The key idea
A compact topic with two halves: the algebra of storing charge (and the ½ in the energy formula), and the exponential decay of discharge — A2's gentlest introduction to the mathematics that nuclear physics will reuse.
Understand it · 01
Treat Q as the magnitude on one plate
A capacitor has two conductors separated by an insulating region. Charging transfers electrons from one conductor to the other, leaving equal and opposite charges when the initially neutral system is isolated overall. In C = Q/V, Q is the magnitude of charge on either plate, not the sum of the two magnitudes. Capacitance is measured in farads, with practical values often expressed in microfarads or nanofarads. A larger capacitance stores more charge at the same voltage.
As charge accumulates, the existing field makes further separation require more energy. For an ideal linear capacitor, charge is proportional to voltage, so a Q–V graph is a straight line through the origin. Its gradient is capacitance. This equation describes the capacitor’s state and is distinct from the rate equation I = ΔQ/Δt. A capacitor can hold a nonzero charge with no steady current flowing through its dielectric.
Understand it · 02
Use the area under the voltage–charge graph
Moving a small additional charge onto a capacitor requires work equal to voltage multiplied by that small charge. Because voltage rises during charging, the total stored energy is the area under a V–Q graph, not simply the final voltage multiplied by all the charge. For a straight-line relation, the triangular area is ½QV, equivalently ½CV² or Q²/(2C). Be clear which quantity is held constant when comparing two capacitors.
If an uncharged ideal capacitor is charged from an ideal fixed-voltage source through a resistor, the source supplies QV while the capacitor stores ½QV. The remaining half is dissipated in the resistance during charging. This result does not mean that every possible charging method must lose half its energy; it follows from the stated fixed-voltage circuit model. Capacitors can retain energy after the supply is disconnected, so practical circuits need appropriate safe discharge procedures.
Understand it · 03
Recognise the time constant on a graph
During discharge through a fixed resistor, the decreasing capacitor voltage produces a decreasing current. Charge and voltage follow exponential decay with time constant τ = RC. After one time constant, either is 1/e, about 37%, of its initial value; after equal time intervals the same fraction remains, rather than the same amount being lost. During charging from zero through a resistor, capacitor voltage reaches about 63% of the supply after one time constant.
To measure a time constant, record capacitor voltage at suitable intervals or use a data logger. A plot of ln V against time should be straight during an ideal discharge, with gradient −1/(RC). Account for the voltmeter’s finite resistance if it is comparable with the discharge resistor, since it provides an additional parallel path. Choose sampling intervals short enough to resolve the initial change, and distinguish residual instrument offset from the ideal curve approaching zero.
Another view of the physics
One time constant leaves 37% of the initial voltage
Key definitions
- Capacitance
- The charge stored per unit potential difference: C = Q ÷ V. Units: farads (F).
- Time constant (τ)
- τ = RC: the time for the charge (or current, or p.d.) of a discharging capacitor to fall to 1/e (about 37%) of its initial value.
Equations, units & conditions
Capacitance
C = Q ÷ V
Units: F
Energy stored
W = ½QV = ½CV²
Units: J
Parallel combination — extension
C = C₁ + C₂
Series combination — extension
1/C = 1/C₁ + 1/C₂
Discharge
Q = Q₀ e^(−t/RC)
Worked examples, step by step
Example 1
A 470 μF capacitor charged to 12 V discharges through a 10 kΩ resistor. Find the time constant and the p.d. after 7.0 s.
Follow the method
- τ = RC = 10,000 × 470 × 10⁻⁶ = 4.7 s.
- V = V₀e^(−t/τ) = 12 × e^(−7.0/4.7) = 12 × e^(−1.49).
τ = 4.7 s; V ≈ 2.7 V
Energy does not decay at the same rate as voltage
A 220 μF capacitor is charged to 10.0 V. Find its initial stored energy and the stored energy when its voltage has fallen to 5.0 V.
Follow the method
- Convert capacitance: 220 μF = 220 × 10⁻⁶ F.
- Initial energy = ½CV² = ½ × 220 × 10⁻⁶ × 10.0² = 0.011 J.
- At 5.0 V, energy = ½ × 220 × 10⁻⁶ × 5.0² = 0.00275 J.
- Halving voltage leaves one quarter of the energy because energy is proportional to V².
Initial energy = 11 mJ; final energy = 2.75 mJ (one quarter of the initial energy).
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Find charge and stored energy
An ideal 330 μF capacitor is connected across 9.0 V. Calculate the charge magnitude on either plate and the energy stored.
Show solution and self-check: question 1
- C = 330 × 10⁻⁶ F.
- Q = CV = 330 × 10⁻⁶ × 9.0 = 2.97 × 10⁻³ C.
- W = ½CV² = ½ × 330 × 10⁻⁶ × 9.0² = 0.013365 J.
Charge magnitude ≈ 3.0 mC on either plate; stored energy ≈ 13 mJ.
Self-check · one mark per point
- Converts μF to F correctly.
- Finds charge using Q = CV.
- Finds energy with the factor ½ and the squared voltage.
Build · 3 suggested marks
2. Stored charge and energy
A 220 μF capacitor is charged to 9.0 V. Find its charge and stored energy.
Show solution and self-check: question 2
- C = 220 × 10⁻⁶ F.
- Q = CV = 220 × 10⁻⁶ × 9.0 = 1.98 × 10⁻³ C.
- Energy = ½CV² = 0.5 × 220 × 10⁻⁶ × 9.0².
1.98 mC and 8.91 mJ.
Self-check · one mark per point
- Convert microfarads to farads.
- Use Q = CV.
- Use ½CV² to obtain 0.00891 J.
Build · 3 suggested marks
3. A discharge after two time constants
A 100 μF capacitor initially at 12 V discharges through a 20 kΩ resistor. Find the time constant and its voltage after 4.0 s.
Show solution and self-check: question 3
- τ = RC = 20 × 10³ × 100 × 10⁻⁶ = 2.0 s.
- V = V₀ exp(−t/τ) = 12 exp(−4.0/2.0).
Time constant 2.0 s; voltage 1.62 V.
Self-check · one mark per point
- Convert resistance and capacitance to SI units.
- Calculate RC = 2.0 s.
- Evaluate the exponential decay.
Build · 4 suggested marks
4. Use a time constant twice
A 100 μF capacitor initially at 8.0 V discharges through 20 kΩ. Find the time constant, the initial discharge-current magnitude and the voltage after 4.0 s.
Show solution and self-check: question 4
- R = 20,000 Ω and C = 100 × 10⁻⁶ F.
- τ = RC = 2.0 s.
- Initial current magnitude = V₀/R = 8.0/20,000 = 4.0 × 10⁻⁴ A.
- At 4.0 s, t/τ = 2, so V = 8.0e⁻² = 1.083 V.
τ = 2.0 s; initial current = 0.40 mA; voltage after 4.0 s ≈ 1.1 V.
Self-check · one mark per point
- Converts kΩ and μF into SI units.
- Finds time constant 2.0 s.
- Finds initial current magnitude 0.40 mA.
- Uses exponential decay to obtain about 1.1 V.
Stretch · 4 suggested marks
5. Find the time to a threshold voltage
A 220 μF capacitor discharges through 15 kΩ from 12 V. Calculate the time for its voltage to fall to 3.0 V and the fraction of its initial stored energy then remaining.
Show solution and self-check: question 5
- τ = RC = 15,000 × 220 × 10⁻⁶ = 3.3 s.
- V/V₀ = exp(−t/τ), so t = τ ln(V₀/V) = 3.3 ln(12/3.0).
- t = 4.575 s, or about 4.6 s.
- W/W₀ = (V/V₀)² = (3.0/12)² = 1/16 = 0.0625.
Time ≈ 4.6 s; energy fraction = 1/16 = 6.25%.
Self-check · one mark per point
- Finds time constant 3.3 s.
- Rearranges using ln(V₀/V).
- Finds time about 4.6 s.
- Squares the voltage ratio to obtain 6.25%.
Stretch · 4 suggested marks
6. Extract capacitance from a straight-line graph
For discharge through a 47 kΩ resistor, a graph of ln(V/V₀) against time in seconds has gradient −0.50 s⁻¹. Calculate the capacitance. State one experimental effect that could make the inferred value unreliable.
Show solution and self-check: question 6
- Taking logs gives ln(V/V₀) = −t/(RC). The gradient is −1/(RC).
- Time constant = −1/gradient = 2.0 s.
- C = τ/R = 2.0/47,000 = 4.255 × 10⁻⁵ F = 42.6 μF.
- Leakage or appreciable current through the voltmeter adds a discharge path, so the effective resistance is smaller than the assumed 47 kΩ. Resistor tolerance or temperature change can also affect the result.
Capacitance ≈ 43 μF. An extra discharge path would invalidate the assumed resistance.
Self-check · one mark per point
- Identifies gradient as −1/(RC).
- Finds time constant 2.0 s.
- Finds capacitance about 43 μF.
- Names and explains a relevant experimental limitation.
Common mistakes & how to avoid them
- Forgetting the ½ in the energy formula. W = QV is the energy to move charge Q through a FIXED p.d.; a capacitor's p.d. grows as it charges, so the stored energy is the area under a voltage-against-charge line: ½QV.
- Using resistor rules for capacitors. They are opposite: capacitors in parallel ADD; in series the reciprocals add. Check your answer's size — series combinations are smaller than the smallest capacitor.
- Misreading the time constant as 'time to fully discharge'. Exponential decay never finishes; τ is the time to fall to 37%, and 5τ is the practical 'almost empty' rule of thumb.
- Logarithm errors when solving for t: from Q = Q₀e^(−t/RC), t = RC ln(Q₀/Q) — the ratio inside the log is initial over final.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why is the energy stored ½QV and not QV?
A capacitor’s voltage rises as charge is added, so each successive small charge transfer requires more work. For a linear capacitor, the average voltage during charging from zero is V/2, giving stored energy ½QV.
The result is the triangular area under a voltage–charge graph, not the rectangle formed by final voltage times total charge. The area under a charge–voltage graph happens to have the same units, but the work argument specifically sums VΔQ. Substituting Q = CV gives ½CV² and Q²/(2C). These equivalent forms are useful for different fixed conditions: changing capacitance at constant charge is not the same comparison as changing it at constant voltage. Real dielectric losses or nonlinear behaviour require more detailed energy accounting than the ideal linear expression.
Energy at fixed voltage
A 330 μF capacitor is charged to 6.0 V. Find its stored energy.
- Convert capacitance to 330 × 10⁻⁶ F.
- W = ½CV² = 0.5 × 330 × 10⁻⁶ × 6.0².
5.94 × 10⁻³ J.
What does the time constant tell you?
The time constant RC determines the pace of an ideal resistor–capacitor response. After one RC of discharge, charge and voltage retain about 37% of their initial values; during charging from zero, voltage reaches about 63% of its final value.
It is not the time to become completely empty or completely full. An exponential approaches its limit progressively, with the same fractional change over equal intervals. Increasing resistance reduces current for a given voltage, while increasing capacitance requires more charge transfer for a given voltage change; both slow the response. Ensure resistance is in ohms and capacitance in farads before multiplying. A voltmeter or leakage path can change the effective discharge resistance, so the measured time constant may differ from the value calculated using only the labelled external resistor.
Why do capacitor combination rules look inverted compared to resistors?
As a circuit extension, parallel capacitances add because they share voltage and their charges add. In series, equal charge magnitudes accompany voltages that add, giving a reciprocal sum for the equivalent capacitance.
These rules can be derived directly from C = Q/V rather than memorised as the opposite of resistor rules. For parallel capacitors, Qtotal = C₁V + C₂V, so Ctotal = C₁ + C₂. For an initially uncharged ideal series arrangement, Vtotal = Q/C₁ + Q/C₂, giving 1/Ctotal = 1/C₁ + 1/C₂. The series voltage division depends on capacitance, so each capacitor need not have the same voltage. The combination formulas are useful additional circuit reasoning, but are not separately listed as core equations in the cited Edexcel IAL Unit 4 specification.
How do I find RC from a logarithmic discharge graph?
For discharge, ln V = ln V₀ − t/(RC). A graph of ln V against time therefore has gradient −1/(RC), allowing the time constant to be found as the negative reciprocal of the gradient.
Use natural logarithms and keep voltage units consistent across the data. The logarithm is applied to the numerical ratio to a fixed reference unit; changing that unit shifts the intercept but leaves the slope unchanged. Readings near zero become strongly affected by offsets and relative uncertainty, so they can distort the fitted line. Check that the circuit is discharging through an effectively constant resistance and that the instrument does not significantly load it. A straight line supports exponential behaviour, while a curved trend may reveal leakage, changing resistance or a measurement offset.
Time constant from log voltage
A graph of ln(V/1 V) against time has gradient −0.25 s⁻¹. The resistor is 50 kΩ. Find RC and capacitance.
- RC = −1/gradient = 4.0 s.
- C = 4.0/(50 × 10³) = 8.0 × 10⁻⁵ F.
Time constant 4.0 s; capacitance 80 μF.
Which plate’s charge is Q in the capacitance equation?
In C = Q/V, Q is the magnitude of charge on one capacitor plate. The two plates carry equal and opposite charges in the simple initially neutral system, so adding their magnitudes would incorrectly double Q.
Charging separates charge rather than creating it. Electrons transferred from one plate to the other leave one side positive and the other equally negative. The net charge of the complete isolated capacitor can remain zero while it stores field energy. Current in the external circuit measures how rapidly this separation changes; it does not require electrons to cross the ideal insulating dielectric. If the capacitor is already charged or connected in a more complicated network, keep track of initial charges and conservation on isolated conductors rather than applying a neutral starting condition automatically.
Why can a voltmeter change the measured capacitor discharge time?
A voltmeter connected across a discharging capacitor provides an additional parallel path for current. If its resistance is not much larger than the intended discharge resistor, the effective resistance decreases and the measured discharge becomes faster.
The relevant resistance in RC is the resistance seen by the capacitor in the actual circuit. Combine the voltmeter input resistance with the external resistor using the parallel rule. A high-input-resistance instrument makes the correction smaller, but leakage within the capacitor can still matter for long time constants. Record the instrument settings because some meters have different input behaviour on different ranges. If the observed decay is faster than predicted, investigate these extra paths before assuming the capacitance label is wrong. Connecting the measuring device can itself change the process being measured.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define capacitance C = Q/V for capacitors and isolated conductors
- Derive and use the energy stored: W = ½QV = ½CV²
- Interpret capacitor charging and discharging curves; combination rules below are an additional circuit extension
- Analyse discharge through a resistor: exponential decay, time constant τ = RC
- Use x = x₀e^(−t/RC) for charge, current and voltage
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Capacitance. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/capacitance
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