# Capacitance and RC circuits: Unit 4.1(a), (c)-(e)

> define capacitance and explain factors affecting a parallel-plate capacitor

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Specification statement (Unit 4.1(a), (c)-(e)): define capacitance and explain factors affecting a parallel-plate capacitor

## Explanation

1. **Capacitance** is the charge stored per unit potential difference: C = Q/V. Unit: farad (F = C V⁻¹). Practical values are µF, nF or pF.
2. For a parallel-plate capacitor: C = ε₀εᵣA/d. Capacitance increases with plate **area** A and the **permittivity** of the dielectric, and decreases with plate **separation** d.
3. A dielectric increases C because its molecules polarise, reducing the field and p.d. between the plates for the same charge.

## Worked example

Two square plates of side 10 cm are 1.0 mm apart in air (ε₀ = 8.85 × 10⁻¹² F m⁻¹). Find the capacitance and the charge at 12 V.

1. A = 0.10 × 0.10 = 0.010 m².
2. C = ε₀A/d = 8.85 × 10⁻¹² × 0.010 / 1.0 × 10⁻³ = 8.85 × 10⁻¹¹ F (88.5 pF).
3. Q = CV = 8.85 × 10⁻¹¹ × 12 = 1.1 × 10⁻⁹ C.

Answer: 88.5 pF; 1.1 nC.

## Check

The plate separation of a parallel-plate capacitor is doubled. Its capacitance:

- Doubles
- Halves
- Is unchanged
- Quadruples

The answer and its explanation are shown in the app once you choose an option.

## Apply it

A slab of dielectric with εᵣ = 4.0 fills the gap of an isolated charged capacitor. Explain what happens to its capacitance and to the p.d. across it.

The model answer is shown in the app after you have tried it.

## Related

- [Capacitance and RC circuits lesson](https://revise-gray.vercel.app/lesson?subject=wjec-alevel-physics&topic=wjec-alevel-physics.capacitance)
- [Capacitance and RC circuits in the library](https://revise-gray.vercel.app/library?topic=wjec-alevel-physics.capacitance)