AQA A Level Physics

Revision Notes

7.6.2 Parallel Plate Capacitor

Relative Permittivity & Dielectric Constant

  • Permittivity is the measure of how easy it is to generate an electric field in a certain material
  • The relativity 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:
    Relative Permittivity Equation
  • 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.

Step 1: Write down the relative permittivity equation

Relative Permittivity Equation

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) = 3.9825 = 4 F m-1

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


  • 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:

Capacitance and Dielectric Equation

  • 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)
  • Capacitor plates are general square, therefore if they have a length L on all sides then their cross-sectional area is A2

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.

Capacitor Dielectric Worked Example (1)Capacitor Dielectric Worked Example (2)

Exam Tip

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!


Join Save My Exams

Download all our Revision Notes as PDFs

Try a Free Sample of our revision notes as a printable PDF.

Join Now
Already a member?
Go to Top