Capacitors
A capacitor stores energy by holding positive and negative charge apart. It stores far less energy than a battery, but it can hand all of it back in a fraction of a second, which is why camera flashes, defibrillators and the smoothing circuit in every phone charger use one. On this page you will see where the ½ in \(W = \tfrac{1}{2}QV\) comes from, why capacitors charge and discharge exponentially, and how a log graph turns that curve into a straight line.
What a capacitor is
The simplest capacitor is two parallel metal plates with an insulator between them (air, paper, plastic or a thin oxide layer). Connect it to a battery and the battery pushes electrons onto one plate and pulls the same number of electrons off the other. No charge ever crosses the gap. One plate ends up with charge \(+Q\), the other with \(-Q\).
As charge builds up, the plates develop a p.d. that opposes the battery. New electrons arriving at the negative plate are repelled by the ones already there. Charging stops when the p.d. across the capacitor equals the e.m.f. of the battery. At that point there is no p.d. left to drive a current, so the current is zero even though the circuit is complete.
A capacitor does not store net charge: \(+Q\) on one plate and \(-Q\) on the other add up to zero. What it stores is energy, by keeping the charges separated. When a question says "the charge stored is \(Q\)" it means the charge on one plate.
Capacitance: C = Q/V
Push twice as much charge onto the plates and the field between them doubles, so the p.d. doubles too. Charge is proportional to p.d., and the constant of proportionality tells you how much charge the capacitor holds for each volt. That is its capacitance.
Capacitance
Capacitance is the charge stored per unit potential difference across the capacitor, \(C = Q/V\). The unit is the farad: a capacitor has a capacitance of 1 F if it stores 1 C of charge when the p.d. across it is 1 V (\(1\ \text{F} = 1\ \text{C V}^{-1}\)).
One farad is huge. Real capacitors are usually measured in microfarads (\(1\ \mu\text{F} = 10^{-6}\ \text{F}\)), nanofarads (\(10^{-9}\ \text{F}\)) or picofarads (\(10^{-12}\ \text{F}\)). A 470 μF capacitor charged to 9.0 V holds \(Q = CV = 470\times10^{-6} \times 9.0 = 4.2\times10^{-3}\ \text{C}\), a few millicoulombs.
Because \(Q \propto V\), a graph of p.d. against charge is a straight line through the origin with gradient \(V/Q = 1/C\). A bigger capacitor gives a shallower line: it takes more charge to raise its p.d. by each volt.
Energy stored in a capacitor
Charging a capacitor takes work. The first bit of charge goes on easily, because the plates are uncharged and the p.d. is zero. Every later bit has to be pushed onto a plate that already repels it, against a p.d. that keeps growing. The last bit is pushed against the full p.d. \(V\).
Moving a small charge \(\Delta Q\) through a p.d. \(V\) takes work \(\Delta W = V\,\Delta Q\). On a graph of \(V\) against \(Q\), that is the area of a thin strip of height \(V\) and width \(\Delta Q\). Add up all the strips and you get the area under the graph, which is a triangle.
To move a small extra charge \(\Delta Q\) onto the plates when the p.d. is \(V\), you do work (p.d. is work done per unit charge):
\[ \Delta W = V\,\Delta Q \]If \(\Delta Q\) is small, \(V\) hardly changes while you add it.
On a graph of \(V\) (vertical) against \(Q\) (horizontal), \(V\,\Delta Q\) is the area of a thin strip under the line. The total work done charging from 0 to \(Q\) is the sum of all the strips, which is the area under the graph:
\[ W = \sum V\,\Delta Q = \text{area under the } V\text{–}Q \text{ graph} \]Because \(V \propto Q\), the graph is a straight line through the origin, so the area is a triangle with base \(Q\) and height \(V\):
\[ W = \tfrac{1}{2} \times \text{base} \times \text{height} \quad\Rightarrow\quad \boxed{W = \tfrac{1}{2}QV} \]The ½ is there because, on average, each coulomb was pushed on against only half of the final p.d.
Start from the result above and the definition of capacitance:
\[ W = \tfrac{1}{2}QV, \qquad C = \frac{Q}{V} \]To get rid of \(Q\), write \(Q = CV\) and substitute:
\[ W = \tfrac{1}{2}(CV)V \quad\Rightarrow\quad \boxed{W = \tfrac{1}{2}CV^2} \]To get rid of \(V\) instead, write \(V = Q/C\) and substitute:
\[ W = \tfrac{1}{2}Q\left(\frac{Q}{C}\right) \quad\Rightarrow\quad \boxed{W = \frac{1}{2}\frac{Q^2}{C}} \]All three are the same energy. Pick the one that uses the two quantities you already know.
Energy stored
Energy stored
Energy stored
For a given capacitor the energy goes as \(V^2\): double the p.d. and you store four times the energy. This is why defibrillator and camera-flash capacitors are charged to hundreds of volts.
\(W = QV\) is the work done by a battery moving charge \(Q\) through a constant p.d. \(V\). For a capacitor the p.d. rises from zero as the charge goes on, so the energy stored is only \(\tfrac{1}{2}QV\). Don't drop the half.
A 220 μF capacitor is charged to a p.d. of 9.0 V. Calculate (a) the charge stored and (b) the energy stored. (c) The p.d. is increased to 18 V. How much extra energy is stored?
Show solution
(a) \(Q = CV = 220\times10^{-6} \times 9.0 = 1.98\times10^{-3}\ \text{C} \approx 2.0\ \text{mC}\)
(b) \(W = \tfrac12 CV^2 = \tfrac12 \times 220\times10^{-6} \times 9.0^2 = 8.9\times10^{-3}\ \text{J}\)
(c) Energy goes as \(V^2\), so doubling \(V\) gives four times the energy: \(4 \times 8.91\times10^{-3} = 3.56\times10^{-2}\ \text{J}\). The extra energy is \(3.56\times10^{-2} - 0.89\times10^{-2} = 2.7\times10^{-2}\ \text{J}\) (three times the original, not one).
Charging and discharging through a resistor
In real circuits a capacitor charges and discharges through a resistor, and the resistor sets how fast it happens. The key to every graph in this topic is to ask: what is the p.d. across the resistor right now? That p.d. decides the current, \(I = V_R/R\).
Charging
At the instant the switch closes, the capacitor is empty, so \(V_C = 0\) and the whole battery e.m.f. is across the resistor. The current starts at its biggest value, \(I_0 = V_0/R\). As charge builds up, \(V_C\) rises. Kirchhoff's second law says \(V_0 = V_C + V_R\), so \(V_R\) has to fall, and so does the current. Less current means charge arrives more slowly, so \(V_C\) rises more slowly. The capacitor creeps towards \(V_0\) but the closer it gets, the slower it goes.
Discharging
Now switch the battery out and connect the charged capacitor across the resistor alone. The capacitor drives a current round the circuit, in the opposite direction to the charging current. The p.d. across the resistor is now the p.d. across the capacitor, so \(I = V_C/R = Q/RC\). The current is proportional to the charge that is left. A full capacitor empties quickly; a nearly empty one has hardly any p.d. to push with and empties slowly.
"The rate of loss is proportional to the amount left" is the signature of exponential decay. You will meet the same pattern in radioactive decay in Unit 5.
The shapes to remember
| Charging | Discharging | |
|---|---|---|
| Charge \(Q\) and p.d. \(V_C\) | rise quickly, then level off at \(Q_0 = CV_0\) and \(V_0\) | fall exponentially towards zero |
| Current \(I\) | starts at \(V_0/R\) and falls exponentially to zero | starts at \(V_0/R\) in the opposite direction and falls exponentially to zero |
| p.d. across \(R\) | \(V_0 - V_C\), falls to zero | equal to \(V_C\) |
\(Q\) and \(V_C\) always have the same shape because \(V_C = Q/C\). The current is the gradient of the \(Q\)–\(t\) graph (\(I = \Delta Q/\Delta t\)), so it is largest where the \(Q\)–\(t\) graph is steepest.
When you sketch the discharge current, check whether the question wants its magnitude (drawn above the axis) or its direction too (drawn below the axis, because it flows the opposite way to the charging current). Also, changing \(R\) does not change the final charge \(Q_0 = CV_0\). It only changes how long it takes to get there.
The time constant RC
How fast all of this happens depends on just one number, the product \(RC\). A bigger resistance means a smaller current, so charge moves more slowly. A bigger capacitance means more charge has to move for every volt of change. Either way, it takes longer. \(RC\) is called the time constant, and it really is a time:
\[ [RC] = \Omega \times \text{F} = \frac{\text{V}}{\text{A}} \times \frac{\text{C}}{\text{V}} = \frac{\text{C}}{\text{A}} = \frac{\text{A s}}{\text{A}} = \text{s} \]The time constant \(RC\) of a capacitor–resistor circuit is the time taken for the charge (or current, or p.d.) of a discharging capacitor to fall to \(1/e\) of its initial value, which is about 37%. While charging from empty, it is the time for the charge to reach \(1 - 1/e\), about 63%, of its final value.
| Time | Fraction left when discharging, \(e^{-t/RC}\) | Fraction reached when charging |
|---|---|---|
| \(RC\) | 0.368 (37%) | 0.632 (63%) |
| \(2RC\) | 0.135 | 0.865 |
| \(3RC\) | 0.050 | 0.950 |
| \(5RC\) | 0.007 | 0.993 |
After about \(5RC\) a capacitor is, for practical purposes, fully charged or fully discharged. An exponential never actually reaches zero, but after \(5RC\) less than 1% is left.
Half-life
An exponential decay loses the same fraction in every equal interval of time. So there is a fixed time, the half-life \(t_{1/2}\), in which the charge always halves: from 100% to 50%, from 50% to 25%, and so on.
After one half-life, the charge is half its starting value:
\[ \tfrac{1}{2}Q_0 = Q_0\, e^{-t_{1/2}/RC} \quad\Rightarrow\quad \tfrac12 = e^{-t_{1/2}/RC} \]Take natural logs of both sides. \(\ln \tfrac12 = -\ln 2\):
\[ -\ln 2 = -\frac{t_{1/2}}{RC} \quad\Rightarrow\quad \boxed{t_{1/2} = RC \ln 2 \approx 0.69\,RC} \]The starting charge cancelled out, which is why the half-life is the same however full the capacitor is.
Two quick ways to read \(RC\) from a discharge graph: find where the curve has fallen to 37% of its starting value, or draw the tangent at \(t = 0\) and see where it cuts the time axis. Both give \(t = RC\). Using the 37% point is usually more accurate, because drawing a tangent by eye is hard.
The exponential discharge equations
For a capacitor discharging through a resistor, the charge left after time \(t\) is given on the formula sheet:
Capacitor discharge
The power \(-t/RC\) is "how many time constants have passed", with a minus sign because the charge is falling. Each time constant multiplies what's left by \(1/e = 0.368\). The p.d. and current behave in exactly the same way, and you need to be able to show why.
Start from the discharge equation and divide both sides by \(C\). Since \(V = Q/C\) at every instant:
\[ \frac{Q}{C} = \frac{Q_0}{C}\, e^{-t/RC} \quad\Rightarrow\quad \boxed{V = V_0\, e^{-t/RC}} \]\(V_0 = Q_0/C\) is the p.d. at the start. \(C\) is a constant, so dividing by it doesn't change the shape.
During discharge the capacitor is connected directly across the resistor, so the p.d. across the resistor equals the p.d. across the capacitor. Ohm's law gives the current:
\[ I = \frac{V}{R} = \frac{V_0}{R}\, e^{-t/RC} \]\(V_0/R\) is the current at the start, \(I_0\), so:
\[ \boxed{I = I_0\, e^{-t/RC}} \]Turning the curve into a straight line
Exponential curves are hard to analyse by eye. Taking the natural log of both sides turns them into straight lines, which is the trick behind Core Practical 11.
Take natural logs of both sides of \(Q = Q_0 e^{-t/RC}\):
\[ \ln Q = \ln\left(Q_0\, e^{-t/RC}\right) \]The log of a product is the sum of the logs, \(\ln(ab) = \ln a + \ln b\):
\[ \ln Q = \ln Q_0 + \ln\left(e^{-t/RC}\right) \]\(\ln\) undoes \(e\), so \(\ln(e^{x}) = x\):
\[ \boxed{\ln Q = \ln Q_0 - \frac{t}{RC}} \]Exactly the same steps on the p.d. and current equations give:
\[ \boxed{\ln V = \ln V_0 - \frac{t}{RC}} \qquad \boxed{\ln I = \ln I_0 - \frac{t}{RC}} \]Compare with \(y = c + mx\): plotting \(\ln V\) against \(t\) gives a straight line with gradient \(-1/RC\) and intercept \(\ln V_0\).
To find a time, rearrange the log equation: \(t = RC \ln(V_0/V)\). Keep \(V_0/V\) bigger than 1 and the answer comes out positive. To find a value after a time, use the exponential form directly, and work out \(t/RC\) first so you don't mistype the power into your calculator.
A 470 μF capacitor is charged to 12.0 V and then discharged through a 2.2 kΩ resistor. Calculate (a) the time constant, (b) the initial current, (c) the p.d. after 2.0 s, (d) the time for the p.d. to fall to 3.0 V and (e) the percentage of the original energy still stored at that moment.
Show solution
(a) \(RC = 2200 \times 470\times10^{-6} = 1.034\ \text{s} \approx 1.03\ \text{s}\)
(b) \(I_0 = V_0/R = 12.0/2200 = 5.45\times10^{-3}\ \text{A} = 5.45\ \text{mA}\)
(c) \(t/RC = 2.0/1.034 = 1.934\), so \(V = 12.0\, e^{-1.934} = 12.0 \times 0.1445 = 1.73\ \text{V}\)
(d) \(t = RC \ln(V_0/V) = 1.034 \times \ln(12.0/3.0) = 1.034 \times 1.386 = 1.43\ \text{s}\)
(e) \(W \propto V^2\), so the fraction left is \((3.0/12.0)^2 = 1/16 = 6.25\%\). Three quarters of the p.d. has gone, but nearly 94% of the energy has gone.
A camera flash uses a 150 μF capacitor charged to 300 V. When the flash fires, 90% of the stored energy is released in 1.0 ms. Calculate the mean power delivered to the flash tube, and explain why the camera cannot simply use its 3 V battery directly.
Show solution
Energy stored: \(W = \tfrac12 CV^2 = \tfrac12 \times 150\times10^{-6} \times 300^2 = 6.75\ \text{J}\)
Energy released: \(0.90 \times 6.75 = 6.08\ \text{J}\), so mean power \(= 6.08 / 1.0\times10^{-3} = 6.1\times10^{3}\ \text{W}\)
The battery has enough energy, but its internal resistance limits the current it can deliver, so it can only supply energy slowly (a few watts). The capacitor is charged slowly from the battery over a few seconds, then delivers the energy very quickly through the low resistance of the flash tube (small \(RC\)).
Core Practical 11: capacitor discharge
Use an oscilloscope or data logger to display and analyse the p.d. across a capacitor as it charges and discharges through a resistor
Method. Connect a capacitor (for example 470 μF), a resistor (for example 47 kΩ), a battery and a two-way switch so that the switch either connects the capacitor to the battery or connects it across the resistor. Connect a voltage sensor and data logger (or an oscilloscope) across the capacitor. Charge the capacitor fully, then flip the switch to discharge it through the resistor, logging the p.d. at regular intervals. Repeat with the switch flipped back to record the charging curve.
Using an oscilloscope. For a small time constant (for example 0.1 μF and 10 kΩ, \(RC\) = 1 ms), replace the battery and switch with a square-wave signal generator. The capacitor charges and discharges every cycle and the oscilloscope shows a steady trace. Read \(V\) at a series of times from the time-base.
Analysis. Since \(\ln V = \ln V_0 - t/RC\), plot \(\ln V\) against \(t\). A straight line confirms the decay is exponential. The gradient is \(-1/RC\), so \(RC = -1/\text{gradient}\), and with \(R\) known, \(C = RC/R\). The intercept is \(\ln V_0\). You can also check by reading the time for \(V\) to fall to 37% of \(V_0\).
- A data logger removes reaction-time error and can take hundreds of readings. If you use a voltmeter and stopwatch instead, choose a large \(RC\) (tens of seconds) and take readings at fixed intervals.
- Stop using readings once \(V\) is very small: the resolution of the meter becomes a large fraction of the reading and the \(\ln V\) points scatter.
- The voltmeter or logger has a finite resistance in parallel with \(R\). If it is not much bigger than \(R\), the capacitor also discharges through the meter, the measured \(RC\) is too small and so is \(C\). This is a systematic error: use a high-resistance logger, or a smaller \(R\).
- Electrolytic capacitors have a large tolerance (often ±10% to ±20%), so the measured \(C\) can differ from the labelled value by this much.
- Safety: connect electrolytic capacitors the right way round and never above their rated voltage, as they can overheat and burst. Discharge large capacitors before handling them.
In Core Practical 11 a student discharges a capacitor through a 47 kΩ resistor and plots \(\ln(V/\text{V})\) against \(t\). The line passes through (0 s, 1.79) and (60 s, 0.29). Determine the capacitance, and the p.d. the capacitor was charged to.
Show solution
Gradient \(= \dfrac{0.29 - 1.79}{60 - 0} = -0.0250\ \text{s}^{-1}\)
Gradient \(= -1/RC\), so \(RC = 1/0.0250 = 40\ \text{s}\), and \(C = \dfrac{40}{47\times10^{3}} = 8.5\times10^{-4}\ \text{F} = 850\ \mu\text{F}\)
Intercept \(= \ln V_0 = 1.79\), so \(V_0 = e^{1.79} = 6.0\ \text{V}\)
Quick checks
A charged capacitor discharges through a resistor. What fraction of its original charge is left after a time 2RC?
\(Q/Q_0 = e^{-2} = 0.135\). Each time constant multiplies the charge by 0.37, so two of them give \(0.37^2 \approx 0.14\). 0.25 would be two half-lives, and a half-life is shorter than \(RC\).
The p.d. across a capacitor is doubled. What happens to the energy stored?
\(W = \tfrac12 CV^2\) and \(C\) is fixed, so \(W \propto V^2\). Doubling \(V\) also doubles \(Q\), and \(W = \tfrac12 QV\) gives the same factor of 4.
Which of these is equivalent to the unit of the time constant, Ω × F?
\(\Omega = \text{V A}^{-1}\) and \(\text{F} = \text{C V}^{-1}\), so \(\Omega\,\text{F} = \text{C A}^{-1} = \text{(A s) A}^{-1} = \text{s}\).
An uncharged capacitor is connected in series with a resistor and a battery. At the instant the switch is closed, which statement is correct?
With no charge on it, the capacitor has no p.d. across it, so the whole e.m.f. is across the resistor. The current is \(V_0/R\), its largest value. It falls as \(V_C\) builds up.
A student plots ln I against t for a capacitor discharging through a resistor. What is the gradient of the line?
\(\ln I = \ln I_0 - t/RC\) has the form \(y = c + mx\) with \(m = -1/RC\). The intercept is \(\ln I_0\).
A capacitor is charged to a fixed p.d. and discharged through a resistor. The resistor is replaced with one of half the resistance and the experiment is repeated. Which statement is correct?
\(I_0 = V_0/R\) doubles and \(RC\) halves, so the discharge is twice as fast. But the capacitor started with the same \(Q_0 = CV_0\) and energy \(\tfrac12 CV_0^2\), and all of it still flows through the resistor (the area under the \(I\)–\(t\) graph is unchanged).