What is the relation between relative permittivity and its electrical susceptibility?Physics

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valentin68 | College Teacher | (Level 3) Associate Educator

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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.

`P =epsilon_0*chi_e*E`

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 =D-P`

`epsilon_0*E = epsilon_0*epsilon_r*E - P`

`epsilon_0*E =epsilon_0*epsilon_r*E -epsilon_0*chi_e*E`

`epsilon_r =1+chi_e`

This is the relation between relative permittivity and electrical susceptibility: `epsilon_r =1+ chi_e`

Sources:

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