What is the relation between relative permittivity and its electrical susceptibility?Physics
Imagine you have a charge `Q_0` located in vacuum. The electric field produced by this charge at distance `R` from it is
`E = Q_0/(4*pi*epsilon_0*R^2)` (1)
Now imagine you choose the value of a charge to be `Q` in a dielectric medium to obtain the same value of electric field:
`E =Q/(4*pi*epsilon_0*epsilon_r*R^2)` (2)
When a charge is located in a dielectric the particles of the medium tend to align themselves to the electric field. Thus a polarisation `P` of the medium is formed. It is natural to consider that this polarisation is proportional to the value of the electric field `E`. The constant of proportionality is named electrical susceptibility.
Since polarisation acts like an additional charge in the medium, it means from the total charge described by (2) in a medium one need to subtract the value of polarisation to obtain the charge described by (1) in vacuum.
`epsilon_0*E = epsilon_0*epsilon_r*E - P`
`epsilon_0*E =epsilon_0*epsilon_r*E -epsilon_0*chi_e*E`
This is the relation between relative permittivity and electrical susceptibility: `epsilon_r =1+ chi_e`