About the electric field in a dielectric and the capacitor plate

Question

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?

Answer 1

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.