Trying to Understand what exactly H B E and D are in the Electromagnetic four field formalism

Question

So I have been introduced to the four field formalism but have not been explained it in a way that i think is logical or makes sense so i need to ask a few questions on here to allow myself to think clearly about this.

I have been told that the following relations hold:

\begin{align} \mathbf D & = \varepsilon\mathbf E \\ \mathbf H & = \frac{\mathbf B}{\mu} \end{align} where $\varepsilon = \varepsilon_0\varepsilon_r$ and $\mu = \mu_0\mu_r$.

Now consider taking an object with $\varepsilon_r = 4.0$ and placing it in an electric field of $E = 10$.

I know that to calculate the Electric field in the dielectric is $10/4 = 2.5$.

However why given the afforementioned relations are we not doing $10/4\varepsilon_0$ instead since $\varepsilon = \varepsilon_0\varepsilon_r$.

Am I wrong in thinking that D is the applied electric field and E is the electric field experienced within the dielectric? Similarly would this suggest that H is an applied magnetic field where B is something that the material feels?

Answer 1

I know that to calculate the Electric field in the dielectric is $10/4 = 2.5$.

This is true only if the dielectric is a slab perpendicular to the electric field. There is no general way to calculate total electric field from the external electric field and permittivity, without taking into account geometry of the object.

Am I wrong in thinking that D is the applied electric field and E is the electric field experienced within the dielectric?

Yes, that is wrong. Displacement field $D$ is not "applied electric field". It is defined by

$$ \mathbf D = \epsilon_0 \mathbf E + \mathbf P $$

where $\mathbf P$ is density of electric moment of the dielectric medium, also called "polarization". This is not related to external electric field (field of free charges) in any simple way. Again, the geometry of the bodies involved has a say.