4.4 Capacitors
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Unit 4 (WPH14), A2 content: specification points 100–104 and Core Practical 11. Combining capacitors in series and in parallel is not part of this specification; the uniform field between charged plates is on the Electric fields page.
What this page covers
8 points
- What a capacitor is, the definition of capacitance \(C = Q/V\) and the farad
- Energy stored as the area under a graph of p.d. against charge: \(W = \tfrac12 QV = \tfrac12 CV^2 = \tfrac12 Q^2/C\)
- Why a supply charging a capacitor transfers twice the energy that ends up stored
- Graphs of charge, p.d. and current against time for charging and discharging through a resistor
- The time constant \(RC\), and what changing \(R\) or \(C\) does
- \(Q = Q_0 e^{-t/RC}\) and its current and p.d. forms, the log forms and the time to halve, \(RC \ln 2\)
- The p.d. across a charging capacitor, \(V_C = V_0(1 - e^{-t/RC})\)
- Core Practical 11: recording a charge and discharge with a data logger or oscilloscope, and analysing it with a ln graph
In one minute
6 points
- A capacitor is two conductors separated by an insulator that stores separated charge (\(+Q\) on one plate, \(-Q\) on the other) and the energy that goes with it, and when it discharges through a resistor its charge, current and p.d. all fall exponentially with time constant \(RC\).
- Capacitance \(C = Q/V\) is the charge stored per unit p.d.; 1 F = 1 C V⁻¹, and practical values are in µF, nF and pF.
- Energy stored = area under the \(V\)–\(Q\) graph: \(W = \tfrac12 QV = \tfrac12 CV^2 = \tfrac12 Q^2/C\). Charging from a fixed supply, the supply transfers \(QV\) and half of it is dissipated in the resistance.
- Through a resistor the current is largest at the start and decays exponentially, in both charging and discharging.
- In one time constant \(RC\), a discharge falls to 37% (\(e^{-1}\)) and a charging capacitor reaches 63% of its final p.d.; after about \(5RC\) the change is effectively complete.
- \(\ln Q = \ln Q_0 - t/RC\): a graph of ln against \(t\) is a straight line of gradient \(-1/RC\). The time to halve is \(RC \ln 2\).
Contents
12 sections
- What is a capacitor, and what does capacitance measure?
- How much energy does a charged capacitor store?
- What happens when a capacitor charges or discharges through a resistor?
- How do you calculate charge, current and p.d. during a discharge?
- How do you calculate the p.d. while a capacitor charges?
- Core Practical 11: how do you record and analyse a discharge?
- Exam tips for 4.4
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is a capacitor, and what does capacitance measure?
A capacitor is two conducting plates separated by an insulator, which may be air, a plastic film, a ceramic or a thin oxide layer. On its own it does nothing. Connect it to a supply, though, and the supply pushes electrons onto one plate and pulls the same number off the other. One plate ends up with charge \(-Q\) and the other is left with \(+Q\).
The capacitor as a whole stays neutral: "the charge stored" means \(Q\), the charge on one plate. No charge ever crosses the gap. Current flows in the wires on both sides while the plates charge, and stops when the p.d. across the capacitor equals the supply p.d.
For a given capacitor, doubling the p.d. doubles the charge it holds. The constant ratio is the capacitance.
A farad is a very large capacitance: lab capacitors are in µF (10⁻⁶ F), nF (10⁻⁹ F) or pF (10⁻¹² F), and only "supercapacitors" reach whole farads. Because \(Q \propto V\), a graph of \(Q\) against \(V\) (or \(V\) against \(Q\)) for a capacitor is a straight line through the origin, and its gradient gives \(C\) (or \(1/C\)).
A 470 µF capacitor is connected to a 9.0 V supply until it is fully charged. (a) Calculate the charge stored. (b) Calculate the number of electrons that have moved from one plate to the other. (c) The supply is changed to 12.0 V. Calculate the extra charge that flows onto the capacitor.
(a) Rearrange \(C = Q/V\), with 470 µF = 470 × 10⁻⁶ F: \(Q = CV = 470\times10^{-6} \times 9.0\) = 4.2 × 10⁻³ C (4.23 mC)
(b) Divide by the charge on one electron: \(N = 4.23\times10^{-3} \div 1.60\times10^{-19}\) = 2.6 × 10¹⁶ electrons
(c) Charge is proportional to p.d., so the extra charge is \(C\Delta V = 470\times10^{-6} \times (12.0 - 9.0)\) = 1.4 × 10⁻³ C (1.41 mC)
The two plates carry \(+Q\) and \(-Q\). The capacitor stores \(Q\), the charge that has been separated, and that is the \(Q\) in \(C = Q/V\). It is neither \(2Q\) nor zero.
How much energy does a charged capacitor store?
Pushing extra charge onto a plate that is already charged takes work, because the charge already there repels it. That work is stored as electric potential energy and released when the capacitor discharges.
To add a small charge \(\Delta Q\) when the p.d. is \(V\), the work done is \(\Delta W = V\Delta Q\). On a graph of \(V\) against \(Q\) that is the area of a thin strip. Adding up all the strips from zero to the final charge gives the key result: the energy stored is the area under the \(V\)–\(Q\) graph. Because \(V \propto Q\), the graph is a straight line through the origin and the area is a triangle, \(\tfrac12 QV\).
Put simply, the charge was not all moved at the final p.d. \(V\): on average it was moved through \(\tfrac12 V\).
The second and third forms come from substituting \(Q = CV\) or \(V = Q/C\) into \(\tfrac12 QV\). Pick the one that uses the quantities you already know. Notice the square: for a fixed capacitor, doubling \(V\) doubles \(Q\) and multiplies the energy by four. In the graph above, the energy at 4.0 V is only 4 mJ, so raising the p.d. from 4.0 V to 8.0 V adds 12 mJ.
The flash unit of a camera uses a 150 µF capacitor charged to 330 V. When the flash fires, the capacitor discharges through the flash tube in 1.2 ms. (a) Calculate the energy stored. (b) Calculate the mean power delivered to the flash tube, assuming all the stored energy is transferred.
(a) \(C\) and \(V\) are known, so use \(\tfrac12 CV^2\): \(W = \tfrac12 \times 150\times10^{-6} \times 330^2\) = 8.2 J (8.17 J)
(b) Power = energy ÷ time: \(P = 8.17 \div 1.2\times10^{-3}\) = 6.8 × 10³ W
Check with another form: \(Q = CV = 0.0495\) C and \(\tfrac12 QV = \tfrac12 \times 0.0495 \times 330 = 8.17\) J. A small battery could not deliver kilowatts; a capacitor can, because it releases its energy very quickly.
Where does the rest of the supply's energy go?
A supply of fixed p.d. \(V\) moves a total charge \(Q\) round the circuit while it charges the capacitor, and every coulomb is given \(V\) joules. So the supply transfers \(QV\). The capacitor stores only \(\tfrac12 QV\). The other half is dissipated as thermal energy in the resistance of the circuit: the resistor, the wires and the internal resistance of the supply.
The split does not depend on the resistance: a smaller resistance gives a larger current for a shorter time, but the energy dissipated is still \(\tfrac12 QV\).
A 1000 µF capacitor is charged to 6.0 V through a resistor from a 6.0 V supply. Calculate (a) the energy transferred by the supply, (b) the energy stored by the capacitor and (c) the energy dissipated in the resistor.
Charge moved by the supply: \(Q = CV = 1000\times10^{-6} \times 6.0 = 6.0\times10^{-3}\) C.
(a) Each coulomb is given 6.0 J: \(W = QV = 6.0\times10^{-3} \times 6.0\) = 0.036 J
(b) \(W = \tfrac12 QV\) = 0.018 J
(c) Dissipated = 0.036 − 0.018 = 0.018 J
Write 470 µF as 470 × 10⁻⁶ F before substituting. A forgotten 10⁻⁶ makes an energy a million times too big, which is usually obvious if you stop to read your answer.
What happens when a capacitor charges or discharges through a resistor?
The circuit below charges the capacitor through \(R\) with the switch at A, and discharges it through the same \(R\) with the switch at B. A voltmeter, or a data logger for fast changes, records the p.d. across the capacitor.
Charging
At the start the capacitor is uncharged, so \(V_C = 0\) and the whole supply p.d. \(V_0\) is across \(R\). The current is at its maximum, \(I_0 = V_0/R\). As charge builds up, \(V_C\) rises, so the p.d. across the resistor, \(V_0 - V_C\), falls, and so does the current. A smaller current means charge arrives more slowly, so \(V_C\) rises more slowly. The result is a graph that rises steeply and then levels off towards \(V_0\) (and the charge towards \(Q_0 = CV_0\)), while the current decays towards zero.
Discharging
Now the capacitor acts as the source. At the start, \(V = V_0\) and the current is largest, \(I_0 = V_0/R\). As charge flows off, \(V = Q/C\) falls, so the current \(V/R\) falls, so the charge flows off more slowly. The rate of loss of charge is proportional to the charge left, and that is the signature of exponential decay: \(Q\), \(V\) and \(I\) all fall by the same fraction in every equal time interval. The discharge current flows in the opposite direction to the charging current.
In one time constant a discharge falls to \(e^{-1} = 0.37\) of its starting value, and a charging capacitor reaches \(1 - e^{-1} = 0.63\) of its final p.d. After about \(5RC\) the process is effectively complete, with less than 1% left to go.
Why do \(R\) and \(C\) set the timing? A larger \(R\) gives a smaller current for the same p.d., so it takes longer to move the charge. A larger \(C\) means more charge has to be moved for the same p.d., so that also takes longer. The final charge \(CV_0\) and the final energy do not depend on \(R\); only the timing does.
Drawing the charging current as a curve that rises, like the p.d. The current is largest at the very start, when the capacitor is empty and the whole supply p.d. is across the resistor. It decays in both charging and discharging.
How do you calculate charge, current and p.d. during a discharge?
Because the rate of discharge is proportional to the charge left, the charge follows an exponential. The same shape appears for \(V\) and \(I\), because \(V = Q/C\) and \(I = V/R\) are both proportional to \(Q\).
Use the \(e^x\) and ln keys on your calculator. The exponent \(-t/RC\) has no unit, so \(t\) and \(RC\) must both be in seconds.
A 220 µF capacitor is charged to 8.0 V and then discharged through a 47 kΩ resistor (the capacitor in the figure above). Calculate (a) the time constant, (b) the initial discharge current and (c) the p.d. across the capacitor and the charge on it 15 s after the discharge starts.
(a) \(RC = 47\times10^{3} \times 220\times10^{-6}\) = 10.3 s (10.34 s)
(b) At \(t = 0\) the full 8.0 V is across the resistor: \(I_0 = V_0/R = 8.0 \div 47\times10^{3}\) = 1.7 × 10⁻⁴ A (0.170 mA)
(c) \(V = 8.0 \times e^{-15/10.34} = 8.0 \times 0.234\) = 1.9 V (1.88 V); then \(Q = CV = 220\times10^{-6} \times 1.88\) = 4.1 × 10⁻⁴ C
Check: 15 s is a little more than two halving times (7.2 s each), so \(V\) should be a little less than 8.0 ÷ 4 = 2.0 V. It is.
Finding a time: the log form and the time to halve
To find when a quantity reaches a value, take logs and rearrange: \(t = RC \ln(V_0/V)\). Putting \(V = \tfrac12 V_0\) gives the time to halve, \(t_{1/2} = RC \ln 2 \approx 0.69\,RC\). This is not in the formulae list, so derive it when you need it: put \(Q = \tfrac12 Q_0\) into \(Q = Q_0 e^{-t/RC}\), which gives \(e^{t/RC} = 2\). It is the same for \(Q\), \(V\) and \(I\), and the same from any starting value, just like the half-life of a radioactive source.
A 0.10 F capacitor keeps the clock of an alarm system running during a power cut. It starts at 5.0 V and the clock behaves as a 25 kΩ resistor. The clock stops working when the p.d. falls below 2.0 V. (a) Calculate how long the clock keeps running. (b) Calculate the time for the p.d. to halve.
Time constant: \(RC = 25\times10^{3} \times 0.10 = 2500\) s.
(a) From \(\ln V = \ln V_0 - t/RC\): \(t = RC \ln(V_0/V) = 2500 \times \ln(5.0 \div 2.0) = 2500 \times 0.916\) = 2.3 × 10³ s (2291 s, about 38 minutes)
(b) \(t_{1/2} = 2500 \times \ln 2\) = 1.7 × 10³ s (1733 s)
The answer to (a) is more than one halving time because 2.0 V is less than half of 5.0 V, which is a quick sense check.
What happens to the energy during a discharge?
Every quantity that decays linearly with \(Q\) (that is \(Q\), \(V\) and \(I\)) falls with the same time constant. Energy does not, because \(W = \tfrac12 CV^2\) depends on \(V^2\): \(W \propto e^{-2t/RC}\). After one time constant only \(e^{-2} \approx 14\%\) of the energy is left, and the energy halves in half the time the p.d. takes.
How do you calculate the p.d. while a capacitor charges?
While charging, the current still decays exponentially: \(I = I_0 e^{-t/RC}\) with \(I_0 = V_0/R\). So the p.d. across the resistor is \(V_R = IR = V_0 e^{-t/RC}\). The supply p.d. is shared between the resistor and the capacitor, \(V_C = V_0 - V_R\), so the capacitor's p.d. is whatever the resistor has not taken.
A 4.7 µF capacitor, initially uncharged, is charged through a 1.0 MΩ resistor from a 9.0 V supply (the circuit in the charging graphs above). (a) Calculate the p.d. across the capacitor and the current 3.0 s after charging starts. (b) A circuit connected across the capacitor switches on a lamp when the p.d. across the capacitor reaches 6.0 V. Calculate the delay.
Time constant: \(RC = 1.0\times10^{6} \times 4.7\times10^{-6} = 4.7\) s.
(a) \(V_C = 9.0 \times (1 - e^{-3.0/4.7}) = 9.0 \times (1 - 0.528)\) = 4.2 V (4.25 V). The rest of the 9.0 V is across the resistor: \(V_R = 9.0 - 4.25 = 4.75\) V, so \(I = 4.75 \div 1.0\times10^{6}\) = 4.8 × 10⁻⁶ A
(b) When \(V_C = 6.0\) V, \(V_R = 3.0\) V, so \(e^{-t/RC} = 3.0 \div 9.0\): \(t = RC \ln(9.0 \div 3.0) = 4.7 \times \ln 3\) = 5.2 s (5.16 s)
To make the delay longer, increase \(R\) or \(C\): the delay is proportional to \(RC\).
\(V = V_0 e^{-t/RC}\) does not give the p.d. across a charging capacitor. During charging, that expression is the p.d. across the resistor. The capacitor's p.d. is what is left: \(V_0\) minus it.
Core Practical 11: how do you record and analyse a discharge?
Aim. Display the p.d. across a capacitor as it charges and discharges through a resistor, and use the discharge to find the time constant, and so \(C\) if \(R\) is known.
Circuit and choices. Use the two-way switch circuit above, with the data logger's voltage sensor (or an oscilloscope) across the capacitor. A time constant of a few seconds or more lets a data logger sample many points; with a voltmeter and stopwatch, aim for tens of seconds. For a time constant of milliseconds, use an oscilloscope with a square-wave supply so that the trace repeats.
Method. Charge fully (wait at least \(5RC\)), start recording, switch to discharge and record \(V\) against \(t\). Set the sampling rate so that there are many readings per time constant.
Analysis. \(\ln V = \ln V_0 - t/RC\) has the form \(y = c + mx\). Plot ln(\(V\)/V) against \(t\). A straight line confirms exponential decay, the gradient is \(-1/RC\) and the intercept is \(\ln V_0\).
Why a data logger or oscilloscope?
It takes many readings at precise, equal intervals, including at the start when the p.d. changes fastest, and removes the reaction-time error of reading a meter and a stopwatch together. It must have a very high resistance; otherwise it provides a second discharge path in parallel with \(R\) and makes the measured time constant too small.
Use the specimen graph above (resistor 10.0 kΩ ± 2%) to determine the capacitance. The gradient has a percentage uncertainty of 3%. Find the uncertainty in \(C\).
Gradient from the large triangle, from \(t\) = 1.0 s to \(t\) = 13.0 s: \((-0.98 - 1.58) \div 12.0 = -2.56 \div 12.0 = -0.213\) s⁻¹.
Gradient \(= -1/RC\), so \(RC = 1 \div 0.213 = 4.69\) s.
\(C = 4.69 \div 10.0\times10^{3}\) = 4.7 × 10⁻⁴ F (469 µF)
\(C = 1/(|\text{gradient}| \times R)\), so the percentage uncertainties add: 3% + 2% = 5%, and \(\Delta C = 0.05 \times 469\) µF ≈ ± 23 µF
The intercept, 1.79, gives \(V_0 = e^{1.79} \approx 6.0\) V, consistent with the 6.0 V charging supply.
Practical issues
- Capacitor tolerance: electrolytic capacitors are often marked ±10% or ±20%, so a measured value a few per cent from the marked value is not necessarily an error.
- Polarity: electrolytic capacitors must be connected with the + terminal to the positive side of the supply, and never above their rated voltage.
- Safety: large capacitors charged to a high p.d. can give a dangerous shock even after the supply is disconnected. Discharge them through a resistor before handling.
- Charging data: for charging, plot ln(\(V_0 - V_C\)) against \(t\); it also has gradient \(-1/RC\).
Exam tips for 4.4
"Charge stored per unit p.d.", with charge meaning the charge on one plate. Writing just "\(Q/V\)" or "how much charge it holds" leaves out the idea of per unit p.d.
All three forms of the energy equation are equivalent. If you know \(C\) and \(V\), use \(\tfrac12 CV^2\); if you are given a \(V\)–\(Q\) graph, find the area, or use \(\tfrac12 QV\) with values read from the graph.
For "explain why the current decreases": charge builds up on the plates → p.d. across the capacitor rises → p.d. across the resistor falls → current falls → charge builds up more slowly. Each arrow is a separate point.
From a discharge curve, read off the time to fall to 37% of the start value (that is \(RC\)), or the time to halve and divide by ln 2. From a ln graph, \(RC = -1/\text{gradient}\). Show your readings on the graph, and label a log axis ln(\(V\)/V): logs are taken of numbers, not units.
When \(C\) is found from \(RC \div R\), add the percentage uncertainties in \(RC\) and \(R\). When a quantity is squared, as \(V\) is in \(\tfrac12 CV^2\), double its percentage uncertainty.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Current flows through a capacitor, across the gap. | The gap is an insulator. | Charge flows onto one plate and off the other; current flows in the wires while the charge separates, but none crosses the gap. |
| A charged capacitor carries a net charge. | The plates carry equal and opposite charges. | The capacitor is neutral overall; it stores separated charge \(Q\), which is why it stores energy. |
| The energy stored is \(QV\). | The p.d. rose from 0 to \(V\) while the charge was added. | \(W = \tfrac12 QV\), the triangle under the \(V\)–\(Q\) graph; the supply transfers \(QV\) and half is dissipated. |
| A larger charging resistor means less energy is stored. | The final p.d. is still the supply p.d. | \(R\) changes only how long charging takes; the stored energy \(\tfrac12 CV_0^2\) is the same. |
| A capacitor is fully discharged after one time constant. | Exponential decay never reaches zero in a finite time. | After \(RC\), 37% remains; after \(5RC\), less than 1%. |
| The time to halve depends on the starting charge. | Each halving takes the same time for an exponential. | \(t_{1/2} = RC \ln 2\), independent of \(Q_0\) or \(V_0\). |
How this topic connects
Current and charge (Unit 2)
\(I = \Delta Q/\Delta t\), so the area under a current–time graph for a charging or discharging capacitor equals the charge that flowed, \(Q_0\). The gradient of a \(Q\)–\(t\) graph is the current.
4.3 Electric fields
Between the charged plates there is a nearly uniform electric field, \(E = V/d\), and the stored energy can be thought of as being in that field.
Radioactive decay (Unit 5)
The same mathematics: a rate proportional to the amount left, exponential decay, a constant half-life (here \(RC \ln 2\)) and a straight-line ln graph. Turning an exponential into a straight line with logs and combining percentage uncertainties are general A2 practical skills (Unit 6).
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(C = Q/V\) | capacitance, charge or p.d. (\(C\) in F, \(Q\) in C, \(V\) in V) | Yes |
| \(W = \tfrac12 QV = \tfrac12 CV^2 = \tfrac12 Q^2/C\) | energy stored (J) | Yes |
| \(W_{\text{supply}} = QV\) | energy transferred by the supply while charging | No — derive it: \(W = QV\) for charge moved through p.d. \(V\) |
| \(\tau = RC\) | time constant (s) | No — learn it |
| \(Q = Q_0 e^{-t/RC}\), \(I = I_0 e^{-t/RC}\), \(V = V_0 e^{-t/RC}\) | discharge | Yes |
| \(\ln Q = \ln Q_0 - t/RC\) (and for \(I\), \(V\)) | finding a time; straight-line graph | Yes |
| \(t_{1/2} = RC \ln 2\) | time to halve | No — derive it |
| \(V_C = V_0(1 - e^{-t/RC})\) | p.d. across a charging capacitor | No — derive it: \(V_0 - V_R\) |
Check your understanding
1. A 100 µF capacitor stores a charge of 1.5 mC. Calculate the p.d. across it.
\(V = Q/C = 1.5\times10^{-3} \div 100\times10^{-6}\) = 15 V.
2. The p.d. across a capacitor is tripled. By what factor does the energy stored change?
\(W = \tfrac12 CV^2\), so the energy increases by a factor of \(3^2\) = 9.
3. Calculate the time constant of a 2.2 µF capacitor discharging through a 330 kΩ resistor.
\(RC = 330\times10^{3} \times 2.2\times10^{-6}\) = 0.73 s (0.726 s).
4. What percentage of its initial charge does a discharging capacitor keep after three time constants?
\(Q/Q_0 = e^{-3} = 0.050\), so 5.0%.
5. In the core practical, a graph of ln(V/V) against t has a gradient of −0.050 s⁻¹. The resistor is 20 kΩ. Calculate the capacitance, and state why ln V is plotted rather than V.
\(RC = 1 \div 0.050 = 20\) s, so \(C = 20 \div 20\times10^{3}\) = 1.0 × 10⁻³ F (1000 µF). ln \(V\) against \(t\) is a straight line for an exponential decay, which tests the relationship and gives \(RC\) from the gradient.
6. Explain why the current in a charging circuit decreases, even though the supply p.d. stays the same.
As charge builds up on the plates, the p.d. across the capacitor rises. The supply p.d. is shared, so the p.d. across the resistor, \(V_0 - V_C\), falls. The current \(I = (V_0 - V_C)/R\) therefore falls, so charge arrives more slowly, and the current tends to zero as \(V_C\) approaches \(V_0\).
Specification checklist
- (100) I can describe how a capacitor stores separated charge, define capacitance as charge stored per unit p.d. and use \(C = Q/V\) with the farad and its sub-multiples.
- (101) I can explain why the energy stored is the area under a graph of p.d. against charge, and use \(W = \tfrac12 QV\), \(\tfrac12 CV^2\) and \(\tfrac12 Q^2/C\).
- (101) I can explain why a supply charging a capacitor transfers \(QV\) but only \(\tfrac12 QV\) is stored.
- (102) I can sketch and explain graphs of charge, p.d. and current against time for a capacitor charging and discharging through a resistor, and explain what the time constant \(RC\) tells me.
- (103) I can describe how to use a data logger or an oscilloscope to record the p.d. across a capacitor as it charges and discharges, and analyse the results with a ln \(V\)–\(t\) graph, including uncertainties.
- (104) I can use \(Q = Q_0 e^{-t/RC}\) and the matching equations for \(I\) and \(V\), their log forms, and the time to halve, \(RC \ln 2\).