$$\mathbf D = \varepsilon \mathbf E$$ I don't understand the difference between $\mathbf D$ and $\mathbf E$. When I have a plate capacitor, a different medium inside will change $\mathbf D$, right? $\mathbf E$ is only dependent from the charges right?


$\mathbf E$ is the fundamental field in Maxwell equations, so it depends on all charges. But materials have lots of internal charges you usually don't care about. You can get rid of them by introducing polarization $\mathbf P$ (which is the material's response to the applied $\mathbf E$ field). Then you can subtract the effect of internal charges and you'll obtain equations just for free charges. These equations will look just like the original Maxwell equations but with $\mathbf E$ replaced by $\mathbf D$ and charges by just free charges. Similar arguments hold for currents and magnetic fields.

With this in mind, you see that you need to take $\mathbf D$ in your example because $\mathbf E$ is sensitive also to the polarized charges inside the medium (about which you don't know anything). So the $\mathbf E$ field inside will be $\varepsilon$ times lower than for the conductor in vacuum.


Like @Marek has said above, the electric field $E$ is the fundamental field, and is in some sense the more physical. However, Maxwell's equations have a neater geometric meaning if you throw in the "auxilary" fields $D$ (and $H$ for $B$). I usually tell my students the following version of electromagnetism:

There are 4 fields in electromagnetism. We call them $E$, $D$, $B$ and $H$. All of these fields are independent and equally important. Furthermore, they actually embody geometric concepts which are manifest in the integral equations: $$\oint_S D \cdot dS = Q(S)$$ $$\oint_S B \cdot dS = 0$$ $$\oint_{\partial S} E \cdot dl + \partial_t \int_S B \cdot dS = 0$$ $$\oint_{\partial S} H \cdot dl - \partial_t \int_S D \cdot dS = \int_S j \cdot dS$$

Note that:

  1. $E$ and $B$ form an independent pair, as do $D$ and $H$.
  2. $E$ and $B$ do not depend on the sources $Q$ and $j$, but $D$ and $H$ do.
  3. $D$ and $B$ are integrated through surfaces, and represent flux through those surfaces. (The correct mathematical gadget to describe these are actually 2-forms.$
  4. $E$ and $H$ are integrated along lines, and end up representing the potential difference across the ends (or circulation in a loop).
  5. The latter pair connect the change of flux through surfaces with certain circulations.

These equation form Maxwell's equations. They do not uniquely determine a physical situation. In particular, they need to be augumented with constitutive relations which describe (macroscopic) material properties. For example, we might have linear, isotropic, homogeneous (LIH) media, in which case we would have $D = \epsilon E$ and $B = \mu H$. But in general, we might have $\epsilon$ and $\mu$ being tensors, varying as functions of time and space, or even depend on the fields $E$, $B$, etc! These constitutive relations could be arbitrarily complicated, and indeed much of the new field of meta-material engineering is all about creating micro-structures which would yield interesting and useful constitutive relations at the macroscopic scale. More commonly, a scenario where the linearity breaks down is in ferromagnets/ferroelectrics.

There is usually another constitutive relation, linking current and electric field. In LIH media this is called Ohm's Law: $J = \sigma E$.

There is one more equation, which is simply always true, which is the conservation of charge; in the notation above, $\partial_t Q(S) - \int_S j \cdot dS = 0$.

Edit: some additional observations:

In a relativistically covariant form, we can merge $E$ and $B$ together to get the 2-form $F$, and $D$ and $H$ to get its Hodge dual $\star F$. The latter in general depends on the metric we choose. For linear materials it's possible to hide the effects of the material polarisation/magnetisation as a background metric. Incidentally, in this form, the energy is given by $F \wedge \star F$, so it is clear that energy/momentum should be "opposing" pairs, i.e. the Poyntin vector is $N = E \times H$.

In numerical simulations, it's doubly important that we obey Maxwell's equations --- failure to do so leads to highly unphysical things like superluminal propagation of waves or failure to conserve energy or momentum. It has been found that the key is to be exact with respect to the integral forms of the equations, and put all of the discretisation error into not meeting the material constitutive properties.

  • $\begingroup$ I think this is an awesome answer, +1! $\endgroup$ – Platypus Lover Feb 17 '11 at 17:06
  • $\begingroup$ Wow, This really is a great answer. I thought I had a pretty good understanding of electrodynamics but you just explained this is a way I've never thought about it before. $\endgroup$ – Colin K Feb 17 '11 at 17:32
  • $\begingroup$ Yes, very good answer. Even if it strays a little away from the question... But personally I don't mind that at all :) $\endgroup$ – Marek Feb 17 '11 at 17:41
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    $\begingroup$ Your point #2 is simply wrong, unless you mean "free charge" by Q. Gauss' law relates the flux of E through a closed surface to the total enclosed charge. By using Q without defining it explicitly as "free charge," you're going to confuse a lot of people. $\endgroup$ – Thucydides411 Sep 9 '17 at 2:41

The electrical field $\mathbf E$ is the fundamental one. In principle, you don't need the electrical displacement field $\mathbf D$, everything can be expressed in terms of the field $\mathbf E$ alone.

This works well for the vacuum. However, to describe electromagnetic fields in matter, it is convenient to introduce another field $\mathbf D$. Maxwells original equations are still valid, but in matter, you have to deal with additional charges and currents that are induced by the electric field and that also induce additional electric fields. (More precisely, one usually makes the approximation that the electric field induces tiny dipoles, which are described by the electric polarization $\mathbf P$.) A little bit of calculation shows that you can conveniently hide these additional charges by introducing the electrical displacement field $\mathbf D$, which then fulfills the equation

$$ \nabla· \mathbf D = \rho_\text{free} .$$

The point is that this equation involves only the "external" ("free") charge density $\rho_\text{free}$. Charges that accumulate inside the block of matter have already been taken into account by the introduction of the $\mathbf D$ field.

  • $\begingroup$ Hm, except for the explicit equation for $\mathbf D$ this is completely identical to my answer, isn't it? $\endgroup$ – Marek Feb 17 '11 at 17:44
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    $\begingroup$ Yes. I like my formulation better, tough. Obviously. ;-) (And I still upvoted your answer) $\endgroup$ – Greg Graviton Feb 17 '11 at 20:21
  • $\begingroup$ all right then. I don't see any difference between this and my answer though (and if there is some problem with formulation in mine, you should've commented on it instead) so I won't up-vote. Nothing personal -- if you posted this first, I'd vote you up. $\endgroup$ – Marek Feb 17 '11 at 21:29
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    $\begingroup$ @Marek: sure, no problem. I was just subtly uncomfortable with your wording, and it's hard to comment meaningfully on that. (For instance, I think you don't make it clear how exactly $\mathbf D$ is introduced, whereas I mention that it arises as a useful device for calculation. The difference is subtle if anything, but I consider it somewhat important.) $\endgroup$ – Greg Graviton Feb 18 '11 at 18:08
  • $\begingroup$ yeah, I considered talking about $\mathbf D$ and $\mathbf P$ little more but decided not too in the name of brevity. Some clarity may have been lost though. $\endgroup$ – Marek Feb 18 '11 at 18:11

To understand what field is "real", write a charge equation of motion. The force in it is determined with the real field there. In a medium it is still E: $m\vec{a} = q\vec{E}$. In case of magnetic field, it is $\vec{B}$ that determines the force: $m\vec{a} = q \vec{v} \times \vec{B}/c$.

  • $\begingroup$ Isn't this just another way of saying that E is the fundamental field? If so, why the -1 vote? $\endgroup$ – Carl Brannen Feb 17 '11 at 0:21
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    $\begingroup$ Carl, for me any vote is good ;-). $\endgroup$ – Vladimir Kalitvianski Feb 17 '11 at 15:03

$D$ is the electric displacement field or commonly the flux density and $E$ is the field intensity. There is a fundamental difference between them which will be understood to certain extent as you go through the following answer. Consider a point charge of $Q$ coulombs. This means that the number of flux lines emitted by the charge is $Q$ coulombs. enter image description here.

Let the hypothetical sphere shown in figure has a radius $r$. Then $D$ is given by \begin{equation} D = \frac{Q}{4\pi r^2}. \end{equation} That is, $D$ is the number of flux lines passing per area. So, to get an intuitive grasp, interpret $Q$ as a number (number of flux lines) and $D$ as a number density (number of flux lines per area). Now, what about $E?$ $E$, which is the electric field intensity, is actually a force ($E$ is defined as force per coulomb) per flux line, that is the force carried by each flux line. So, the relation $D = \varepsilon E$ connects the number density of flux lines, D, with a force per flux line term, $E$. Now, the permittivity $\varepsilon$ is defined as the ability to pass lines of electric flux through it. This is a qualitative way of saying. Quantitatively, it can be seen as the ratio $\frac{D}{E}$, that is, $\varepsilon$ is the number of electric flux lines (unit is coulomb, as mentioned earlier) passing through unit area for unit force/flux (which is unit field intensity). That is, say $\varepsilon = 5$ (this value of $\varepsilon$ is hypothetical and considered only for the sake of explanation) means, there are 5 flux lines in a unit area considered normal to an electric field with each flux line carrying $1 N$ force.

  • $\begingroup$ Sir I've been searching for last few hours now but I couldn't get the reason that we choose to avoid bound charges in calculation of D. The surface integral of D yields us only the free charge. I can't understand how bound charges don't contribute to electric flux density. Can you please explain. $\endgroup$ – Deep Sep 1 at 12:52
