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How does a capacitor store charge and energy, and how do charge and current decay when it discharges?

Capacitance: the definition of capacitance, energy stored on a capacitor, capacitors in series and parallel, and the exponential charge and discharge through a resistor with time constant RC.

A focused answer to the Eduqas A-Level Physics Component 2 capacitance content, covering the definition of capacitance, the energy stored on a capacitor, combining capacitors in series and parallel, and the exponential charge and discharge through a resistor with the time constant RC.

Generated by Claude Opus 4.813 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Eduqas wants you to define capacitance, calculate the energy stored on a charged capacitor, combine capacitors in series and parallel, and analyse the exponential charge and discharge of a capacitor through a resistor using the time constant RCRC.

The answer

The definition of capacitance

Energy stored

Capacitors in series and parallel

Exponential charge and discharge

Examples in context

Capacitors smooth the output of rectified power supplies, store energy for camera flashes and defibrillators, and set the timing of oscillators and timers through their RCRC time constant. The exponential discharge is the same mathematics as radioactive decay, so the technique of taking logarithms to find a decay constant transfers directly. Touchscreens and many sensors detect tiny changes in capacitance.

Try this

Q1. Define capacitance and state its unit. [2 marks]

  • Cue. The charge stored per unit potential difference, C=QVC = \frac{Q}{V}, measured in farads.

Q2. A 100 μF100\ \mu\text{F} capacitor is charged to 20 V20\ \text{V}. Find the energy stored. [2 marks]

  • Cue. E=12CV2=12(100×10−6)(20)2=0.020 JE = \frac{1}{2}CV^2 = \frac{1}{2}(100 \times 10^{-6})(20)^2 = 0.020\ \text{J}.

Q3. A capacitor discharges through a resistor with time constant RC=2.0 sRC = 2.0\ \text{s}. State the fraction of the initial charge remaining after 2.0 s2.0\ \text{s}. [1 mark]

  • Cue. 1e≈0.37\frac{1}{e} \approx 0.37, about 37 per cent.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20194 marksA 470 μF470\ \mu\text{F} capacitor is charged to 12 V12\ \text{V}. Calculate the charge stored and the energy stored.
Show worked answer →

Charge from Q=CVQ = CV: Q=470×10−6×12=5.64×10−3 CQ = 470 \times 10^{-6} \times 12 = 5.64 \times 10^{-3}\ \text{C}, about 5.6 mC5.6\ \text{mC}.

Energy from E=12CV2E = \tfrac{1}{2}CV^2: E=12(470×10−6)(12)2=12(470×10−6)(144)=0.034 JE = \tfrac{1}{2}(470 \times 10^{-6})(12)^2 = \tfrac{1}{2}(470 \times 10^{-6})(144) = 0.034\ \text{J}, about 34 mJ34\ \text{mJ}.

Markers reward Q=CV=5.6 mCQ = CV = 5.6\ \text{mC}, E=12CV2E = \frac{1}{2}CV^2, and the energy about 34 mJ34\ \text{mJ}.

Eduqas 20225 marksA 2200 μF2200\ \mu\text{F} capacitor charged to 9.0 V9.0\ \text{V} is discharged through a 15 kΩ15\ \text{k}\Omega resistor. Calculate the time constant and the voltage across the capacitor 20 s20\ \text{s} after discharge begins.
Show worked answer →

Time constant: τ=RC=(15×103)(2200×10−6)=33 s\tau = RC = (15 \times 10^{3})(2200 \times 10^{-6}) = 33\ \text{s}.

Voltage decays exponentially: V=V0e−t/RC=9.0 e−20/33=9.0 e−0.606V = V_0 e^{-t/RC} = 9.0\,e^{-20/33} = 9.0\,e^{-0.606}.

e−0.606=0.545e^{-0.606} = 0.545, so V=9.0×0.545=4.9 VV = 9.0 \times 0.545 = 4.9\ \text{V}.

Markers reward τ=RC=33 s\tau = RC = 33\ \text{s}, using V=V0e−t/RCV = V_0 e^{-t/RC}, and the voltage about 4.9 V4.9\ \text{V} after 20 s20\ \text{s}.

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