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## Dielectric Constant / Relative Permittivity

The **dielectric constant** or Relative Permittivity is a dimensionless physical constant (Dielectric constant has no units) that describes how an electric field affects a material. The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space.

Capacity of a capacitor depends on the dielectric constant. It is known that the value of the capacity of a capacitor is given by the following formula: C = Q / V. Where:

- C: Capacitor capacity
- Q: capacitor charge
- V: Potential difference (voltage) between the capacitor plates

The capacitance of a capacitor can also be obtained differently by including the value of the dielectric constant.

## How to obtain the Capacity, knowing the dielectric constant and its physical dimensions?

A capacitor is formed by two parallel plates. If there is a vacuum between these plates, the value of the capacity is: C = εo a / d. where:

- a = area of each plate in m
^{2} - d = distance between plates in meters
- εo = dielectric constant in vacuum, whose value is: 8.85 x 10
^{-12}farad / meter

If a dielectric is introduced between the plates, the capacitance will increase by a factor εr. So the capacity is:

C = εo εr a / d or C = ε a / d, where:

- ε = εo εr
- εr is the relative dielectric constant and depends on the physical properties of the medium used.
- ε is the absolute dielectric constant.

There is a great difference between values of the dielectric constants of different materials. Some important examples of dielectric constants are shown in the following table.

## Table of dielectric constants

Table of dielectric constants (20 °C)

Example of capacity and charge calculation of parallel plate’s capacitor.

We have a parallel plates capacitor separated by vacuum. The plates are 1 mm apart and have an area of 2 x 10^{-6} meters. If the capacitor is connected to a voltage of 250 volts:

- What is the capacitance of the capacitor?
- What is the charge on each plate?
- What is the capacitance to be if it has a paper dielectric?

1. Capacitance:

Using the formula: C = εo εr a / d.

C = (8.85 x 10^{-12}) (1) (2 x 10^{-6}) / (1 x 10^{-3}) = 1.77 x 10^{-14} Farads.

2. Charge on each plate:

Using the formula: C = Q / V, we obtain that Q = C x V.

Q = (1.77 x 10^{-14}) x 250 = 4.425 x 10^{-12} Coulombs.

3. Capacitance with paper dielectric (εr = 2.5)

Using the formula: C = εo εr a / d

C = (8.85 x 10^{-12}) (2.5) (2 x 10^{-6}) / (1 x 10^{-3}) = 4.425 x 10^{-14} Farads.

You can see the increase in capacitor capacity with a higher value dielectric.

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