About the electric field in a dielectric and the capacitor plate Why is the electric field caused by the induced surface charges in the dielectric less than the electric field caused by the conductor plates in a capacitor?
 A: 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.
