If an electric field passing through a dielectric medium, back into the original medium, is it "back to normal"?

This can be compared to the setup of a parallel plate capacitor with three layers of mediums inbetween, parallel to the plates: layer one with $epsilon _1$, layer two with $epsilon _2$ and layer three again with $epsilon _1$. The electric field is created by charges on the two plates.

Now if you separate this one capacitor into three, each with one layer, the electric field in each layer will remain the same. The charges on the inserted new plates will have the same absolute value but inverse signs. Therefore the electric fields in layer 1, layer 2 and layer 3 will be caused by the same amount of electric charge and differ only based on the distance between the plates and the dielectric material between them.

Hence the electric field in layer 1 and 3 are the same and the electric field is the same after passing through layer 2.

Andreas Schuldei makes a great point about the parallel plate capacitors and different media between the plates.

On the micro level there are some important factors. An applied field distorts the electron cloud around atoms in molecules yielding a dipole moment. However, this happens among adjacent molecules/atoms. So if you have two molecules stacked on top of each other, with electrons being pointed in the same direction, the negative part of the resulting dipole will overlap with the positive part of the dipole below, cancelling out. Consequently, the net bound charge difference ends up being on the surface of the media.

With equal and opposite bound charge collected on the surfaces of a media, they cancel each other out so the media below that doesn't "see" the effect of the bound charge.

an electric point charge causing an electric field E in a medium with a dielectric constant ?1 ϵ 1 . You can calculate ... the gradient field ? E .Now imagine you insert a dielectric ... (like a piece of glass, a plastic board, etc

That's just a thin layer, right? And, it's uncharged, so the field it generates is dependent on dipole moment (displaced equal (+) and (-) charges) across that small thickness. Rather than that, consider a point charge having a radial-directed field around it, and place a glass marble next to it. The glass marble will become a dipole (by induction from the point charge) and the field around the ensemble is no longer radially symmetric, but has both point-charge and dipole character.

In some cases (parallel-plate capacitor) we have symmetry and Gauss's law to predict field insensitivity, but not for that marble. There are no general rules that make an electric field simpler than the full solution of a bunch of boundary-value additions to a differential equation.

At best, we can consider a thin sheet of material to have little displaced charge, and mainly only affecting the volume inside the sheet. That, however, is an approximation or a use of some perceived symmetry.

Am not an expert but I think there might, in some cases, be a difference in the net force on a test particle depending on whether the dielectric is present or absent. Consider two cases A,B. Assume that electric fields are superposed.

---

Case (A) Imagine a thin planar sheet which acts as the (+ve) charged (primary) source and, parallel to it, but offset some distance d$x$ in the $x$ direction, a thin planar dielectric sheet. Let both sheets have infinite extent in the $y$ and $z$ directions.

In this case the induced polarisation of the dielectric will shift dielectric electrons towards the primary (+) source and dielectric protons away from the primary (+) source.

Now consider three (+ve) test particles at different positions P,Q,R, as follows:-

(i) The (+) test particle positioned at a point P between the source and the dielectric inner surface will experience a net secondary force directed away from the (+) source. This is because the field from the (closer) dielectric electrons is slightly stronger than the field from the (more distant) dielectric protons.

(ii) The (+) test particle positioned at a point Q in the middle of the dielectric will experience a net secondary force directed towards the source.

(iii) The (+) test particle positioned at a point R beyond the far side of the dielectric will experience a net secondary force directed away from the source.

So, for Case A, the presence of the dielectric will cause additional forces on test particles.

---

Case (B) Imagine a point (+ve) charge source surrounded by a thin spherical spherical shell of (electrically neutral) dielectric material.

The induced polarisation of the dielectric will shift dielectric electrons towards the primary (+) source and shift dielectric protons away from the primary source. We can imagine two spherical arrays: an outer array of protons and an inner array of electrons. There are equal numbers of electrons and protons.

Now according to Newton's Shell Theorem, the net gravitational force exerted by a spherical shell upon an external test particle is the same as the force that would be exerted if all the mass was positioned at the centre of the sphere. We can apply the same principle to the forces caused by electric fields from point sources, (being careful with the force directions). In this case the secondary effects, upon a test particle (R) located externally (outside the dielectric shell), of the shifted dielectric electrons and protons are equal and opposite and so the external test particle will not experience anything more than the effect of the primary field. It can likewise be shown that the net secondary force on a test particle (P) which is located internally between the centre and the dielectric sphere will be zero. And, assuming the Shell Theorem, the same applies to a test particle (Q) located within the dielectric shell.

In reality the shifted charges do not form perfectly uniform shells and so there will be complex variations in the net force vector for any test particles (P,Q or R) located very close to individual electrons and protons in the dielectric.