Lesson 12: Capacitance

Capacitors store charge — and with it, energy — and their charging and discharging follow exponential curves that mirror radioactive decay and cooling. This topic (an extension of Theme B.5 work) appears in Paper 2 with a predictable structure: combinations of capacitors, energy calculations, and time-constant analysis from graphs.

Capacitance

The capacitance of a capacitor is the charge stored per unit potential difference: C = Q/V, measured in farads (F) — 1 F is enormous, so expect μF and nF in questions. The charge on one plate is +Q and on the other −Q; the stored energy is the area under the Q–V graph: E = ½QV = ½CV² = Q²/(2C). Note the factor ½: as charge is added, the pd rises from zero, so the average pd during charging is V/2. (The battery supplying it does work QV in total; half becomes stored energy and half is dissipated in the circuit resistance — a beautiful examiner's discussion point.)

Combinations: in parallel, capacitances add (C = C₁ + C₂) because the pd is common and charges add. In series, reciprocals add (1/C = 1/C₁ + 1/C₂) because the charge on each is the same and the pd's add; the series total is smaller than the smallest capacitor.

Charging and discharging

Through a resistor R, discharge follows Q = Q₀ e^(−t/RC) and V = V₀ e^(−t/RC); charge growth during charging is Q = Q₀(1 − e^(−t/RC)). The product RC is the time constant τ: in one time constant the quantity falls to 1/e ≈ 37% of its value; after 5τ it is effectively zero (0.7%). The initial discharge current is V₀/R and falls with the same exponential.

Worked problem 1 — combinations, charge and energy

A 6.0 μF and a 3.0 μF capacitor are connected in series, and the pair in parallel with a 4.0 μF capacitor, across a 24 V supply. Find the total capacitance, total charge supplied, and total stored energy.

Series pair: 1/C = 1/6.0 + 1/3.0 = 1/2.0 → C = 2.0 μF
Total: 2.0 + 4.0 = 6.0 μF
Charge from supply: Q = CV = 6.0 μF × 24 = 144 μC
Energy: E = ½CV² = ½ × 6.0 × 10⁻⁶ × 24² = 1.7 × 10⁻³ J

Inside the series pair the charge on each is the same: Q = 2.0 μF × 24 V = 48 μC, giving pd's of 8.0 V (across 6.0 μF) and 16 V (across 3.0 μF) — the smaller capacitor takes the larger share of the voltage. Checking that 8 + 16 = 24 V is the kind of self-verification examiners reward.

Worked problem 2 — discharge

A 470 μF capacitor charged to 12 V discharges through a 2.2 kΩ resistor. Find (a) the time constant, (b) the charge remaining after 2.0 s.

(a) τ = RC = 2.2 × 10³ × 470 × 10⁻⁶ = 1.03 s
(b) Q₀ = CV = 470 × 10⁻⁶ × 12 = 5.64 mC
    Q = Q₀ e^(−t/τ) = 5.64 × e^(−2.0/1.03) = 5.64 × e^(−1.94) = 5.64 × 0.144 = 0.81 mC

Practical worked example — measuring C from a discharge graph

A capacitor discharges through a 10.0 kΩ resistor while a datalogger records V every 0.2 s. A plot of ln V against t is a straight line of gradient −0.48 ± 0.02 s⁻¹.

ln V = ln V₀ − t/(RC) → gradient = −1/(RC) → RC = 1/0.48 = 2.08 s
C = 2.08 / 10 000 = 2.08 × 10⁻⁴ F = 208 μF, uncertainty from gradient: 0.02/0.48 ≈ 4%
C = 208 ± 8 μF (manufacturer's label: 220 μF — within tolerance, and the discrepancy invites comment)

Linearising the exponential (Lesson 18) converts curve-fitting into gradient-reading — the standard, robust route in the IA as well.

Nature of Science

Exponential laws recur across science — capacitor discharge, radioactive decay, Newton's cooling, atmospheric pressure with altitude — because each describes a rate proportional to the quantity itself. Recognising the same mathematics in new clothing is one of the deepest pattern-matching skills physics teaches.

Real-world application

A camera flash charges a capacitor slowly from a small battery (low power over seconds) and discharges it through the flash tube in a millisecond — power amplification by time compression. Defibrillators do the same at life-saving scale: about 200 J stored at ~2 kV delivered in a controlled pulse through the chest.

Common mistakes (examiner style)

• Adding series capacitances directly (that rule belongs to parallel; series capacitors combine like parallel resistors).
• Dropping the ½ in stored energy, or confusing the battery's work QV with the stored ½QV.
• Using R in kΩ and C in μF and losing the powers of ten in τ = RC (work in ohms and farads; 2.2 × 10³ × 470 × 10⁻⁶ ≈ 1.0 s is a useful anchor).
• Assuming charge decays linearly — a constant "half-life" pattern instead: every τ multiplies by 0.37.
• Stating the discharge current is constant while the pd falls (I = V/R follows V down).

IB Exam Experience

Paper 1-style questions. 1. Two 4.0 μF capacitors in series have a total capacitance of: A. 2.0 μF B. 4.0 μF C. 8.0 μF D. 16 μF. Key: A — series: half of one (for equal values, C/n). 2. A capacitor discharges through a resistor. After one time constant, the charge remaining is: A. 50% B. 37% C. 63% D. 100/e² %. Key: B — e⁻¹ = 0.37. 3. A 10 μF capacitor is charged to 5.0 V. The stored energy is: A. 25 μJ B. 125 μJ C. 250 μJ D. 125 mJ. Key: B — E = ½ × 10 × 10⁻⁶ × 25 = 1.25 × 10⁻⁴ J.

Paper 2-style structured question (7 marks). A 2200 μF capacitor is charged from a 9.0 V supply and then discharged through a 33 Ω resistor. (a) Calculate the initial charge stored. [2] (b) Calculate the time constant of the discharge. [2] (c) Sketch, with axes values, the variation of pd with time during the first 0.20 s of discharge. [2] (d) Calculate the pd after 0.15 s. [1] *Marking notes:* (a) Q₀ = CV = 2200 × 10⁻⁶ × 9.0 = 0.0198 C ≈ 0.020 C. (b) τ = RC = 33 × 2200 × 10⁻⁶ = 0.0726 s ≈ 0.073 s. (c) Exponential decay from 9.0 V, passing through ≈ 3.3 V at t = τ and tending to zero; marks for correct starting value and exponential shape reaching < 1 V by 0.20 s (≈ 2.8τ). (d) V = 9.0 e^(−0.15/0.0726) = 9.0 × e^(−2.07) = 9.0 × 0.127 = 1.1 V. Common error: using 0.15/RC with RC computed in ms — unit discipline is explicitly credited.

IA idea: Measure the capacitance of an unmarked capacitor by two independent methods (discharge timing and charge integration) and evaluate which has the smaller total uncertainty.

Dielectrics: what the material between the plates does

Inserting an insulating dielectric between the plates raises the capacitance by the material's relative permittivity εr: the field polarises the material, its surface charges partially cancel the plates' field, and more charge can be stored at the same pd. Mica (εr ≈ 6) and ceramic (εr up to thousands) make compact capacitors possible. If the dielectric is inserted while the capacitor stays connected to a battery, V is fixed and Q and stored energy rise; if inserted while isolated, Q is fixed and V and stored energy fall — the energy difference is the work done by the field pulling the slab in. This connected-versus-isolated distinction is a perennial HL trap even in SL-style papers.

A two-capacitor energy audit

Charge a 4.0 μF capacitor to 12 V (stored energy ½CV² = 288 μJ, charge 48 μC), disconnect it, and connect it across an identical uncharged capacitor. Charge conservation: the total 48 μC now spreads over 8.0 μF, so the common pd is 6.0 V and the stored energy is ½ × 8.0 × 10⁻⁶ × 36 = 144 μJ — exactly half has vanished, dissipated as heat (and a spark, and electromagnetic radiation) in the connecting wires, whatever their resistance. Halving the resistance doubles the current and halves the time; the integral of the loss is the same. Few results in the course reward careful thinking about conservation laws as richly as this one.

Capacitors as timers

Because the discharge is a clock with timescale RC, capacitors time everything from camera flashes to 555-timer circuits to the delay between windscreen-wiper sweeps. Choosing R = 1.0 MΩ and C = 10 μF gives τ = 10 s — a human-scale delay from two cheap components. In the exam, any question mixing a capacitor with "time delay", "flash rate" or "smoothing" is inviting RC reasoning first and formula-hunting second.

Choosing a capacitor by its time constant and voltage rating

Practical selection runs on two numbers: the capacitance sets the timescale (with the circuit's resistance), and the working voltage must exceed the largest pd the capacitor will ever see, including transients. An electrolytic rated 16 V on a 12 V rail has a sensible margin; the same part on a rectified 12 V AC line sees peaks of 17 V and will fail early. Leakage current — a real capacitor slowly self-discharges — sets the longest usable timescale (hours, not days). Exam questions occasionally probe the rating logic verbally: exceeding the working voltage breaks down the dielectric, and the failure is destructive, unlike a resistor merely running hot.