Parallel Plate Capacitor (AQA A Level Physics): Revision Note
Exam code: 7408
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 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:
Where:
= relative permittivity
= permittivity of a material (F m-1)
= 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
Step 2: Rearrange for permittivity of the material
Step 3: Substitute in the values
[1 mark]
[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

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:
Where:
= capacitance (F)
= cross-sectional area of the plates (m2)
= separation of the plates (m)
= relative permittivity of the dielectric between the plates
= permittivity of free space (F m-1)

Capacitor plates are generally square, so if they have a length on all sides, then their cross-sectional area is
Worked Example
A parallel-plate capacitor has square plates of length separated by distance and is filled with a dielectric. A second capacitor has square plates of length separated by distance 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
Step 2: Substitute the known values into the capacitance equation
The cross-sectional area of capacitor 1 is
The cross-sectional area of capacitor 2 is
Since the dielectric for capacitor 2 is air, and air has a permittivity of
[1 mark]
Step 3: Equate the capacitances
Step 4: Find the value of
[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!
Unlock more, it's free!
Was this revision note helpful?
Build on this topic