Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
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.
What is the value of the permittivity of free space, and in what unit is it measured?
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
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
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
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
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
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
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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