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How does a capacitor store charge and energy, and how does it charge and discharge?

Capacitance and the charge stored, the energy stored in a capacitor, capacitors in series and parallel, and the exponential charging and discharging through a resistor.

A CCEA A-Level Physics answer on capacitance and the charge stored, the energy stored in a capacitor, combining capacitors in series and parallel, and the exponential charging and discharging of a capacitor through a resistor with the time constant.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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

What this dot point is asking

CCEA wants you to define capacitance and use the charge stored, calculate the energy stored in a capacitor, combine capacitors in series and parallel, and describe and use the exponential charging and discharging of a capacitor through a resistor. Exponential-decay calculations with the time constant are a staple.

The answer

Capacitance and charge stored

A capacitor stores equal and opposite charge on its two plates, separated by an insulator (the dielectric). The farad is a large unit, so practical values are usually microfarads or smaller.

Energy stored and combinations

The factor of one half appears because the voltage rises from zero to VV as charge accumulates, so the average voltage during charging is 12V\tfrac{1}{2}V. Half the energy supplied by a battery while charging is dissipated in the circuit resistance, which is why the stored energy is only 12QV\tfrac{1}{2}QV.

Exponential charging and discharging

A larger RCRC means a slower discharge. After one time constant 63%63\% of the charge has gone; after five time constants the capacitor is effectively fully discharged.

Worked example: finding the time constant from data

Examples in context

Example 1. A camera flash. A capacitor is slowly charged from the camera battery over a second or two, storing energy 12CV2\tfrac{1}{2}CV^2. When the shutter fires, it discharges through the flash tube in a few milliseconds (small RR, so small RCRC), delivering a high power burst of light that the battery alone could never supply.

Example 2. A smoothing capacitor in a power supply. After a rectifier converts AC to a bumpy DC, a large capacitor across the output charges on each peak and discharges slowly between peaks. A large RCRC compared with the mains period keeps the output voltage almost steady, smoothing the ripple seen by the load.

Try this

Q1. A 200 μF200\ \mu\text{F} capacitor is charged to 12 V12\ \text{V}. Find the energy stored. [2 marks]

  • Cue. E=12CV2=12(200×10−6)(12)2=1.44×10−2 JE = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(200 \times 10^{-6})(12)^2 = 1.44 \times 10^{-2}\ \text{J}.

Q2. A 100 μF100\ \mu\text{F} capacitor discharges through a 50 kΩ50\ \text{k}\Omega resistor. Find the time constant. [2 marks]

  • Cue. τ=RC=(50×103)(100×10−6)=5.0 s\tau = RC = (50 \times 10^{3})(100 \times 10^{-6}) = 5.0\ \text{s}.

Q3. A capacitor discharges with a time constant of 4.0 s4.0\ \text{s}. What fraction of the initial charge remains after 8.0 s8.0\ \text{s}? [2 marks]

  • Cue. Q/Q0=e−8.0/4.0=e−2=0.135Q/Q_0 = e^{-8.0/4.0} = e^{-2} = 0.135, so about 14%14\% remains.

Exam-style practice questions

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

CCEA 20196 marksA 470 microfarad capacitor is charged to 9.0 V and then discharged through a 22 kilohm resistor. Calculate the initial charge and the initial energy stored, the time constant of the circuit, and the charge remaining after 15 s.
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The initial charge is

Q0=CV=470×10−6×9.0=4.23×10−3Q_0 = CV = 470 \times 10^{-6} \times 9.0 = 4.23 \times 10^{-3} C.

The initial energy stored is

E=12CV2=12×470×10−6×9.02=12×470×10−6×81=1.90×10−2E = \tfrac{1}{2}CV^2 = \tfrac{1}{2} \times 470 \times 10^{-6} \times 9.0^2 = \tfrac{1}{2} \times 470 \times 10^{-6} \times 81 = 1.90 \times 10^{-2} J.

The time constant is

τ=RC=22×103×470×10−6=10.3\tau = RC = 22 \times 10^{3} \times 470 \times 10^{-6} = 10.3 s.

After 15 s the charge remaining is

Q=Q0e−t/RC=4.23×10−3×e−15/10.3=4.23×10−3×e−1.456=4.23×10−3×0.233=9.9×10−4Q = Q_0 e^{-t/RC} = 4.23 \times 10^{-3} \times e^{-15/10.3} = 4.23 \times 10^{-3} \times e^{-1.456} = 4.23 \times 10^{-3} \times 0.233 = 9.9 \times 10^{-4} C.

Markers reward Q=CVQ = CV, the half in the energy, the correct time constant, and a clean substitution into the exponential.

CCEA 20214 marksTwo capacitors of 4.0 microfarad and 12 microfarad are connected first in parallel and then in series across a 6.0 V supply. Determine the combined capacitance in each arrangement, and calculate the total charge drawn from the supply when they are connected in parallel.
Show worked answer →

In parallel the capacitances add:

Cpar=4.0+12=16C_{\text{par}} = 4.0 + 12 = 16 microfarad.

In series the reciprocals add:

1Cser=14.0+112=312+112=412\frac{1}{C_{\text{ser}}} = \frac{1}{4.0} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12}, so Cser=3.0C_{\text{ser}} = 3.0 microfarad.

The total charge drawn in the parallel arrangement is

Q=CparV=16×10−6×6.0=9.6×10−5Q = C_{\text{par}}V = 16 \times 10^{-6} \times 6.0 = 9.6 \times 10^{-5} C.

Markers reward the correct (opposite to resistor) combination rules and the charge from Q=CVQ = CV.

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