Syllabus coverage
- Define capacitance, as applied to both isolated spherical conductors and to parallel plate capacitors
- Recall and use C = Q / V
- Derive, using C = Q / V, formulae for the combined capacitance of capacitors in series and in parallel
- Use the capacitance formulae for capacitors in series and in parallel
- Determine the electric potential energy stored in a capacitor from the area under the potential-charge graph
- Recall and use W = 1/2 Q V = 1/2 C V^2
- Analyse graphs of the variation with time of potential difference, charge and current for a capacitor discharging through a resistor
- Recall and use tau = RC for the time constant for a capacitor discharging through a resistor
- Use equations of the form x = x0 e^-(t / RC), where x could represent current, charge or potential difference, for a capacitor discharging through a resistor
Part of the AS & A Level Physics syllabus 9702.
Capacitors and capacitance: charge stored per unit potential difference
A capacitor is a device for storing charge: two conductors separated by an insulator, which is called the dielectric. When you connect a capacitor to a supply, electrons are pushed off one plate and onto the other, so one plate ends up with a charge of -Q and the other with +Q. The net charge on the whole capacitor is still zero, which is why 'the charge stored on a capacitor' always means the magnitude Q on either plate.
The capacitance of a capacitor is defined as the charge stored on one plate divided by the potential difference across the capacitor: C = Q / V. Divide a charge in coulomb by a potential difference in volts and the answer is a capacitance in farad (F), where 1 F = 1 C V-1. The farad is a very large unit for practical circuits, so capacitors are sold in microfarad (1 microfarad = 1 x 10-6 F) and picofarad (1 picofarad = 1 x 10-12 F). A capacitor that stores 4.8 x 10-4 C when the potential difference across it is 12 V has C = 4.8 x 10-4 / 12 = 4.0 x 10-5 F, which is 40 microfarad.
For a particular capacitor Q is proportional to V, so a graph of charge against potential difference is a straight line through the origin whose gradient is the capacitance. That gradient is the working meaning of capacitance: a large capacitance stores a lot of charge for a small potential difference, and a small capacitance stores only a little. The same definition applies to a single isolated conductor. An isolated sphere carrying a charge Q sits at a potential V relative to infinity, and its capacitance is Q / V.
- Capacitance C = Q / V, measured in farad (F), where 1 F = 1 C V-1.
- 1 microfarad = 1 x 10-6 F and 1 picofarad = 1 x 10-12 F, so 40 microfarad is 4.0 x 10-5 F.
- For a fixed capacitor Q is proportional to V, so the charge-potential difference graph is a straight line through the origin with gradient C.
- A capacitor stores equal and opposite charges on its two plates; the net charge on the capacitor is zero.
The capacitance of an isolated sphere and of a parallel plate capacitor
Take an isolated conducting sphere of radius r carrying a charge Q. The potential of its surface, measured relative to infinity, is V = Q / (4 pi epsilon0 r), where epsilon0 is the permittivity of free space. Substituting that expression into C = Q / V gives C = 4 pi epsilon0 r: the capacitance of an isolated sphere depends only on its radius. Doubling the radius doubles the capacitance, and the charge on the sphere makes no difference at all, because putting more charge on it raises the potential in proportion.
A parallel plate capacitor is two flat plates of area A separated by a distance d, with a uniform electric field between them. For a given charge on the plates the field strength is E = V / d, so a larger plate area can hold more charge for the same field, and a larger separation needs a larger potential difference to hold the same charge. The capacitance therefore rises as the plates are made larger and falls as they are moved further apart.
Filling the gap with a dielectric raises the capacitance. The molecules of the dielectric polarise, turning their negative ends towards the positive plate and their positive ends towards the negative plate, so they set up a field of their own that opposes the field of the plates. The potential difference across the capacitor falls while the charge on the plates is unchanged, so C = Q / V rises. A material with a large relative permittivity, such as mica or a ceramic, gives a large increase, which is why real capacitors have a dielectric between the plates instead of air.
- Isolated sphere: C = 4 pi epsilon0 r, fixed by the radius alone.
- Parallel plate: capacitance increases with the plate area and decreases with the separation.
- A dielectric polarises and opposes the field of the plates, reducing the potential difference for the same charge and so increasing C.
- All of these change the capacitance itself; changing the charge on the capacitor does not.
Capacitors in series and in parallel: the derivations
In parallel, the plates you connect together are at the same potential difference, so each capacitor has the same V across it while the charge supplied splits between them: Q = Q1 + Q2. Using Q = C V for each, C V = C1 V + C2 V, and dividing through by V gives C = C1 + C2. Capacitances in parallel add, so two 470 microfarad capacitors in parallel give 940 microfarad.
In series, the capacitors are joined end to end, so the same charge Q is transferred onto each of them, and the total potential difference is the sum of the potential differences across each: V = V1 + V2. Writing each potential difference as V = Q / C gives Q / C = Q / C1 + Q / C2, and dividing through by Q gives 1 / C = 1 / C1 + 1 / C2. The reciprocals of the capacitances add: two 470 microfarad capacitors in series give 235 microfarad, and a 6.0 microfarad and a 3.0 microfarad in series give 1 / C = 1 / 6.0 + 1 / 3.0 = 0.50 per microfarad, so C = 2.0 microfarad.
Notice that this runs against the rule for resistors, and the reason is which quantity is shared. Capacitors in series store the same charge while resistors in series carry the same current; capacitors in parallel share the same potential difference while resistors in parallel share the same current. If you work in reciprocals, always finish by taking the reciprocal back: an answer of 0.50 is the value of 1 / C, so the combined capacitance is 2.0 microfarad. In a series combination the combined capacitance is always smaller than the smallest capacitance in the combination.
- Parallel: same potential difference across each capacitor, and C = C1 + C2.
- Series: same charge on each capacitor, and 1 / C = 1 / C1 + 1 / C2.
- The rules are the opposite way round to resistors, because it is the shared quantity that swaps over.
- Take the reciprocal back at the end: 1 / C = 0.50 per microfarad means C = 2.0 microfarad.
| Connection | What is shared | Formula for the combined capacitance | Worked example |
|---|---|---|---|
| Parallel | The same potential difference across each capacitor | C = C1 + C2 | 470 microfarad + 470 microfarad = 940 microfarad |
| Series | The same charge stored on each capacitor | 1 / C = 1 / C1 + 1 / C2 | 470 microfarad and 470 microfarad give 235 microfarad |
| Series, unequal capacitors | The same charge stored on each capacitor | 1 / C = 1 / C1 + 1 / C2 | 6.0 microfarad and 3.0 microfarad give 1 / C = 0.50 per microfarad, so C = 2.0 microfarad |
Energy stored in a capacitor: the area under the potential-charge graph
Charging a capacitor stores energy, because work has to be done to push charge onto the plates against the potential difference that is already there. That potential difference is not constant: it starts at zero and rises steadily to V as the charge builds up to Q. The work done in transferring a small charge delta Q while the potential difference is V is V delta Q, which is the area of a narrow vertical strip on the potential-charge graph. Adding up all the strips from zero to Q gives the total work done, so the energy stored is the whole area under that graph.
The graph is a straight line from the origin to the point (Q, V), so the area is that of a triangle: W = 1/2 x base x height = 1/2 Q V. Substituting Q = C V gives the form you will use most often, W = 1/2 C V^2, and substituting V = Q / C gives W = 1/2 Q^2 / C. All three are the same energy written in different variables, so choose the one that matches the quantities the question gives you.
The energy is stored in the electric field between the plates, and it is released again when the capacitor discharges. Because the energy depends on the square of the potential difference, doubling the potential difference stores four times the energy, which is why very high voltages are used when a large amount of energy has to be delivered quickly. Worked numbers: a 470 microfarad capacitor charged to 6.0 V stores Q = C V = 470 x 10-6 x 6.0 = 2.8 x 10-3 C and W = 1/2 C V^2 = 0.5 x 470 x 10-6 x 6.0^2 = 8.5 x 10-3 J.
The factor of one half is worth understanding rather than memorising. The charge Q passes through an average potential difference of V / 2, not the full V, because the potential difference only reaches its final value at the very end of the charging process. Multiplied out, Q x V / 2 is the same as 1/2 Q V.
- Energy stored = area under the potential-charge graph, which is a triangle of area 1/2 Q V.
- W = 1/2 Q V = 1/2 C V^2 = 1/2 Q^2 / C; pick the form that suits the data you are given.
- The energy is stored in the electric field between the plates.
- 470 microfarad at 6.0 V stores 2.8 x 10-3 C and 8.5 x 10-3 J.
Discharging a capacitor through a resistor: the time constant RC
Connect a charged capacitor across a resistor and it discharges: the charge on the plates drives a current back round the circuit until the potential difference has fallen to nearly zero. The current obeys I = V / R, so a large potential difference gives a large current, charge leaves the plates quickly, and the potential difference falls quickly. As the potential difference falls the current falls with it, so the discharge is fastest at the beginning and becomes slower and slower. That feedback produces the exponential decay curve.
The decay is described by x = x0 e^-(t / RC), where x0 is the initial value of the quantity x, which may be the potential difference, the charge or the current, and t is the time since the discharge started. The product RC is the time constant tau of the circuit, in seconds when R is in ohm and C is in farad. Because Q = C V and V = I R, the potential difference, the charge and the current are all proportional to one another throughout the discharge, so all three follow the same exponential with the same time constant, differing only by the constant factors C and 1 / R.
At t = tau, x = x0 / e = 0.37 x0, so one time constant is the time for the value to fall to about 37 per cent of its starting value. After five time constants the value is down to 0.0067 of its starting value, less than one per cent, and for practical purposes the capacitor is then fully discharged. Worked values for a 47 microfarad capacitor discharging through 200 kohm, so tau = 2.0 x 10^5 x 47 x 10-6 = 9.4 s: after 9.4 s the potential difference has fallen from 12 V to 12 x e^-1 = 4.4 V, and after 28.2 s it is 12 x e^-3 = 0.60 V.
The time constant can be changed by changing either component. A larger resistance reduces the current for a given potential difference and slows the discharge down; a larger capacitance stores more charge for the same potential difference, so it takes longer to empty. That is why a smoothing capacitor in a power supply is made large: it must hold its charge for longer than the gap between the peaks of the rectified output.
| Time t / s | t / tau | Potential difference / V | Fraction of the initial 12 V remaining |
|---|---|---|---|
| 0 | 0 | 12.0 | 1.00 |
| 9.4 | 1 | 4.4 | 0.37 |
| 18.8 | 2 | 1.6 | 0.14 |
| 28.2 | 3 | 0.60 | 0.050 |
Graphs of potential difference, charge and current against time
The three graphs of potential difference, charge and current against time all have exactly the same shape, because the quantities are tied together by Q = C V and I = V / R. Every value falls exponentially with the same time constant RC, and the three curves differ only by the constant multipliers C and 1 / R. The current is largest at the instant the discharge begins, I0 = V0 / R, and it falls in step with the potential difference because the resistor is fixed.
You can read the time constant straight off a graph in two ways. Either find the time at which the value has fallen to 0.37 of its starting value, which is tau = RC, or find the time taken for the value to halve, which is 0.69 RC because ln 2 = 0.693. Both give the same time constant, so use whichever reading is easier on the axes you have been given.
The strongest evidence that a discharge really is exponential comes from taking natural logarithms. Starting from x = x0 e^-(t / RC), taking ln of both sides gives ln x = ln x0 - t / RC, so a graph of ln x against t is a straight line with gradient -1 / RC and intercept ln x0. A straight line on this graph proves the relationship is exponential, and from its gradient the time constant follows as RC = -1 / gradient. Dividing that by the resistance gives the capacitance, C = tau / R.
- Potential difference, charge and current all decay exponentially with the same time constant RC.
- The current starts at I0 = V0 / R and falls in step with the potential difference.
- Time constant from a graph: the time to fall to 0.37 of the initial value, which is the same as the time to fall to 1 / e.
- Half-value time = 0.69 RC, because ln 2 = 0.693.
- A graph of ln x against t is a straight line of gradient -1 / RC: a straight line proves the decay is exponential.
Capacitors at work: the camera flash, the defibrillator and the smoothing capacitor
A camera flash charges a capacitor slowly and then releases the energy in a single burst. A typical flash capacitor has a capacitance of about 100 microfarad and is charged to 300 V, so it stores W = 1/2 C V^2 = 0.5 x 100 x 10-6 x 300^2 = 4.5 J. The battery charges it through a resistor over a few seconds, and the capacitor then discharges through the xenon tube in about a millisecond, which is an average power of roughly 4.5 kW. No battery of that size could deliver 4.5 kW, which is exactly why the energy is stored in a capacitor first and released afterwards.
A defibrillator does the same thing on a larger scale. A 100 microfarad capacitor charged to 2.0 kV stores 1/2 x 100 x 10-6 x (2.0 x 10^3)^2 = 200 J, and that energy is delivered through the patient's chest in a few milliseconds, an average power of tens of kilowatts. The short pulse is the whole point: the same 200 J spread over minutes would do nothing useful, and the ability of the capacitor to give its energy back quickly is what makes the treatment work. Both devices use C = Q / V and W = 1/2 C V^2 in exactly the form you meet in questions.
A smoothing capacitor in a direct current power supply works the other way round, storing energy to fill in the gaps in a varying supply. A rectifier produces a potential difference that rises to a peak and falls back to zero many times a second, and the capacitor is connected across the output, so it charges up near each peak and then discharges through the load between them. The output is then a nearly steady direct potential difference with a small ripple, and the size of the ripple is set by the time constant: for good smoothing, RC must be much longer than the time between successive peaks, which is 10 ms for a 50 Hz supply that has been full-wave rectified. A 4700 microfarad capacitor with a 1.0 kohm load gives RC = 4.7 s, so in the 10 ms between peaks the output falls by only about 0.2 per cent.
- Camera flash: a 100 microfarad capacitor at 300 V stores 4.5 J and gives it up through the xenon tube in about a millisecond.
- Defibrillator: a 100 microfarad capacitor at 2.0 kV stores 200 J and delivers it to the patient in a few milliseconds.
- Smoothing capacitor: charged near each peak of the rectified output, discharged into the load between peaks, so a large RC keeps the ripple small.
- A 4700 microfarad capacitor with a 1.0 kohm load has RC = 4.7 s, which is hundreds of times longer than the 10 ms between peaks.
Key terms you must be able to define
- Capacitance
- The charge stored on one plate of a capacitor divided by the potential difference across it, C = Q / V; it is a property of the capacitor itself, measured in farad.
- Farad
- The unit of capacitance, 1 F = 1 C V-1: a capacitance of one farad stores one coulomb when the potential difference across it is one volt. Practical capacitors are rated in microfarad (1 x 10-6 F) or picofarad (1 x 10-12 F).
- Parallel plate capacitor
- Two flat conducting plates of area A separated by a distance d, with a uniform electric field between them; its capacitance increases when the plates are made larger or the separation is reduced, and when a dielectric is placed between them.
- Dielectric
- The insulating material between the plates of a capacitor. Its molecules polarise in the field of the plates and set up an opposing field, which reduces the potential difference for a given charge and so increases the capacitance.
- Time constant
- For a capacitor discharging through a resistor, tau = RC: the time taken for the charge, potential difference or current to fall to 1 / e, about 0.37, of its initial value. R in ohm and C in farad give tau in seconds.
- Half-value time
- The time taken for the potential difference, charge or current of a discharging capacitor to fall to half of its initial value, equal to RC ln 2 = 0.69 RC.
- Exponential discharge
- The way charge, potential difference and current fall with time when a capacitor discharges through a resistor: x = x0 e^-(t / RC), a curve that falls by the same fraction in every equal interval of time and never quite reaches zero.
- Energy stored in a capacitor
- The work done in charging the capacitor, stored in the electric field between its plates: W = 1/2 Q V = 1/2 C V^2 = 1/2 Q^2 / C, equal to the area under the potential-charge graph.
Applied to the real world
Cambridge sets 9702 in unfamiliar contexts - a fairground ride, a power station, a hospital scanner, a satellite, an astronomical observatory. Use these as models for how to apply the physics, then write your own.
A 47 microfarad capacitor is charged to 12 V and then discharged through a 200 kohm resistor while a data logger records the potential difference across it every 2 s. The time constant is tau = RC = 2.0 x 10^5 x 47 x 10-6 = 9.4 s, so the reading should have fallen to 4.4 V after 9.4 s and to 0.60 V after 28.2 s. Plotting ln V against time gives a straight line of gradient -1 / RC, which shows that the discharge is exponential and gives the capacitance from C = tau / R.
The flash unit of a camera stores its energy in a capacitor of about 100 microfarad charged to 300 V. The stored energy is W = 1/2 C V^2 = 0.5 x 100 x 10-6 x 300^2 = 4.5 J, and it is released through the xenon tube in roughly a millisecond, an average power of about 4.5 kW. The capacitor is charged from the battery through a resistor over a few seconds, which is why there is a delay between one flash and the next.
A smoothing capacitor turns the pulsating output of a rectifier into a steady direct potential difference. It charges near each peak of the rectified waveform and discharges through the load between peaks, and good smoothing requires the time constant RC to be much longer than the 10 ms between peaks of a full-wave rectified 50 Hz supply. With a 4700 microfarad capacitor and a 1.0 kohm load, RC = 4.7 s, so the output falls by only about 0.2 per cent before the next peak arrives.
A defibrillator charges a capacitor of about 100 microfarad to 2.0 kV, storing W = 1/2 C V^2 = 0.5 x 100 x 10-6 x (2.0 x 10^3)^2 = 200 J. That energy is delivered through the patient's chest in a few milliseconds, which is an average power of tens of kilowatts. A battery could supply the same energy but not in that time, and it is the very short pulse that makes the treatment effective.
Hybrid buses and some stop-start cars store the energy captured during braking in a bank of supercapacitors, which can be charged and discharged hundreds of thousands of times without wearing out and accept charge far faster than a battery. The energy recovered is stored as 1/2 C V^2 in the capacitor bank and returned to the motor when the vehicle pulls away again, instead of being wasted as heat in the brakes.
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Exam-style questions
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Exam technique for this topic
- Convert the units before you substitute anything. 470 microfarad is 470 x 10-6 F, 200 kohm is 2.0 x 10^5 ohm and 2.0 kV is 2.0 x 10^3 V. Quote the final answer with its unit: F or microfarad for capacitance, C for charge, J for energy and s for a time constant.
- The time constant is tau = RC, in seconds, and it is not the time to discharge completely: at t = tau the value is 0.37 of its starting value, and the graph only gets to about one per cent of it after five time constants. Read the half-value time as 0.69 RC, not RC.
- When you use x = x0 e^-(t / RC), write the exponent out before evaluating it. The exponent t / RC is dimensionless, so the time and the time constant must be in the same unit; an exponent of 1 gives 0.37 of the initial value, which is a useful check on the arithmetic.
- Check the series and parallel rules against the shared quantity. Capacitors in series store the same charge and their reciprocals add; capacitors in parallel share the same potential difference and their capacitances add. This is the opposite way round to resistors, so deciding from the circuit diagram rather than from memory is safer.
- Use the energy form that matches the data: W = 1/2 C V^2 when the potential difference is given, W = 1/2 Q V when the charge is given, and W = 1/2 Q^2 / C when the capacitance and charge are given. In every case the factor is one half, and the energy is stored in the electric field between the plates.
- In a derivation question, begin with the quantity that is shared: V for a parallel connection and Q for a series connection. Substitute into C = Q / V, then divide the whole equation by the shared quantity. State the final equation in reciprocal form for series, because C = C1 + C2 in series is the mark most often lost.