# Can Gauss law can be used at this situation?

We define the following dielectric and the plate.

The dielectric of the permeability $$\epsilon$$ exists.

The plate which has the infinite length and the plate was made of the dielectric.

It has been assumed that the plate is placed in the vacuum vertically against the x-axis of cartesian coordinate system.

We define the $$3$$ domains.

domain $$1:=$$ domain inside the left side vacuum.

domain $$2:=$$ domain inside the dielectric.

domain $$3:=$$ domain inside the right side vacuum.

$$0<\sigma:=$$ amount of the uniform density of the polarization charges which appears uniformly at each surface of the dielectric(borders between the vacuum and the dielectric).

The left side surface has been charged with $$-\sigma$$ per unit area and the right side surface has been charged with $$+\sigma$$ per unit area.

We want to deduce the each electric field in the each domain.

My textbook states that none of electric fields exists at the domain$$1$$ and the domain$$3$$ and only the domain$$2$$ has the electric field from the principle of superposition.

If the border(surface) between the vacuum and the dielectric is a conductor,then Gauss law can be used since there is no electric field which can penetrate a conductor(surface) so easily assuming the differential volume which surrounds the border leads to deduce the equations of electric fields.

However actually the border is a dielectric so penetrations of electric fields may be occurred.

So currently I'm unable to get the meaning of the statement of the textbook.

Is it able to assume that a surface of a plate of a bunch of uniform polarization charges as a conductor?

If so,then Gauss law can be used and my problem is solved.

Can anyone tell me what I've been missing?, or the website(s) which describe(s) about it.

• This would really benefit from a diagram. Feb 27 at 19:51
• @BobD Could you tell me some recommended diagram maker? Mar 13 at 1:34

Gauss’ law actually uses the free charge whereas induced charges leading to polarization are bound charges. The effect of these bound charges on the field inside the dielectric is accounted for by the permittivity $$\epsilon$$ of the dielectric. The bound charges have no effect in the vacuum.