← All notesPhysics · Capacitance⬇ Download PDF
A Level Physics 9702 · A Level · Topic 19 · ~4 min read

Capacitance

Capacitance C = Q/V, capacitors in series and parallel, the energy stored as ½QV, the capacitance of an isolated sphere, and the exponential discharge of a capacitor through a resistor with time constant RC.

Capacitance C = Q/VIsolated sphere 4πε₀rSeries & parallelEnergy W = ½QV = ½CV²Area under Q–V graphDischarge Q = Q₀e^{−t/RC}Time constant τ = RC
RSby Riyaz Sir
mentorriyaz.in
Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Capacitance C = Q/V: the charge stored per unit potential difference. Unit: farad (F) = C V⁻¹.
  • Parallel: C = C₁ + C₂ + …. Series: 1/C = 1/C₁ + 1/C₂ + …. (The opposite of resistors.)
  • Energy stored W = ½QV = ½CV²: the area under a graph of V against Q.
  • Discharge through R: Q, V and I all fall exponentially: x = x₀ e−t/RC. The time constant τ = RC is the time to fall to 1/e (37%).

By the end you can

  • Define capacitance and use C = Q/V, including for an isolated sphere C = 4πε₀r.
  • Derive and use the formulae for capacitors in series and in parallel.
  • Determine the energy stored from the area under a potential–charge graph, and use W = ½QV = ½CV².
  • Analyse discharge using x = x₀ e−t/RC and the time constant τ = RC.

1Capacitance

C = Q/V

Capacitance — the charge stored per unit potential difference: C = Q/V.

  • Parallel-plate capacitor: equal and opposite charges +Q and −Q on the plates; the net charge is zero.
  • Isolated sphere: V = Q/(4πε₀r), so C = 4πε₀r. A sphere of radius 0.10 m has C = 1.1 × 10⁻¹¹ F.
Parallel: C = C₁ + C₂ + … ; Series: 1/C = 1/C₁ + 1/C₂ + … DATA & FORMULAE
parallel share V; series share Q
Worked example

Find the total capacitance of a 2 μF and a 3 μF in series, with a 4 μF in parallel with that pair.

  1. series: 1/C = 1/2 + 1/3 ⇒ C = 1.2 μF C1
  2. parallel: C = 1.2 + 4 = 5.2 μF A1
Common mistakeUsing the resistor rules for capacitors. Capacitors in parallel add directly.

2Energy stored

W = ½QV

W = ½QV = ½CV² = Q²/(2C) DATA & FORMULAE
area under the V–Q graph (a triangle)
  • Why the ½: as charge builds up, the p.d. rises from 0 to V, so the average p.d. while charging is V/2.
Worked example

A 470 μF capacitor is charged to 12 V. Find the charge and the energy stored.

  1. Q = CV = 470 × 10⁻⁶ × 12 = 5.6 × 10⁻³ C A1
  2. W = ½CV² = ½ × 470 × 10⁻⁶ × 144 = 0.034 J A1

3Discharge

Exponential decay

tQ, V or IQ₀0.37Q₀τ = RC2τ3τQ = Q₀ e^(−t/RC)after one τ: 37% leftafter 5τ: < 1% leftgradient = currentDischarge through a resistor: the same exponential shape for Q, V and I.
Discharge: Q, V and I fall by the same fraction in each equal time interval.
x = x₀ e−t/RC ; τ = RC DATA & FORMULAE
x stands for Q, V or I
Worked example

The 470 μF capacitor (12 V) discharges through 10 kΩ. Find (a) the time constant, (b) the p.d. after 3.0 s, (c) the time for the p.d. to fall to 3.0 V.

  1. (a) τ = RC = 10 × 10³ × 470 × 10⁻⁶ = 4.7 s A1
  2. (b) V = 12e−3.0/4.7 = 6.3 V C1 A1
  3. (c) 3 = 12e−t/4.7 ⇒ t = 4.7 ln 4 = 6.5 s C1 A1
  • Finding RC from a graph: plot ln V against t; the gradient is −1/RC.
  • Charging (not needed in detail): Q rises towards CE with the same time constant.
Common mistakeTaking logs of the wrong side, or forgetting that ln(V/V₀) is negative. Rearrange to t = RC ln(V₀/V).
Linking questionCharge and current are in Electricity; potential from a point charge is in Electric fields; and exponential decay reappears in Nuclear physics. Smoothing capacitors are in Alternating currents.

Test yourself — cover the answers first

Quick check

  1. Define the farad.
  2. Find C for 3 μF and 6 μF in series.
  3. What fraction of the charge remains after one time constant?
  4. A capacitor's p.d. doubles. What happens to the energy stored?
  5. What does the area under a V–Q graph represent?

Exam-style

  1. A 2200 μF capacitor at 9.0 V discharges through a lamp of 50 Ω. Find the initial current and the charge left after 0.20 s.
  2. Find the time for the energy stored to halve during discharge.
  3. Three 6 μF capacitors are in series across 12 V. Find the charge on each and the total energy.
Show answers
  1. one coulomb per volt
  2. 2 μF
  3. 1/e ≈ 0.37
  4. it quadruples
  5. energy stored
  6. I₀ = 0.18 A; Q = 0.0198 e−0.20/0.11 = 3.2 × 10⁻³ C
  7. W ∝ V² ∝ e−2t/RC: t = (RC/2) ln 2 = 0.35RC
  8. C = 2 μF; Q = 24 μC each; W = ½ × 2 × 10⁻⁶ × 144 = 1.44 × 10⁻⁴ J
Practise this chapter
A Level practice papers
Next chapterMagnetic fields →
Stuck on a question?

Send it to me on WhatsApp. I read every message.

Want regular help? Send this to your parent →
Share with a friendAsk Riyaz Sir

Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.