What a capacitor actually stores
Capacitance — The charge stored per unit potential difference: C = Q/V, measured in farads (1 F = 1 C V⁻¹).
A capacitor is two conducting plates separated by an insulator. Connect it to a supply and charge accumulates on the plates — equal and opposite, so the device as a whole stays neutral. It is the separation of charge that is stored, not a net charge.
Capacitance measures how much charge the arrangement holds per volt applied. A large capacitance means a lot of charge for a modest voltage. The unit, the farad, is enormous in practice: real components are usually measured in microfarads or picofarads.
- C
- capacitancein farads, though µF and pF are what you meet
- Q
- charge on one platethe other carries −Q
- V
- potential differenceacross the plates
The combination rules are inverted
Capacitors in parallel simply add, because the plate area effectively increases. Capacitors in series combine reciprocally, so the total is always smaller than the smallest one. This is exactly the reverse of resistors, and confusing the two is the most common error in the topic. A quick check: if your series answer is bigger than any individual capacitor, it is wrong.
The energy stored, and the factor of one half
Charging a capacitor takes work, because each additional charge must be pushed onto a plate that already repels it. The first charge arrives easily; the last arrives against the full voltage.
That is precisely why the energy is ½QV rather than QV. The voltage rises linearly from zero to V as the charge accumulates, so the average voltage during charging is V/2. On a graph of V against Q the energy is the area under a straight line through the origin — a triangle, hence the half.
- ½QV
- the area under the V–Q grapha triangle, not a rectangle
- ½CV²
- most useful formwhen C and V are the known quantities
- Q²/2C
- the third formwhen the charge is fixed and C changes
A 470 µF capacitor is charged to 12 V. Find the charge stored and the energy stored. It is then discharged through a resistor — state what happens to the energy.
- Q = CV = 470 × 10⁻⁶ × 12 = 5.64 × 10⁻³ C.Converting microfarads to farads before substituting.
- W = ½CV² = ½ × 470 × 10⁻⁶ × 144.Using the form containing the two known quantities avoids finding Q first.
- W = 0.0338 J = 33.8 mJ.A small amount, which is why capacitors store far less energy than batteries of the same size.
- On discharge the stored energy is dissipated as heat in the resistor.Energy is conserved: it leaves the electric field between the plates and appears as internal energy in the resistor.
Q = 5.64 mC, W = 33.8 mJ, dissipated as heat
Discharge: the same fraction in equal times
Connect a charged capacitor across a resistor and it discharges. The current is largest at the start, because the voltage driving it is largest, and it falls as the capacitor empties. Since the rate of loss is proportional to how much is left, the decay is exponential.
This is the same mathematical shape as radioactive decay, and it has the same defining property: equal intervals of time remove equal fractions, not equal amounts. After one time constant the voltage is always 37% of its starting value, whatever that starting value was.
Strictly, the capacitor never finishes discharging — the exponential approaches zero without reaching it. In practice it is treated as empty after about five time constants, by which point less than 1% remains.
- τ = RC
- the time constantthe time to fall to 1/e ≈ 37% of the start
- V₀
- the initial voltagethe value at t = 0
- e^(−t/RC)
- the decay factornever reaches zero, only approaches it
The marks at 1τ, 2τ and 3τ sit at 37%, 13.5% and 5%. Each step down multiplies by the same factor, which is what makes the decay exponential — and changing RC stretches the whole curve without changing those percentages.
Things that follow from the exponential shape
- A larger R or a larger C means a slower discharge — τ = RC.
- The half-life is
0.693RCand never changes as the capacitor empties. - To find a time, take logarithms:
t = −RC ln(V/V₀). - The same equation with the same time constant governs Q, V and I, since all three are proportional.
- Charging follows
V = V₀(1 − e^(−t/RC))— the mirror image, rising towards the supply voltage. - Ohms × farads really does give seconds, which is worth checking when a question looks dimensionally odd.
What capacitors are actually used for
Because a capacitor releases its charge far faster than a battery can supply it, the practical uses cluster around situations needing a large current for a very short time, or a store that must survive a brief interruption.
A camera flash is the clearest example: a battery charges the capacitor slowly over several seconds, and the capacitor then dumps that energy into the flash tube in about a thousandth of a second. The peak power is enormous even though the stored energy is modest, because power is energy divided by time.
The same behaviour underlies smoothing in a power supply. A rectifier produces a bumpy voltage, and a capacitor across the output charges at each peak and discharges gently between them, filling the gaps. A larger capacitance, or a larger load resistance, gives a longer time constant and therefore smoother output — which is the standard exam question on the topic.
- Camera flash — charged slowly, discharged in milliseconds for a huge peak power.
- Smoothing — filling the troughs in a rectified supply, with more smoothing for a larger RC.
- Backup power — holding memory contents for the seconds a supply is interrupted.
- Timing circuits — the predictable RC decay used to measure an interval.
- Filters — passing or blocking signals according to their frequency.
Why not just use a battery?
A battery stores far more energy for its size, but it can only release it slowly, because the chemical reaction limits the current. A capacitor stores much less energy but has no chemistry to wait for, so it can deliver it almost instantly. The two are complementary rather than competing: capacity of storage against speed of delivery.