is that a single component spends a second or two quietly hoarding electric charge, then dumps the whole lot in one blinding instant. That component is a capacitor.

A capacitor is beautifully simple: two conducting plates held close together but separated by an insulator (called the dielectric) so that no charge can cross the gap. Connect it to a battery and electrons are pulled off one plate and piled onto the other: one plate ends up positive, the other equally negative. Charge Q builds on the plates, and a potential difference V grows across them, until that voltage matches the supply and the flow stops. The capacitor is now charged — it holds a store of charge, and with it a store of energy, ready to be released the moment you give it a path.

This page asks the two questions that make capacitors useful: how much charge does a given capacitor hold at a given voltage, and how much energy is packed into it — and then, how that store leaks away when you finally let it flow.

Capacitance: charge stored per volt

Push a capacitor to a higher voltage and it grabs proportionally more charge — double the voltage, double the charge on the plates. The constant of proportionality, the charge held for each volt across the plates, is the capacitor's capacitance, written C:

C = \dfrac{Q}{V}.

Here Q is the charge in coulombs (\text{C}), V the potential difference in volts (\text{V}), and C the capacitance in farads (\text{F}). Read the meaning straight off the units:

1\ \text{F} = 1\ \dfrac{\text{C}}{\text{V}} = \text{one coulomb of charge stored for every volt across the plates.}

Capacitance is a property of the capacitor itself — its plate area, their spacing, and the dielectric between them — not of how much you happen to have charged it. A bigger C stores more charge at the same voltage: a roomier bucket for charge.

Worked example 1 — finding the capacitance. A capacitor holds 60\ \mu\text{C} of charge when there is 12\ \text{V} across it. What is its capacitance?

C = \dfrac{Q}{V} = \dfrac{60\times10^{-6}\ \text{C}}{12\ \text{V}} = 5\times10^{-6}\ \text{F} = 5\ \mu\text{F}.

Worked example 2 — finding the charge. That same 5\ \mu\text{F} capacitor is now charged to 20\ \text{V}. Rearranging C = Q/V gives Q = CV:

Q = CV = 5\times10^{-6}\ \text{F} \times 20\ \text{V} = 100\times10^{-6}\ \text{C} = 100\ \mu\text{C}.

Look at a real capacitor and it is stamped with something like 100\ \mu\text{F}, 10\ \text{nF} or 22\ \text{pF} — never a plain "2 F". That is because the farad is a colossal unit. A one-farad capacitor would hold a whole coulomb of charge at just one volt, and a coulomb is an enormous pile of electrons; to build one out of flat plates you would need a sheet the size of a sports field. So everyday capacitors are measured in microfarads (1\ \mu\text{F} = 10^{-6}\ \text{F}), nanofarads (10^{-9}) and picofarads (10^{-12}). The chunky "supercapacitors" now used to back up memory chips and smooth electric-car power really do reach several farads — and they are marvels of packed-up surface area to manage it.

The energy stored: a triangle under the Q–V graph

Charging a capacitor takes work: you are forcing more and more charge onto a plate that is already charged up and pushing back. Crucially, the voltage does not stay put while you do it — it rises from zero as the charge builds, because V = Q/C. Plot charge Q against voltage V and you get a straight line through the origin (its gradient is the capacitance C). The energy stored is the area underneath that line — and the area of a triangle is \tfrac12 \times \text{base} \times \text{height}:

E = \tfrac12 QV.

Because Q = CV, the very same energy can be written three equivalent ways — reach for whichever pair of quantities you happen to know:

E = \tfrac12 QV = \tfrac12 CV^2 = \dfrac{Q^2}{2C}.

Worked example 3 — the energy stored. The 5\ \mu\text{F} capacitor from before is charged to 20\ \text{V}. How much energy does it hold?

E = \tfrac12 CV^2 = \tfrac12 \times 5\times10^{-6}\ \text{F} \times (20\ \text{V})^2 = 1\times10^{-3}\ \text{J} = 1\ \text{mJ}.

Check it against the other formula — with Q = 100\ \mu\text{C}, E = \tfrac12 QV = \tfrac12 \times 100\times10^{-6} \times 20 = 1\ \text{mJ}. The two forms agree, as they must.

Drag the sliders below to build a capacitor's charge up to a chosen voltage. The straight line is Q = CV; the shaded triangle under it is the energy \tfrac12 QV you are storing. Notice that pushing to a higher voltage grows the triangle in two directions at once — taller and wider — which is why the energy climbs with V^2, not just with V.

Discharging: an exponential fall, and the time constant

Now give the charged capacitor somewhere to go. Connect it across a resistor R and it drives a current that carries its charge away. But here is the twist: as charge leaks off, the voltage drops, so the current it can drive drops too — which means the charge leaks away more slowly. A quantity whose rate of change is proportional to how much is left always falls in the same shape: an exponential decay.

Q = Q_0\,e^{-t/RC}.

The voltage and current fall in exactly the same way — V = V_0\,e^{-t/RC} and I = I_0\,e^{-t/RC} — because each is just a fixed multiple of the charge. The rate of the fall is set entirely by the combination RC, and this combination is so important it gets its own name and symbol, the time constant \tau (Greek "tau"):

\tau = RC.

Put t = \tau into the formula and the exponent becomes -1, so Q = Q_0\,e^{-1} \approx 0.37\,Q_0. The time constant is the time for the charge (or voltage, or current) to fall to 1/e \approx 37\% of its starting value. A bigger R or C makes \tau bigger, so the capacitor charges and discharges more slowly. After about 5\tau the charge has dropped below 1\% — the capacitor is, for all practical purposes, fully discharged.

Worked example 4 — the time constant. A 470\ \mu\text{F} capacitor discharges through a 10\ \text{k}\Omega resistor. Its time constant is

\tau = RC = 10\,000\ \Omega \times 470\times10^{-6}\ \text{F} = 4.7\ \text{s}.

Worked example 5 — charge left after a time. A capacitor is charged to Q_0 = 100\ \mu\text{C} and discharges with time constant \tau. How much charge remains after two time constants, t = 2\tau?

Q = Q_0\,e^{-t/RC} = 100\ \mu\text{C} \times e^{-2} = 100 \times 0.135 = 13.5\ \mu\text{C}.

After one \tau it was 37\ \mu\text{C}; after two, 13.5\ \mu\text{C}; after five, well under one microcoulomb. Each time constant knocks the charge down by the same factor, never by the same amount — that is the signature of exponential decay.

The curve below is a live discharge, Q = Q_0\,e^{-t/RC}. Turn the resistance and capacitance sliders and watch the whole decay stretch or squeeze: the marked point is one time constant \tau = RC, where the charge has fallen to the 37\% line, and the far marker shows where — after about 5\tau — the capacitor is essentially empty.

What capacitors are for

This little charge-and-release trick turns out to be everywhere:

That rising whine before a camera flash is the sound of a small circuit frantically pumping the battery's 1.5\ \text{V} up to a few hundred volts to charge the flash capacitor; when it is full, the whine peaks and the flash is armed. Fire it, and the capacitor empties through the flash tube in a few thousandths of a second — a tiny stored energy delivered at huge power, which is what makes the light so bright.

The flip side is a genuine hazard. A capacitor does not forget its charge when you switch the power off — with nothing to discharge through, it can sit fully charged for a very long time. The large capacitors inside an old television or a microwave oven can hold a dangerous, even lethal, charge for minutes to hours after the plug is pulled. That is exactly why engineers deliberately short the plates out with a resistor before poking around inside — the capacitor is patiently waiting, and it does not care that the set is unplugged.