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 = ½ QV = ½ CV² = ½ Q²/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 = Q₀ 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 = V₀(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

1. 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.

2. 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.

3. Energy stored = area under the V–Q graph: W = ½ QV = ½ CV² = ½ Q²/C. Charging from a fixed supply, the supply transfers QV and half of it is dissipated in the resistance.

4. Through a resistor the current is largest at the start and decays exponentially, in both charging and discharging.

5. 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.

6. ln Q = ln Q₀ - t/RC: a graph of ln against t is a straight line of gradient -1/RC.

The time to halve is RC ln 2.

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 ∝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).

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) Rearrange C = Q/V, with 470 µF = 470 × 10⁻⁶ F: Q = CV = 470×10⁻⁶ ×9.0 = 4.2 × 10⁻³ C (4.23 mC)

(b) Divide by the charge on one electron: N = 4.23×10⁻³ ÷1.60×10⁻¹⁹ = 2.6 × 10¹⁶ electrons

(c) Charge is proportional to p.d., so the extra charge is CΔV = 470×10⁻⁶ ×(12.0 - 9.0) = 1.4 × 10⁻³ C (1.41 mC)

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 ΔQ when the p.d. is V, the work done is ΔW = VΔQ. 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 ∝Q, the graph is a straight line through the origin and the area is a triangle, ½ QV.

Put simply, the charge was not all moved at the final p.d. V: on average it was moved through ½ V.

The second and third forms come from substituting Q = CV or V = Q/C into ½ 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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) C and V are known, so use ½ CV²: W = ½ ×150×10⁻⁶ ×330² = 8.2 J (8.17 J)

(b) Power = energy ÷ time: P = 8.17 ÷1.2×10⁻³ = 6.8 × 10³ W

Check with another form: Q = CV = 0.0495 C and ½ QV = ½ ×0.0495 ×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 ½ 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 ½ QV.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Charge moved by the supply: Q = CV = 1000×10⁻⁶ ×6.0 = 6.0×10⁻³ C.

(a) Each coulomb is given 6.0 J: W = QV = 6.0×10⁻³ ×6.0 = 0.036 J

(b) W = ½ QV = 0.018 J

(c) Dissipated = 0.036 − 0.018 = 0.018 J

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. V₀ is across R. The current is at its maximum, I₀ = V₀/R. As charge builds up, V_C rises, so the p.d. across the resistor, V₀ - 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 V₀ (and the charge towards Q₀ = CV₀), while the current decays towards zero.

Discharging

Now the capacitor acts as the source. At the start, V = V₀ and the current is largest, I₀ = V₀/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 CV₀ and the final energy do not depend on R; only the timing does.

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 eˣ and ln keys on your calculator. The exponent -t/RC has no unit, so t and RC must both be in seconds.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

(a) RC = 47×10³ ×220×10⁻⁶ = 10.3 s (10.34 s)

(b) At t = 0 the full 8.0 V is across the resistor: I₀ = V₀/R = 8.0 ÷47×10³ = 1.7 × 10⁻⁴ A (0.170 mA)

(c) V = 8.0 ×e^(-15/10.34) = 8.0 ×0.234 = 1.9 V (1.88 V); then Q = CV = 220×10⁻⁶ ×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(V₀/V). Putting V = ½ V₀ gives the time to halve, t_(1/2) = RC ln 2 ≈0.69 RC. This is not in the formulae list, so derive it when you need it: put Q = ½ Q₀ into Q = Q₀ 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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Time constant: RC = 25×10³ ×0.10 = 2500 s.

(a) From ln V = ln V₀ - t/RC: t = RC ln(V₀/V) = 2500 ×ln(5.0 ÷2.0) = 2500 ×0.916 = 2.3 × 10³ s (2291 s, about 38 minutes)

(b) t_(1/2) = 2500 ×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 = ½ CV² depends on V²: W ∝e^(-2t/RC). After one time constant only e⁻² ≈14% of the energy is left, and the energy halves in half the time the p.d. takes.

While charging, the current still decays exponentially: I = I₀ e^(-t/RC) with I₀ = V₀/R. So the p.d. across the resistor is V_R = IR = V₀ e^(-t/RC). The supply p.d. is shared between the resistor and the capacitor, V_C = V₀ - V_R, so the capacitor's p.d. is whatever the resistor has not taken.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Time constant: RC = 1.0×10⁶ ×4.7×10⁻⁶ = 4.7 s.

(a) V_C = 9.0 ×(1 - e^(-3.0/4.7)) = 9.0 ×(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 ÷1.0×10⁶ = 4.8 × 10⁻⁶ A

(b) When V_C = 6.0 V, V_R = 3.0 V, so e^(-t/RC) = 3.0 ÷9.0: t = RC ln(9.0 ÷3.0) = 4.7 ×ln 3 = 5.2 s (5.16 s)

To make the delay longer, increase R or C: the delay is proportional to RC.

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 V₀ - 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 V₀.

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.

WORKED EXAMPLE · QUESTION

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.

SOLUTION

Gradient from the large triangle, from t = 1.0 s to t = 13.0 s: (-0.98 - 1.58) ÷12.0 = -2.56 ÷12.0 = -0.213 s⁻¹.

Gradient = -1/RC, so RC = 1 ÷0.213 = 4.69 s.

C = 4.69 ÷10.0×10³ = 4.7 × 10⁻⁴ F (469 µF)

C = 1/(|gradient| ×R), so the percentage uncertainties add: 3% + 2% = 5%, and ΔC = 0.05 ×469 µF ≈ ± 23 µF

The intercept, 1.79, gives V₀ = e^(1.79) ≈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(V₀ - V_C) against t; it also has gradient -1/RC.

Current and charge (Unit 2)

I = ΔQ/Δt, so the area under a current–time graph for a charging or discharging capacitor equals the charge that flowed, Q₀. The gradient of a Q–t graph is the current.

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).

1. A 100 µF capacitor stores a charge of 1.5 mC. Calculate the p.d. across it.

2. The p.d. across a capacitor is tripled. By what factor does the energy stored change?

3. Calculate the time constant of a 2.2 µF capacitor discharging through a 330 kΩ resistor.

4. What percentage of its initial charge does a discharging capacitor keep after three time constants?

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.

6. Explain why the current in a charging circuit decreases, even though the supply p.d. stays the same.

1. V = Q/C = 1.5×10⁻³ ÷100×10⁻⁶ = 15 V.

2. W = ½ CV², so the energy increases by a factor of 3² = 9.

3. RC = 330×10³ ×2.2×10⁻⁶ = 0.73 s (0.726 s).

4. Q/Q₀ = e⁻³ = 0.050, so 5.0%.

5. RC = 1 ÷0.050 = 20 s, so C = 20 ÷20×10³ = 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. 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, V₀ - V_C, falls. The current I = (V₀ - V_C)/R therefore falls, so charge arrives more slowly, and the current tends to zero as V_C approaches V₀.