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Does Electric flux depend upon medium?

Physics Asked by mia on April 12, 2021

Wikipedia or any other site that matters, expresses electric flux as dependent on the permitivity of vacuum and not on the permitivity of the medium. This is confusing, to me the electric flux should depend upon the material media as electric field intensity depends upon material media, but the Gauss law seems to say elsewise.

I tried to look up on this site and the closest explanation I could get was:

We will apply Gauss’ law in a unique way: neglect effects of the dielectric. Using this way, we will consider the effect of polarization of the dielectric which would have otherwise been neglected if Gauss’ Law was applied with the correct constant (ϵ instead of ϵ0)

I am not sure If I understand it entirely, why would using ϵ neglect the effect of polarization? If we used the incorrect constant for the sake of derivation,why did we not write the correct one in the original equation?

Also, none of the textbooks I have seen so far refer to medium at all when talking about Gauss’ law. The concluding remarks usually express electric flux as dependent on the charge enclosed and independent of the shape and the location of the charge. Does Gauss law have really nothing to do with it?

One Answer

The $epsilon$ is defined with polarization and Maxwell's equations in mind such that the equations look as similar as possible to the vacuum case: that is, $epsilon$ incorporates polarization.

Take the first equation, or Gauss' law, like you mentioned. The vacuum-case equation is $$nabla cdot mathbf{E} = frac{rho}{epsilon},$$ where $rho$ is the (free) charge density. In the case of a polarizable medium, there will be bound charges as well as free charges, so we can write $rho = rho_f + rho_b$ (you can infer the subscripts easily). Gauss' law then becomes $$nabla cdot mathbf{E} = frac{rho_b + rho_f}{epsilon_0}.$$

It can be shown that the polarization $mathbf{P}$ is related to the bound charge density as $$nabla cdot mathbf{P} = -rho_f,$$ so, replacing that in Gauss' law and rearranging, we get

$$nabla cdot (epsilon_0mathbf{E} + mathbf{P}) = rho_f.$$

To simplify a bit, consider the case of a 'linear' medium, one where the polarization is directly proportional to the external electric field: $$mathbf{P} = epsilon_0 chi mathbf{E},$$ where $chi$ is a constant called the 'electric susceptibility' of the medium. Inserting that into the previous equation, we get $$nabla cdot epsilon_0 (1+ chi) mathbf{E}= rho_f.$$ We then define the 'electric displacement' or $$mathbf{D} = epsilon_0 (1+ chi )mathbf{E},$$ which is the analog of the electric field in vacuum, with polarization taken into account. Finally, we define the general electric permittivity as $$epsilon = epsilon_0 (1+ chi ),$$ with which, Gauss' law in media looks similar to that in vacuum:

$$nabla cdot mathbf{D} = rho_f implies nabla cdot mathbf{E} = frac{rho_f}{epsilon},$$ since $mathbf{D} = epsilon mathbf{E}.$

Correct answer by Yejus on April 12, 2021

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