Dielectric effects confined inside the dielectric material?

[Ocirne94]

Hi all,
I am wondering how is it possible that the polarization effects of a dielectric material remain confined inside the material itself.
That is: for a LIH dielectric, the equations state that the electric field inside the material is reduced by \epsilon_r. But outside the material, no matter how close to it, the electric field is back to normal.
Now consider a dielectric such as the one in the image: there is a net \sigma_p > 0 on the right face and a net -\sigma_p < 0 on the left one, and there is no net charge (free or polarization) between the two faces.
If we measure the electric field at the position of the \sigma_p label to the right, how can it be unaffected by the near positive charge density? I would instead say - by Coulomb's theorem - that the \sigma_p produces an electrical field E_p = \frac{\sigma_p}{\epsilon_0} which should be summed to the external field. And, if the dielectric is wide enough, the effects of the charged left side are negligible. But this conflicts with Maxwell's equations for dielectrics.
Where is the mistake?
Thank you in advance for your patience and your time,
Ocirne

 

[mfb]

That is: for a LIH dielectric, the equations state that the electric field inside the material is reduced by \epsilon_r. But outside the material, no matter how close to it, the electric field is back to normal.

Which equations say that? It is not true, but it could be a good approximation.

If we measure the electric field at the position of the \sigma_p label to the right, how can it be unaffected by the near positive charge density? I would instead say - by Coulomb's theorem - that the \sigma_p produces an electrical field E_p = \frac{\sigma_p}{\epsilon_0} which should be summed to the external field. And, if the dielectric is wide enough, the effects of the charged left side are negligible.

Only if the dielectric is small compared to its thickness.

But this conflicts with Maxwell's equations for dielectrics.

How?

 

[Ocirne94]

Which equations say that? It is not true, but it could be a good approximation.

Consider this setup:

I was told that here the fields are
E_1=\frac{\sigma}{\epsilon_0}=E_3 (outside the dielectric),
E_2=\frac{\sigma}{\epsilon_0 \epsilon_r} inside it.
The fields outside the dielectric are the same as if there were no dielectric: they ignore any possible contribution from the surface polarization densities.
Is it because the field's expression comes from the infinite plane approximation (E=\frac{\sigma}{2\epsilon_0}), which could imply that the dielectric's thickness is negligible?

 

[mfb]

That is an approximation for infinite size of the plates and the dielectric (or a very narrow gap). It does not stay valid if you consider other shapes.

Is it because the field's expression comes from the infinite plane approximation ($E=\frac{\sigma}{2\epsilon_0})$, which could imply that the dielectric's thickness is negligible?

Right.

 

[Ocirne94]

Thank you.
So for other setups without this approximation the polarization density does indeed affect the electric field outside the dielectric's volume, which likely means some double integral to calculate these effects...

 

[mfb]

Or even messier methods like numerical simulations, yes.