Question

What is the relation between permittivity of medium and dielectric constant?

Answer 1

The permittivity of a medium and its dielectric constant are closely related concepts, especially when it comes to how a material responds to an electric field.

1. Permittivity (ε):
   

   It is a property of a material that describes how much electric field (E) is reduced when it passes through the material. Essentially, it tells us how easily a material can be polarized by an electric field. Permittivity is usually represented as \( \varepsilon \), and it is measured in Farads per meter (F/m).

   There are two types of permittivity:
   - Vacuum permittivity (\( \varepsilon_0 \)): The permittivity of free space, which is a constant value of approximately \( 8.854 \times 10^{-12} \, \text{F/m} \).
   - Relative permittivity (\( \varepsilon_r \)): This is the permittivity of the material relative to the permittivity of free space.

1. Dielectric Constant (K):
   

   The dielectric constant of a material is simply the relative permittivity of the material. It is a dimensionless quantity and represents the factor by which the material can increase the capacitance of a capacitor compared to when there is a vacuum or air between the plates. So, the dielectric constant \( K \) is equivalent to \( \varepsilon_r \), the relative permittivity.

Relation:
\[
K = \varepsilon_r = \frac{\varepsilon}{\varepsilon_0}
\]

Where:

1. \( K \) or \( \varepsilon_r \) is the dielectric constant (dimensionless).
   

1. \( \varepsilon \) is the absolute permittivity of the material.
   

1. \( \varepsilon_0 \) is the permittivity of free space (approximately \( 8.854 \times 10^{-12} \, \text{F/m} \)).
   

In simple terms, the dielectric constant (K) tells you how much a material can "resist" the electric field in comparison to a vacuum, and it's calculated by dividing the material's permittivity by the permittivity of free space.