AQA A-Level Physics 7408
Measurements and Their Errors
Particles and Radiation
Waves
Mechanics and Materials
Electricity
Further Mechanics and Thermal Physics
Fields and Their Consequences
Nuclear Physics
AstrophysicsOptional topic
Medical PhysicsOptional topic
Engineering PhysicsOptional topic
Turning Points in PhysicsOptional topic
ElectronicsOptional topic
Fields and Their Consequences
Parallel plate capacitor
- 4 things to know
- Papers 2 and 3
Overview
Permittivity and relative permittivity, dielectrics, and how a polar molecule responds to a field.
What you need to know
Answer each one before you reveal it.
What does d stand for in the capacitance formula, and what is a dielectric?
How does a polar molecule respond to an applied field, and how is relative permittivity measured?
What instrument can be used to investigate how capacitance varies with plate area and plate separation?
Questions from this subtopic
Each one with the model answer underneath.
- 1
In the formula for capacitance of a parallel plate capacitor, what does the symbol ε₀ represent?
Model answer
ε₀ represents the permittivity of free space.
- 2
In the equation C = ε₀εᵣA/d, what quantity does εᵣ denote?
Model answer
εᵣ denotes the relative permittivity of the dielectric material between the plates.
- 3
What does the symbol A represent in the parallel plate capacitor equation C = ε₀εᵣA/d?
Model answer
A represents the overlapping area of the plates.
- 4
In the capacitance formula C = ε₀εᵣA/d, what does d stand for?
Model answer
d stands for the separation between the plates.
- 5
State the formula for the capacitance of a parallel plate capacitor and identify each term.
Model answer
C = ε₀εᵣA/d, where C is capacitance, ε₀ is permittivity of free space, εᵣ is relative permittivity of the dielectric, A is plate area, and d is plate separation.
- 6
Calculate the capacitance when permittivity of free space = 8.85 × 10⁻¹² F m⁻¹, relative permittivity = 2.6, plate area = 0.026 m², plate separation = 0.0043 m.
Model answer
Capacitance = (8.85 × 10⁻¹² × 2.6 × 0.026) ÷ 0.0043 = 1.39 × 10⁻¹⁰ F.
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