Electric displacement field density equation

[lys04]

Homework Statement

I want some clarification of what each of the terms in these two equations mean

Relevant Equations

$$\vec{D}=\epsilon_0 \vec{E} + \vec{P}$$
$$\vec{D}=\epsilon_r \epsilon_0 \vec{E}$$

From my understanding,

The electric field, $\vec{E}$ in both of these equations are referring to the same electric field, the electric field produced by free charges in vacuum?

$\vec{D}$ refers to the electric displacement field density INSIDE the dielectric and $$\epsilon_0 \vec{E}$$ is the electric displacement field density in vacuum?

A high relative permittivity, $\epsilon_r$ means that the material polarises a lot in response to an external electric field and therefore the electric field inside the dielectric should be less than that in a vacuum, given by $\frac {1} {\epsilon_r} E$?

But what does the electric displacement field density represent and why is it greater in a dielectric than it is in a vacuum since $\vec{D}= \epsilon _0 \epsilon _r \vec{E}$ and $\epsilon_r = \frac {\epsilon}{\epsilon_0}$ which is always greater than 1, whereas its the opposite for the electric field?

 

[ergospherical]

Applying an electric field to a dielectric material induces a polarization in the material. You can imagine lots of little dipoles having formed in the dielectric in response to the applied field. The +ve and -ve "bound" charges separate from each other slightly, and the polarization is resulting dipole moment per unit volume. The polarization is related to the "bound" charge density by the relation $\rho_{\mathrm{b}} = - \nabla \cdot \mathbf{P}$ (see this diagram for an explanation).

The displacement field ($\mathbf{D} := \epsilon_0 \mathbf{E} + \mathbf{P}$) is defined in such a way that it lets us isolate the effect of "free" charges. In other words, by using $\mathbf{D}$, we "don't have to worry" about what's going on with the complicated polarization effects inside the material. Express the total charge density as the sum of the "bound" and "free" charges:

$\rho = \rho_{\mathrm{b}} + \rho_{\mathrm{f}} = -\nabla \cdot \mathbf{P} + \rho_{\mathrm{f}}$

Gauss' law $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$ in its original form refers to the total charge density $\rho$. If you stick this into the equation above, you get

$\epsilon_0 \nabla \cdot \mathbf{E} = - \nabla \cdot \mathbf{P} + \rho_{\mathrm{f}}$
$\implies \nabla \cdot (\epsilon_0 \mathbf{E} + \mathbf{P}) = \rho_{\mathrm{f}}$
$\implies \nabla \cdot \mathbf{D} = \rho_{\mathrm{f}}$

You've "bundled" the effect of the polarization into a new quantity, $\mathbf{D}$, and now you have a way to handle situations where you know only know about the free charge density (e.g. useful for capacitors containing insulating dielectrics, where you might only know about the "free" charge density on the metal plates). You can use the same mathematical techniques that you already learned for pure conductors.

The "true" electric field is still always $\mathbf{E}$. It's probably best to think of $\mathbf{D}$ as a helpful book-keeping term which absorbs polarization effects. Note that $\mathbf{D}$ isn't a field strength (it has units of $\mathrm{C} / \mathrm{m}^2$).