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Notes/Physics/Paper 4/Capacitance
CAIEA Level9702§19.1–19.3

Capacitance

What a capacitor is, C = Q/V, the isolated sphere, series and parallel networks, the energy stored, exponential discharge through a resistor, the time constant, and capacitors at work.

130 min read 7 sub-topics
118
question parts
2021–2025 · 34 papers
7 marks
per paper
≈ 7% of the paper
2.2/3
avg difficulty
moderate
#9
most examined
of 16 topics by marks

A defibrillator has a fraction of a second to deliver hundreds of joules to a chest; a camera flash needs a burst of current far larger than a battery can supply at once; a power supply must hand out steady voltage from a wobbling input. All three solve the problem the same way: bank it up slowly, pay it out fast. The component that does the banking is the capacitor — two conductors separated by an insulator, a device that stores charge, and therefore energy, in the electric field between its plates. This note defines capacitance, prices the energy stored, works out how capacitor networks combine, and then tackles the part of the topic that earns it its fearsome reputation: the exponential decay of a capacitor discharging through a resistor.

The bank agrees the topic is worth the effort: 253 marks across 2021–2025 put Capacitance 9th of the 16 A2 topics, carried by 118 question parts in 34 paper sittings — roughly 7.4 marks on every Paper 4. The yearly haul climbs, 44, 48, 36, 57, 68, with a record 68 in 2025. The number to take seriously is the mean difficulty of 2.19 — the 2nd-highest of all 16 A2 topics (only Thermodynamics, at 2.25, is harder). The difficulty is not in the definitions, which the schemes reprint word-for-word year after year; it lives in the discharge family, where rearranging x=x0e−t/RCx = x_0 e^{-t/RC} with natural logs separates the prepared from the hopeful. That family carries 103 marks in the window — the heaviest single strand — which is why §05 and §06 give it two full sections.

The route through is: §01 what a capacitor is, the word-perfect definition, and the µF conversion drill — §02 the isolated spherical conductor and C=4πε0RC = 4\pi\varepsilon_0 R — §03 series and parallel combinations, derived from C=Q/VC = Q/V as the syllabus demands, plus charge-sharing — §04 the energy stored and the area under the V–Q graph — §05 why discharge is exponential at all, and the time constant τ=RC\tau = RC — §06 the ln machinery: solving for tt, reading ln-graphs, and the gradient tricks examiners love — and §07 capacitors in action: smoothing, flash energy, average current, and the examiner toolkit.

Before you start you should be able to

  • Handle p.d. and charge from AS: p.d. is work done per unit charge, charge flows driven by p.d., and V=IRV = IR for a resistor

  • Combine resistors in series and parallel — because capacitor networks obey the mirror-image rules, and the contrast only makes sense against the resistor versions

  • Use electric potential of a point charge V=Q/(4πε0r)V = Q/(4\pi\varepsilon_0 r) from the A2 Electric Fields note — it builds the isolated-sphere capacitance in §02

  • Rearrange exponentials with natural logarithms: ln⁡(ea)=a\ln(e^a) = a, ln⁡\ln of a product as a sum, and solving a=ebta = e^{bt} for tt — the entire engine of §06

  • Convert prefixes without ceremony: pico (10−1210^{-12}), nano (10−910^{-9}), micro (10−610^{-6}), milli (10−310^{-3}), kilo (10310^{3}) — and multiply powers of ten in standard form

  • Read values off graphs honestly: intercepts, gradients over large triangles, and pairs of coordinates inside one smooth segment

By the end of this page you can

  • Define capacitance as applied to BOTH isolated spherical conductors and parallel-plate capacitors; recall and use C = Q/V

  • Derive, using C = Q/V, the formulae for combined capacitance in series and in parallel, and use those formulae for capacitor networks

  • Determine the electric potential energy stored in a capacitor from the area under the potential–charge graph; recall and use W = ½QV = ½CV²

  • Analyse graphs of the variation with time of p.d., charge and current for a capacitor discharging through a resistor

  • Recall and use τ = RC for the time constant, and use equations of the form x=x0 e−t/RCx = x_0\,e^{-t/RC} where x is current, charge or p.d.

  • Recall standard applications of capacitors, including smoothing and the storage of energy — cross-listed with the Alternating Currents note, where rectification and smoothing are analysed in full

01

What a capacitor is; capacitance defined

Syllabus requirement · §19.1

“

define capacitance, as applied to both isolated spherical conductors and to parallel plate capacitors; recall and use C = Q / V

”

A component that banks charge

A battery is a steady but feeble pump: it delivers energy at a rate its chemistry can sustain, no faster. Plenty of real jobs need the opposite — a slow trickle saved up, then paid out in one violent burst. A defibrillator charges for several seconds and then dumps hundreds of joules through the chest in milliseconds. A camera flash fires a current spike no battery could deliver directly. A touch sensor detects the tiny capacitor formed by your fingertip and the screen. And in almost every mains-powered device, a capacitor smooths a wobbly rectified voltage into something steady (§07).

The component doing all of this is the capacitor: two conductors separated by an insulator (air, paper, plastic, or just a vacuum — the insulator itself is not on your syllabus; only its job of keeping the charges apart matters). Nothing crosses the gap. What a capacitor stores is separated charge, and with it energy in the electric field between the plates.

Building the parallel-plate picture

Connect a battery across two parallel metal plates. The battery does not create charge — it pumps it. Electrons are pushed onto one plate, making it negative, and pulled off the other, leaving it positive. The pump runs until the p.d. between the plates equals the battery's e.m.f., at which point the flow stops and the capacitor is fully charged: plate one carries −Q-Q, plate two carries +Q+Q, and the p.d. between them is VV.

Two details here are worth their weight in marks:

  • The charge stored QQ means the charge on ONE plate — not the two added together. The plates carry equal and opposite charges +Q+Q and −Q-Q, and adding them gives zero, which would be a useless definition of "stored". What is stored is the separation: the work the battery did moving charge from one plate to the other.
  • The p.d. VV is measured between the plates — the potential of one plate minus that of the other.
+−Eelectron flow+Q−Qchargeseparation

Charging a parallel-plate capacitor. The battery pumps electrons onto one plate and pulls them off the other, leaving charges −Q and +Q. The stored charge Q is the magnitude on ONE plate; V is the p.d. between the plates; a uniform electric field fills the gap.

The definition, word-perfect

Different capacitors hold different amounts of charge at the same p.d., just as different buckets hold different volumes of water at the same level. The property that measures the bucket size is the capacitance CC, defined by the syllabus for both parallel-plate capacitors and isolated spherical conductors:

capacitance = charge / potential difference (M1) — where the charge is the charge on one plate, and the potential is the p.d. between the plates (A1).

Both clauses earn marks separately. Writing only "charge per unit voltage" scores the first mark and quietly loses the second, because it never says which charge or which p.d. The scheme's own model answer (O/N 2024) is exactly: charge is the charge on one plate, and the p.d. is between the plates. Rearranged into its working form:

C=QVC = \frac{Q}{V}

The unit is the farad (F): a capacitor of 1 farad stores 1 coulomb at a p.d. of 1 volt. One farad is enormous — a single farad capacitor is the size of a drink can. Real capacitors are rated in microfarads (µF, 10−610^{-6} F), nanofarads (nF, 10−910^{-9} F) or picofarads (pF, 10−1210^{-12} F).

C=QVC = \frac{Q}{V}

Capacitance = charge on ONE plate ÷ p.d. BETWEEN the plates. Unit: farad, F = C V⁻¹. Practical capacitors: µF (×10⁻⁶), nF (×10⁻⁹), pF (×10⁻¹²).

The prefix conversion, done the same way every time
  1. 1

    Spot the prefix and write its power of ten: µ → ×10⁻⁶, n → ×10⁻⁹, p → ×10⁻¹².

    The single most-tagged technique in this topic (a dozen-plus scheme tags in the 2021–2025 window) — and the most common silent killer, because the arithmetic that follows a missed conversion still produces a plausible-looking number.

  2. 2

    Replace the prefixed value in the formula with value × power of ten, e.g. 470 µF → 470 × 10⁻⁶ F.

    Write the substitution with the power of ten visible. A bare '470' in a formula is where wrong units hide.

  3. 3

    Multiply the number parts and ADD the exponents at the end, quoting the answer in standard form.

    Prefactors and exponents handled separately never go wrong: 24 × 470 = 11 280, then ×10⁻⁶, then round to 2–3 sf.

Invented demo — one capacitor, three questions

A 100 μF100\ \mu\text{F} capacitor is charged until the p.d. between its plates is 12 V12\ \text{V}.
(i) Calculate the charge on one plate.
(ii) A student says the capacitor must now store 2.4 mC2.4\ \text{mC}, "because both plates are charged". Explain the error.
(iii) Express the capacitance in farads and in nanofarads.

Show full working
  1. 1

    (i) Name the relation and convert first: Q=CVQ = CV with C=100 μF=100×10−6 FC = 100\ \mu\text{F} = 100\times10^{-6}\ \text{F} and V=12 VV = 12\ \text{V}.

    Definition rearranged for charge; the µF becomes ×10⁻⁶ BEFORE anything is multiplied — the conversion habit from the method box above.

  2. 2

    Substitute and simplify:

    Q=(100×10−6)×12=1.2×10−3 C=1.2 mCQ = (100\times10^{-6}) \times 12 = 1.2\times10^{-3}\ \text{C} = 1.2\ \text{mC}

    Prefactor 100 × 12 = 1200, then ×10⁻⁶. Quoting the mJ-style shorthand (1.2 mC) keeps the size of the answer visible.

  3. 3

    (ii) The plates carry charges +1.2 mC+1.2\ \text{mC} and −1.2 mC-1.2\ \text{mC}. Their sum is zero — adding them measures nothing stored. The stored charge QQ is defined as the charge on one plate: the separated charge the battery pumped across.

    The 'both plates' error doubles every charge in the topic. What is stored is separation, not net charge — the capacitor as a whole is still neutral.

  4. 4

    (iii) 100 μF=100×10−6 F=1.0×10−4 F100\ \mu\text{F} = 100\times10^{-6}\ \text{F} = 1.0\times10^{-4}\ \text{F}.
    Going down to nano means multiplying by 10310^{3}: 1.0×10−4 F=1.0×105 nF1.0\times10^{-4}\ \text{F} = 1.0\times10^{5}\ \text{nF}.

    µ→F shifts six places down; F→nF shifts nine places up, i.e. three up from µ. Prefix ladders are climbed in steps of 10³.

Answer

(i) Q = CV = 1.2×10⁻³ C. (ii) Q is defined as the charge on ONE plate — the plates carry +Q and −Q, which sum to zero; the stored quantity is the separated charge. (iii) 1.0×10⁻⁴ F = 1.0×10⁵ nF.

Whenever a question says 'the charge on the capacitor', it means the magnitude on one plate. Every formula in this note uses that Q.

Charging a real capacitor on a real circuit

9702/41 O/N 2022 Q5(a)(i)2 marks

A capacitor of capacitance 470 μF470\ \mu\text{F} is connected to a battery of electromotive force (e.m.f.) 24 V24\ \text{V} in the circuit shown. The two-way switch S is initially at position X, so the capacitor is fully charged.
Calculate the charge Q0Q_0 on the capacitor at time t=0t = 0.

Fig. 5.1 from the question paper: a 24 V battery and a 470 µF capacitor connected through two-way switch S. Position X charges the capacitor; position Y connects it to two long wires P and Q (each 5.6 kΩ), with Q wired to a voltmeter.

Fig. 5.1 from the question paper: a 24 V battery and a 470 µF capacitor connected through two-way switch S. Position X charges the capacitor; position Y connects it to two long wires P and Q (each 5.6 kΩ), with Q wired to a voltmeter.

Show full working
  1. 1

    Name the relation: the defining equation solved for charge,

    Q=CVQ = CV

    Quoting the definition first keeps the substitution mark safe even if the arithmetic slips — and names which idea does the work.

  2. 2

    Identify the pieces with units converted: C=470 μF=470×10−6 FC = 470\ \mu\text{F} = 470\times10^{-6}\ \text{F} and, because the capacitor is fully charged across the battery, V=24 VV = 24\ \text{V}.

    'Fully charged' means the plate p.d. has risen to equal the e.m.f. — that is the licence to use 24 V here. The µF conversion is written out, never done in the head.

  3. 3

    Substitute and evaluate:

    Q0=24×470×10−6=1.128×10−2 CQ_0 = 24 \times 470\times10^{-6} = 1.128\times10^{-2}\ \text{C} Q0=0.011 C(2 s.f.)Q_0 = 0.011\ \text{C}\quad\text{(2 s.f.)}

    24 × 470 = 11 280, then ×10⁻⁶. Two significant figures match the data — and match the scheme's own answer, 0.011 C.

Answer

Q₀ = CV = 24 × 470×10⁻⁶ F × V = 0.011 C.

Later parts of this same question discharge the capacitor through the wires and time it — this Q₀ is the x₀ that starts every exponential in §05.

Common mistakes

  • Doubling Q "because both plates are charged": Q = 2CV.

    Q is the charge on ONE plate; the plates carry +Q and −Q.

    The definition's first clause names the charge on one plate. Doubling it poisons C = Q/V, every energy formula and every discharge calculation downstream.

  • Substituting 470 for a 470 µF capacitor, as if the unit were farads.

    470 µF = 470 × 10⁻⁶ F, converted before substituting.

    The number that survives a missed conversion still looks plausible — which is exactly why schemes tag the conversion separately and refuse the final mark without it.

  • Defining capacitance as "charge per unit voltage".

    Charge ÷ p.d., where the charge is on one plate and the p.d. is between the plates.

    The M1/A1 split in the scheme pays one mark for the ratio and a separate one for the one-plate/between-plates precision. Loose wording strands the second mark.

Your turn

One routine Q = CV drill, one conversion workout, and the real two-mark definition from O/N 2024.

  1. 13 marks

    A 470 nF470\ \text{nF} capacitor is charged to a p.d. of 9.0 V9.0\ \text{V}.
    (i) Calculate the charge on one of its plates, giving your answer in µC.
    (ii) The p.d. is doubled to 18 V18\ \text{V}. State the new charge, without recomputing from scratch.

    Stuck? Show hint

    Convert nF first; then for (ii) ask how Q responds when V doubles at fixed C.

    Show solution
    1. 1

      (i) Convert: 470 nF=470×10−9 F470\ \text{nF} = 470\times10^{-9}\ \text{F}.

      nano → ×10⁻⁹, written before the multiplication.

    2. 2

      Apply Q = CV:

      Q=(470×10−9)×9.0=4.23×10−6 C≈4.2 μCQ = (470\times10^{-9}) \times 9.0 = 4.23\times10^{-6}\ \text{C} \approx 4.2\ \mu\text{C}

      470 × 9.0 = 4230, then ×10⁻⁹. Converting the exponent to µC at the end (×10⁶) keeps the answer at a readable size.

    3. 3

      (ii) Q=CVQ = CV with CC unchanged: Q∝VQ \propto V, so doubling VV doubles QQ:

      Q=2×4.2=8.4 μCQ = 2 \times 4.2 = 8.4\ \mu\text{C}

      Proportional reasoning finishes in one line — and shows understanding, which 'I typed it into the calculator again' does not.

    Answer

    (i) 4.2 µC. (ii) 8.4 µC — Q ∝ V at fixed C, so it doubles.

  2. 24 marks

    (i) Express in farads: 22 μF22\ \mu\text{F}, 100 nF100\ \text{nF}, 4.7 pF4.7\ \text{pF}.
    (ii) A 10 μF10\ \mu\text{F} capacitor carries a charge of 50 μC50\ \mu\text{C}. Calculate the p.d. across it.

    Stuck? Show hint

    For (ii), the definition rearranged: V = Q/C — with both values in farads and coulombs.

    Show solution
    1. 1

      (i) 22 μF=2.2×10−5 F22\ \mu\text{F} = 2.2\times10^{-5}\ \text{F}; 100 nF=1.0×10−7 F\quad 100\ \text{nF} = 1.0\times10^{-7}\ \text{F}; 4.7 pF=4.7×10−12 F\quad 4.7\ \text{pF} = 4.7\times10^{-12}\ \text{F}.

      Three prefixes, three powers of ten: −6, −9, −12. This ladder is worth memorising outright.

    2. 2

      (ii) Rearrange the definition: V=Q/CV = Q/C with Q=50×10−6 CQ = 50\times10^{-6}\ \text{C} and C=10×10−6 FC = 10\times10^{-6}\ \text{F}:

      Both values are prefixed, so both get converted — a µ over a µ would have cancelled, but only by luck; convert by habit, not by luck.

    3. 3
      V=50×10−610×10−6=5.0 VV = \frac{50\times10^{-6}}{10\times10^{-6}} = 5.0\ \text{V}

      The powers of ten cancel exactly, leaving 50/10. A clean answer like this is a sign the conversions were right.

    Answer

    (i) 2.2×10⁻⁵ F; 1.0×10⁻⁷ F; 4.7×10⁻¹² F. (ii) V = Q/C = 5.0 V.

  3. 39702/42 O/N 2024 Q7(a)2 marks

    Define the capacitance of a parallel-plate capacitor.

    Stuck? Show hint

    Two clauses, two marks: the ratio, then the precision about which charge and which p.d.

    Show solution
    1. 1

      Clause 1: capacitance = charge ÷ potential difference (M1).

      The ratio itself is the first mark — charge per unit p.d.

    2. 2

      Clause 2: the charge is the charge on one plate, and the potential is the p.d. between the plates (A1).

      The scheme's model answer is exactly this: 'charge is charge on one plate, and potential is p.d. between the plates'. The precision is the second mark.

    Answer

    Capacitance = charge ÷ p.d., where the charge is the charge on one plate and the p.d. is between the plates.

Practise C = Q/V definition and charge calculationsReal past-paper questions · Capacitance C = Q/V for conductors and capacitors

The rest of this note

Checking your access…

Can you do all of these?

  • Define capacitance word-perfectly: charge ÷ p.d., where the charge is on ONE plate and the p.d. is BETWEEN the plates — both clauses are marked separately

  • Convert µF/nF/pF to farads BEFORE any substitution — the most-tagged technique in the topic

  • Derive the series and parallel rules from C = Q/V (same charge + p.d.s add; same p.d. + charges add), not just quote them

  • Remember the combination rules are the OPPOSITE of the resistor rules — and that two equal capacitors in series give C/2

  • Track charge-sharing: charge is conserved, the final p.d.s are equal, and stored energy DROPS (heat in the wires)

  • Pick the right energy form: ½CV² when the p.d. is fixed, ½Q²/C when the charge is fixed; for a change use ΔE = ½C(V₂²−V₁²)

  • Explain the discharge shape as a feedback chain: V_C = V_R, I ∝ V, I is the rate charge leaves, so as Q falls the current falls

  • Derive dQ/dt ∝ −Q ⇒ exponential, and state τ = RC as 'falls to x₀/e ≈ 0.37 x₀ in one time constant'

  • Solve for time with t = −RC ln(x/x₀) — ln of the ratio, never ln of individual values first

  • Read ln x–t graphs: straight line, gradient −1/RC, intercept ln x₀; use a large triangle for the gradient

  • For ripple read-offs, take both coordinate pairs from INSIDE one discharge segment

  • Quote 'show that' answers to 3+ significant figures before comparing with the printed value

Now do the questions
118 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes
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