Capacitance

Capacitors are components that store charge and energy in electric fields. They are ubiquitous in electronic circuits and are a major topic in AQA specification 3.7.4. This lesson covers the definition of capacitance, energy storage, the parallel plate model, dielectrics, and capacitor combinations.

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Definition of Capacitance

Key Definition: Capacitance (C) is the charge stored per unit potential difference across a capacitor.

C = Q/V

where C is the capacitance (F, farads), Q is the charge stored (C), and V is the potential difference (V).

1 farad = 1 coulomb per volt (1 F = 1 C V⁻¹). A 1 F capacitor is extremely large; typical capacitors range from picofarads (pF = 10⁻¹² F) to millifarads (mF = 10⁻³ F).

Prefix	Symbol	Value
picofarad	pF	10⁻¹² F
nanofarad	nF	10⁻⁹ F
microfarad	μF	10⁻⁶ F
millifarad	mF	10⁻³ F

A capacitor consists of two conducting plates (or surfaces) separated by an insulator (the dielectric). When connected to a voltage supply, charge flows onto the plates: +Q on one plate and −Q on the other. The net charge on the capacitor is zero, but we refer to Q as "the charge stored" (meaning the charge on each plate).

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The Parallel Plate Capacitor

For a parallel plate capacitor with vacuum (or air) between the plates:

C = ε₀A/d

where ε₀ = 8.85 × 10⁻¹² F m⁻¹ (permittivity of free space), A is the overlapping area of the plates (m²), and d is the separation between the plates (m).

This shows that capacitance:

• Increases with plate area A (more space to store charge)
• Decreases with plate separation d (a smaller gap means a stronger field for the same voltage, allowing more charge to be held)

Derivation of C = ε₀A/d

Starting from the uniform field between the plates: E = V/d, and the field due to a surface charge density σ = Q/A on a plate: E = σ/ε₀ = Q/(ε₀A).

Setting these equal: V/d = Q/(ε₀A)

Rearranging: Q/V = ε₀A/d

Since C = Q/V: C = ε₀A/d ✓

Worked Example 1 — Capacitance of a Parallel Plate Capacitor

Question: A parallel plate capacitor has square plates of side length 20 cm, separated by an air gap of 1.5 mm. Calculate its capacitance.

Solution:

A = (0.20)² = 0.040 m²

C = ε₀A/d = (8.85 × 10⁻¹² × 0.040) / (1.5 × 10⁻³)

C = 3.54 × 10⁻¹³ / 1.5 × 10⁻³

C = 2.36 × 10⁻¹⁰ F = 236 pF

This is a very small capacitance — practical capacitors achieve larger values by using very thin dielectrics and large effective areas (e.g., by rolling the plates into a cylinder).

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Permittivity and Dielectrics

When an insulating material (a dielectric) is placed between the plates, the capacitance increases:

C = εᵣε₀A/d

where εᵣ is the relative permittivity (also called the dielectric constant) of the material. εᵣ is dimensionless and always ≥ 1 (for vacuum, εᵣ = 1).

How Dielectrics Increase Capacitance

1. The electric field between the plates polarises the dielectric molecules — their positive and negative charges are displaced slightly in opposite directions.
2. This creates a field inside the dielectric that opposes the applied field, reducing the net electric field between the plates.
3. A reduced field means a reduced voltage V for the same charge Q.
4. Since C = Q/V, a smaller V for the same Q means a larger C.

Dielectric Material	Relative Permittivity (εᵣ)
Vacuum	1.00
Air	1.0006
Paper	3.5
Mica	5–7
Glass	4–10
Water	80
Barium titanate ceramic	1 000–10 000

Worked Example 2 — Effect of a Dielectric

Question: A capacitor has a capacitance of 470 pF with air between the plates. A sheet of mica (εᵣ = 6.0) is inserted between the plates, completely filling the gap. Calculate the new capacitance.

Solution:

C_new = εᵣ × C_air = 6.0 × 470 × 10⁻¹² = 2.82 × 10⁻⁹ F = 2.82 nF

The capacitance has increased by a factor of 6.

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Energy Stored in a Capacitor

A charged capacitor stores energy in the electric field between its plates. The energy can be calculated using three equivalent expressions:

W = ½QV = ½CV² = Q²/(2C)

Derivation of W = ½QV