Parallel Plate Capacitor (AQA A Level Physics): Revision Note

Exam code: 7408

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Relative permittivity & dielectric constant

  • Permittivity is the measure of how easy it is to generate an electric field in a certain material

  • The relative permittivity εr is sometimes known as the dielectric constant

  • For a given material, it is defined as:

The ratio of the permittivity of a material to the permittivity of free space

  • This can be expressed as:

εr = εε0

  • Where:

    • εr = relative permittivity

    • ε = permittivity of a material (F m-1)

    • ε0 = permittivity of free space (F m-1)

  • The relative permittivity has no units because it is a ratio of two values with the same unit

Worked Example

Calculate the permittivity of a material that has a relative permittivity of 4.5 × 1011. State an appropriate unit for your answer.

[2]

Answer:

Step 1: Write down the relative permittivity equation

εr = εε0

Step 2: Rearrange for permittivity of the material ε

ε = εrε0

Step 3: Substitute in the values

ε = (4.5 × 1011) × (8.85 × 10−12) [1 mark]

ε = 3.98 F m−1 [1 mark]

Dielectric action in a capacitor

  • When the polar molecules in a dielectric align with the applied electric field from the plates, they each produce their own electric field

    • This electric field opposes the electric field from the plates

Polar molecules in a dielectric between two charged plates, producing an electric field in the opposite direction to the field between the plates. An upward arrow shows the plates’ electric field, opposed by each molecule’s downward field.
The electric field of the polar molecules opposes that of the electric field produced by the parallel plates
  • The larger the opposing electric field from the polar molecules in the dielectric, the larger the permittivity

    • In other words, the permittivity is how well the polar molecules in a dielectric align with an applied electric field

  • The opposing electric field reduces the overall electric field, which decreases the potential difference between the plates

    • Therefore, the capacitance of the plates increases

  • The capacitance of a capacitor can also be written in terms of the relative permittivity:

C = Aε0εrd

  • Where:

    • C = capacitance (F)

    • A = cross-sectional area of the plates (m2)

    • d = separation of the plates (m)

    • εr = relative permittivity of the dielectric between the plates

    • ε0 = permittivity of free space (F m-1)

A parallel-plate capacitor has two conductive plates of area A carry +Q and −Q, separated by distance d, with a dielectric of permittivity ε between them.
A parallel plate capacitor consists of conductive plates each with area A, a distance d apart and a dielectric ε between them
  • Capacitor plates are generally square, so if they have a length L on all sides, then their cross-sectional area is L2

Worked Example

A parallel-plate capacitor has square plates of length L separated by distance d and is filled with a dielectric. A second capacitor has square plates of length 3L separated by distance 3d and has air as its dielectric. Both capacitors have the same capacitance.

Determine the relative permittivity of the dielectric in the first capacitor.

[2]

Answer:

Step 1: Write down the capacitance equation with the relativity permittivity

C = Aε0εrd

Step 2: Substitute the known values into the capacitance equation

  • The cross-sectional area of capacitor 1 is A = L2

C1 = L2ε0εrd

  • The cross-sectional area of capacitor 2 is A = (3L)2 = 9L2

  • Since the dielectric for capacitor 2 is air, and air has a permittivity of ε0

εr = εε0 = ε0ε0 = 1

C2 = 9L2ε03d [1 mark]

Step 3: Equate the capacitances

L2ε0εrd = 9L2ε03d

Step 4: Find the value of εr

εr = 9L2ε0d3dL2ε0 = 93 = 3 [1 mark]

Examiner Tips and Tricks

Remember that A, the cross-sectional area, is only for one of the parallel plates. Don't multiply this by 2 for both the plates for the capacitance equation!

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Ashika

Author: Ashika

Expertise: Physics Content Creator

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

Tim

Reviewer: Tim

Expertise: Content Creator

Timothy graduated with a first class degree in Mathematics and Physics from the University of Warwick. After working as a postgraduate researcher, Timothy has worked as a content creator for various online revision platforms, creating physics resources for a range of levels and exam boards.