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Capacitance

The capacitance of a parallel plate capacitor

  • 3 things to know
  • Component 2

Overview

Recall questions on C = ε₀A/d with every symbol, the permittivity of free space, how capacitance depends on the plate area and separation, worked calculations, and why a practical capacitor is rolled from long strips of foil.

What you need to know

Answer each one before you reveal it.

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Key term

State the equation for the capacitance of a parallel plate capacitor with nothing but vacuum between the plates, defining every symbol and its unit.

Key term

What is the value of the permittivity of free space, and in what unit is it measured?

Key term

How does the capacitance of a parallel plate capacitor depend on the plate area?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the equation for the capacitance of a parallel plate capacitor with nothing but vacuum between the plates, defining every symbol and its unit.

    Model answer

    The equation is C = ε₀A/d, where C is the capacitance in farads (F), ε₀ is the permittivity of free space in F m⁻¹, A is the overlapping area of one plate in m² and d is the plate separation in m.

  2. 2

    What is the value of the permittivity of free space, and in what unit is it measured?

    Model answer

    The permittivity of free space is ε₀ = 8.85 × 10⁻¹² F m⁻¹.

  3. 3

    How does the capacitance of a parallel plate capacitor depend on the plate area?

    Model answer

    The capacitance is directly proportional to the overlapping area of the plates, so doubling the area doubles the capacitance.

  4. 4

    How does the capacitance of a parallel plate capacitor depend on the plate separation?

    Model answer

    The capacitance is inversely proportional to the separation, so halving the gap doubles the capacitance.

  5. 5

    Calculate the capacitance of a parallel plate capacitor of plate area 0.020 m² with an air gap of 0.50 mm.

    Model answer

    The capacitance is C = (8.85 × 10⁻¹² × 0.020) ÷ (5.0 × 10⁻⁴) = 3.5 × 10⁻¹⁰ F, or 354 pF. Substituting back, A = Cd/ε₀ = 3.54 × 10⁻¹⁰ × 5.0 × 10⁻⁴ ÷ 8.85 × 10⁻¹² = 0.020 m².

  6. 6

    A parallel plate capacitor of plate area 0.10 m² is to have a capacitance of 1.0 nF. Calculate the plate separation required.

    Model answer

    The separation is d = ε₀A/C = (8.85 × 10⁻¹² × 0.10) ÷ (1.0 × 10⁻⁹) = 8.9 × 10⁻⁴ m, or about 0.89 mm. Substituting back, C = 8.85 × 10⁻¹² × 0.10 ÷ 8.85 × 10⁻⁴ = 1.0 nF.

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