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EnglandPhysicsSyllabus dot point

How does a capacitor store charge and energy, and what limits how much it can hold?

The definition of capacitance, the energy stored on a capacitor, the effect of a dielectric and relative permittivity, and parallel plate capacitors.

A focused answer to AQA A-Level Physics 3.7.4, covering the definition of capacitance, the energy stored by a capacitor, the role of dielectrics and relative permittivity, and the parallel plate capacitor.

Generated by Claude Opus 4.89 min answer

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

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  1. What this dot point is asking
  2. Defining capacitance
  3. Energy stored
  4. Dielectrics and relative permittivity
  5. The parallel plate capacitor
  6. Try this

What this dot point is asking

AQA specification point 3.7.4 wants you to define capacitance, calculate the energy stored on a capacitor in three equivalent ways, explain the effect of a dielectric using relative permittivity, and use the parallel plate capacitor formula.

Defining capacitance

A capacitor stores equal and opposite charges on its two plates; the charge QQ in the equation is the magnitude on either plate. The farad is a very large unit, so practical capacitors are usually rated in microfarads (μF\mu\text{F}, 10−6 F10^{-6} \text{ F}), nanofarads (nF\text{nF}, 10−9 F10^{-9} \text{ F}) or picofarads (pF\text{pF}, 10−12 F10^{-12} \text{ F}). A larger capacitance means more charge can be stored for the same voltage.

Energy stored

The work done charging a capacitor is stored as electrical potential energy. As charge is transferred from one plate to the other, the plates become increasingly charged and the potential difference rises, so each successive small charge requires more work to move against the growing field. Because charge builds up gradually, the total energy is the area under a graph of charge QQ against potential difference VV, which is a triangle of area 12QV\tfrac{1}{2}QV.

The three forms are equivalent: substitute Q=CVQ = CV to move between them. Use 12CV2\tfrac{1}{2}CV^2 when the voltage is known, Q22C\dfrac{Q^2}{2C} when the charge is known.

Dielectrics and relative permittivity

Typical values are around 2 to 7 for common plastics and oils, and much higher for specialised ceramics, which is why capacitors are made compact by filling them with high-permittivity dielectrics.

The parallel plate capacitor

The capacitance is increased by using larger plates, a smaller separation, or a higher-permittivity dielectric. This explains practical capacitor design: long strips of foil separated by a thin plastic film are rolled up to pack a large area into a small volume.

Try this

Q1. Define capacitance. [1 mark]

  • Cue. The charge stored per unit potential difference, C=QVC = \dfrac{Q}{V}.

Q2. A capacitor stores 0.50 J0.50 \text{ J} when charged to 50 V50 \text{ V}. Calculate its capacitance. [2 marks]

  • Cue. E=12CV2E = \tfrac{1}{2}CV^2, so C=2EV2=2×0.50502=4.0×10−4 FC = \dfrac{2E}{V^2} = \dfrac{2 \times 0.50}{50^2} = 4.0 \times 10^{-4} \text{ F}.

Q3. State two ways to increase the capacitance of a parallel plate capacitor. [2 marks]

  • Cue. Increase the plate area, decrease the separation, or use a higher-permittivity dielectric.

Exam-style practice questions

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

AQA 20184 marksA 470 μF470 \text{ }\mu\text{F} capacitor is charged to a potential difference of 9.0 V9.0 \text{ V}. Calculate the charge stored and the energy stored.
Show worked answer →

Use Q=CVQ = CV for the charge, converting the capacitance to farads first: Q=(470×10−6)(9.0)=4.23×10−3 CQ = (470 \times 10^{-6})(9.0) = 4.23 \times 10^{-3} \text{ C}.

Use E=12CV2E = \tfrac{1}{2}CV^2 for the energy: E=12(470×10−6)(9.0)2=12(470×10−6)(81)=1.9×10−2 JE = \tfrac{1}{2}(470 \times 10^{-6})(9.0)^2 = \tfrac{1}{2}(470 \times 10^{-6})(81) = 1.9 \times 10^{-2} \text{ J}.

Markers reward converting microfarads to farads, the correct charge, and the factor of one half in the energy equation. A common error is to omit the one half or to forget to square the voltage.

AQA 20224 marksExplain how inserting a dielectric between the plates of a parallel plate capacitor increases its capacitance, and state how relative permittivity is defined.
Show worked answer →

The molecules of the dielectric become polarised in the field between the plates: their charges shift slightly so that bound negative charge faces the positive plate and bound positive charge faces the negative plate. This sets up an opposing field inside the dielectric.

For a fixed charge on the plates, the polarisation reduces the net field and hence the potential difference between the plates. Since C=QVC = \dfrac{Q}{V}, reducing VV for the same QQ increases the capacitance.

The relative permittivity εr\varepsilon_r is the factor by which the capacitance increases compared with a vacuum (or air), equal to the ratio of the capacitance with the dielectric to the capacitance without it.

Markers reward polarisation of the dielectric, the reduced field or pd for fixed charge, and defining εr\varepsilon_r as the capacitance ratio.

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