About the electric field in a dielectric and the capacitor plate

You can work in the framework of the Electric Displacement Field. (See https://en.wikipedia.org/wiki/Electric_displacement_field)

Suppose you have capacitor with two infinitely long top plates oriented horizontally and spaced vertically with "free" charge $+Q$ in the top plate and $-Q$ in the bottom plate. Your electric displacement field is given by:

$$int{vec{D}bullet vec{dA}}=Q$$

You can take a Gaussian Pill Box and you can do this integral and obtain:

$$vec{D}=-sigma hat{y}$$

where $sigma$ is your surface charge density in your capacity (i.e. $Q/A$)

Your electric field and electric displacement field are related by the following relation:

$$vec{D}=epsilon_{0}vec{E}+vec{P}$$

where $P$ is your polarisability, which relates to all the dipoles that exist in your material and their tendency to line up with the applied field. The polarisability is given empirically by $$vec{P}=epsilon_{0} chi_{e}vec{E}$$ where $chi_{e}$ is called the electric susceptibility.

Putting everything together you get $$vec{D}=epsilon_{0}vec{E}+epsilon_{0} chi_{e}vec{E}=epsilon_{0}(1+chi_{e})vec{E}=epsilon_{0} epsilon_{r}vec{E}$$

where $epsilon_{r}$ is your relative permittivity.

This factor is what gives the different strengths of electric field in different dielectric materials.

$$vec{E}=-frac{sigma}{epsilon_{0} epsilon_{r}} hat{y}$$

For vacuum $epsilon_{r}=1$. This will give the largest field. When you insert a dielectric into your capacitor since $epsilon_{r}>=1$ your electric field in your dielectric will be smaller.

Note that if you have your capacitor connected to a constant voltage source. You can show that electric field in the dielectric must be the same as the electric field with no dielectric. This is because $E=frac{V}{d}$, what is happening in this case, is there are more surface and/or volume bound charges in your dielectric which are induced to compensate for the drop in electric field.