Capacitance

A capacitor stores electrical charge (and energy) by building up equal and opposite charges, $+Q$ and $-Q$, on two conducting plates separated by an insulating dielectric. Connect it to a supply and charge flows onto the plates until the p.d. across the capacitor equals the supply p.d.

The capacitance $C$ measures how much charge a capacitor stores for each volt of p.d. across it: $C = \frac{Q}{V}$

• $Q$ is the charge on one plate (the plates carry $+Q$ and $-Q$; the "charge stored" is $Q$, not the net charge, which is zero).
• $V$ is the potential difference between the plates.
• Capacitance is measured in farads (F), where $1\ \text{F} = 1\ \text{C V}^{-1}$.

The farad is a very large unit — a 1 F capacitor would store one coulomb per volt — so real capacitors are usually quoted in microfarads ($\mu\text{F} = 10^{-6}\ \text{F}$), nanofarads ($\text{nF} = 10^{-9}\ \text{F}$) or picofarads ($\text{pF} = 10^{-12}\ \text{F}$). Because $Q = CV$, for a fixed capacitor the charge stored is directly proportional to the p.d.: double the voltage and you double the charge. A graph of $Q$ against $V$ is therefore a straight line through the origin whose gradient is the capacitance.

Combining capacitors. The combination rules are the opposite of the resistor rules:

• In parallel (same p.d. across each): capacitances add, $C_{\text{total}} = C_1 + C_2 + \dots$
• In series (same charge on each): reciprocals add, $\dfrac{1}{C_{\text{total}}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dots$, so the total is smaller than the smallest capacitor.

Charging a capacitor takes work: you have to push charge onto a plate against the repulsion of the charge already there. As charge builds up, the p.d. rises in proportion to the charge ($V = Q/C$), so the work needed to add each extra bit of charge increases.

The energy stored equals the area under the graph of p.d. $V$ against charge $Q$. Because $V \propto Q$, this graph is a straight line from the origin, and the area up to the final point $(Q, V)$ is a triangle: $W = \tfrac12 \times \text{base} \times \text{height} = \tfrac12 QV$

Using $Q = CV$, this can be written three equivalent ways: $W = \tfrac12 QV = \tfrac12 CV^2 = \frac{Q^2}{2C}$

Choose whichever form matches the quantities you are given. The crucial feature is the factor of $\tfrac12$: it is not $QV$ (the rectangle), because the p.d. rises from zero, so on average only half the final p.d. does work.

This stored energy can be released quickly: a discharging capacitor delivers a short burst of power (as in a camera flash or a defibrillator), which is why capacitors are used where a large, brief pulse of energy is needed.

When a capacitor discharges through a resistor $R$, the current is driven by the capacitor's own p.d. As charge leaves the plates the p.d. falls, so the current falls, so the charge leaves more slowly — the rate of decrease is always proportional to the charge remaining. This gives exponential decay: $Q = Q_0\,e^{-t/RC}$ and, because $V = Q/C$ and $I = V/R$, the p.d. and current decay identically: $V = V_0\,e^{-t/RC}, \qquad I = I_0\,e^{-t/RC}$

Charging through a resistor from a supply is the mirror image: the charge and p.d. rise towards their final values, quickly at first and then levelling off: $Q = Q_0\left(1 - e^{-t/RC}\right), \qquad V = V_0\left(1 - e^{-t/RC}\right)$

Here $Q_0$ (or $V_0$) is the final charge (or p.d.) the capacitor approaches. The $(1 - e^{-t/RC})$ bracket is the signature of charging — always ask whether the quantity should be rising or falling before you pick the equation.

When you sketch these curves, get the endpoints right: charging starts at $(0, 0)$ and levels off at $Q_0$; discharging starts at $(0, Q_0)$ and levels off at $0$. Neither curve ever exactly reaches its limit — they approach it asymptotically — but after about $5RC$ the capacitor is treated as effectively fully charged or discharged (over 99% / under 1%).

The product $RC$ has units of seconds (check: $\Omega \times \text{F} = \dfrac{\text{V}}{\text{A}} \times \dfrac{\text{C}}{\text{V}} = \dfrac{\text{C}}{\text{A}} = \text{s}$) and is called the time constant, $\tau = RC$. It sets the timescale over which a capacitor charges or discharges.

Putting $t = RC$ into the discharge equation: $Q = Q_0\,e^{-RC/RC} = Q_0\,e^{-1} = 0.37\,Q_0$

So after one time constant the charge (or p.d., or current) has fallen to $\dfrac{1}{e} \approx 37\%$ of its initial value on discharge. Equivalently, when charging, after one time constant the charge has risen to $63\%$ of its final value (since $1 - e^{-1} = 0.63$).

• A large $RC$ (large $R$ or large $C$) means slow charging/discharging.
• A small $RC$ means the capacitor charges/discharges quickly.

The time constant is not a half-life. The time to fall to $50\%$ is $t = RC\ln 2 \approx 0.69\,RC$, which is shorter than one time constant. Confusing $37\%$ with $50\%$ is one of the most common errors in this topic.

Reading $RC$ off a curved exponential graph is imprecise. The exam skill is to linearise the equation by taking natural logs. Starting from $Q = Q_0\,e^{-t/RC}$ and taking $\ln$ of both sides: $\ln Q = \ln Q_0 - \frac{t}{RC}$

Compare this with the straight-line form $y = mx + c$, taking $y = \ln Q$ and $x = t$:

• the gradient is $m = -\dfrac{1}{RC}$;
• the y-intercept is $c = \ln Q_0$.

So a plot of $\ln Q$ (or $\ln V$) against $t$ is a straight line with a negative gradient. From it: $RC = -\frac{1}{\text{gradient}}, \qquad Q_0 = e^{\text{intercept}}$

Then, if $R$ is known, $C = \dfrac{RC}{R}$ (or if $C$ is known, $R = \dfrac{RC}{C}$). The straight line also confirms the decay is genuinely exponential.

The classic trap is to write gradient $= -RC$. Look at the equation: $t$ is divided by $RC$, so the coefficient of $t$ is $-\dfrac{1}{RC}$, making the gradient $-\dfrac{1}{RC}$ and $RC$ the negative reciprocal of the gradient.

Core Practical 11 investigates the discharge of a capacitor through a resistor and determines $RC$ (and hence the capacitance).

Circuit. Connect the capacitor in series with a known resistor $R$ and a two-way switch, charged from a d.c. supply. A voltmeter or datalogger with very high input resistance is connected across the capacitor to record its p.d. $V$. Using a datalogger (or an oscilloscope) lets you capture many readings automatically and quickly.

Method. Fully charge the capacitor, then switch it to discharge through $R$. Record the p.d. $V$ across the capacitor at regular time intervals as it falls, until it is close to zero.

Analysis. From $V = V_0\,e^{-t/RC}$, take logs: $\ln V = \ln V_0 - \dfrac{t}{RC}$. Calculate $\ln V$ for each reading and plot $\ln V$ against $t$. This gives a straight line of gradient $-\dfrac{1}{RC}$. Measure the gradient from a large triangle, then $RC = -\dfrac{1}{\text{gradient}}$ and $C = \dfrac{RC}{R}$. Compare with the manufacturer's stated value.

Systematic error — the voltmeter's resistance. The voltmeter is connected in parallel with the capacitor, so it offers an extra path for charge to leak away. Its finite internal resistance $R_V$ is in parallel with $R$, giving a smaller effective discharge resistance $R_{\text{eff}} = \dfrac{R R_V}{R + R_V} < R$. The measured time constant is then $R_{\text{eff}}C$, so dividing by the nominal $R$ makes the calculated capacitance come out too small — a systematic error that repeating cannot remove. Use a voltmeter/datalogger with a very high resistance ($R_V \gg R$) so that $R_{\text{eff}} \approx R$ and the leakage is negligible.

Reducing random error. Take many readings with a datalogger, repeat the discharge and average, and use a best-fit line through the $\ln V$–$t$ points so a single anomalous reading does not distort the gradient.