Does the electric field change only inside the dielectric, or even after exiting the dielectric?

Question

I know that the field inside a dielectric is changed and is $\epsilon$ times less the original field (some books use $K$ for this constant).

This is because the induced field inside the dielectric is in opposite direction to the original field.

But what happens to the field after it passes the dielectric?

For example:

Assume a point charge $q$ is somewhere on the $z$ axis, above the $x-y$ plane, and that the $x-y$ plane is made out of a dielectric material with $\epsilon>1$, does the field below the $x-y$ plane is still given by $E=\frac{q}{r^{3}}\vec{r}$?

Answer 1

First off, the dielectric material on the $x$-$y$ plane ought to have some thickness, say $\Delta$, so that the slab occupies the region $-\Delta/2 < z < \Delta/2$.

Now we assume a point charge above the slab on the $z$ axis. This charge causes a polarization $P$ in the dielectric slab. Thus there will be a bound surface charge density of $\sigma = P \cdot \hat{n}$ at either surface of the slab.

These surface charges and the original point charge are the only unbalanced charges in the system; the charge density is zero everywhere else. Thus we can forget we have a dielectric material and just take the $\sigma$ for granted.

Now we have a new system with a point charge on the $z$-axis and planar charge density $\sigma$ where the surfaces of the dielectric slab used to be. The planar charge creates an electric field which is felt in all of space: it is felt not only between the two surfaces but also on either side.

Going back to the original problem with the dielectric, we have found that the dielectric alters the electric field not only in the slab, and after the field "passes through" the slab, but it also alters the field in the region above the slab where the point charge is.