0
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ Please do not post formulae as screenshots, but use MathJax instead. $\endgroup$ – Emilio Pisanty Apr 22 '18 at 19:15
  • $\begingroup$ "I know that to calculate the Electric field in the dielectric is 10/4=2.5" - what textbook are you using, and where did you take that procedure from? Are you sure you're not mixing up resources built for CGS units with relations that use the SI system? The two systems are incompatible for macroscopic electrodynamics (i.e. anything that involves D or H) and you need to pick one system and stick with it. $\endgroup$ – Emilio Pisanty Apr 22 '18 at 19:20
  • $\begingroup$ physics.stackexchange.com/q/340705 $\endgroup$ – DJA Apr 22 '18 at 21:02
  • $\begingroup$ I would agree with your last two statements: D is the applied field, E is the field in the medium. H is the applied field, B is the field in the medium. As @EmilioPisanty said, you may be confusing two different unit systems. D and E in your example don't have the same units, which isn't necessarily an issue as long as your system is consistent in its definitions. Of course in natural units ($c$=1), $\mu_0 = \epsilon_0 = 1$ and then E and D have the same units. $\endgroup$ – Kai Apr 22 '18 at 22:58
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.