The electric displacement field is defined as $$\mathbf{D}=\epsilon_0\mathbf{E}+\mathbf{P}$$

But these equalities hold as well:

$$\mathbf{P}=\epsilon_0\chi \mathbf{E}$$

$$\mathbf{D}=\epsilon_0(1+\chi)\mathbf{E} $$

I've taken my electromagnetism course a long ago, but this doubt suddenly came to my mind. I recall that there were some cases in which $\mathbf{D}=0$, for example if a non-conductive material was polarized. But here, $\mathbf{E}$ and $\mathbf{P}$ would have a non-zero value. However, given the equations above, if one of the three fields is $0$, then the other two should equal $0$ as well.

Where am I making the mistake?


First of all note that the fact that $\vec{P}$ and $\vec{D}$ are proportional to $\vec{E}$ is valid only when dealing with linear dielectrics, which is what is usually taught in classical electromagnetism classes. Under this assumption, $\vec{P}$ is non-zero only in the presence of polarization charges. This is strictly equivalent to saying that $\vec{P}$ is non-zero only if $\chi \ne 0$ or, equivalently, if the dielectric relative permittivity $\epsilon = \chi + 1$ is larger than 1.

Therefore, in the vacuum (or in any other material for which $\epsilon = 1$) $\vec{P} = 0$. However, in this case $\vec{E}$ and $\vec{D}$ can still be non-zero, and are linked by the relation $\vec{D} = \epsilon_0 \vec{E}$.

By contrast, if $\vec{E}$ or $\vec{D}$ are null then all three fields are identically zero, since $\chi > 0$.

  • $\begingroup$ But isn't it possible that $\mathbf{D}=0$ and $\mathbf{E} \neq 0$, for example in the case of a polarized non-conductive sphere? I think this is possible, but nevertheless the relationship $\mathbf{D}=\epsilon_0 \mathbf{E}$ states that it is not. $\endgroup$ – Tendero Mar 22 at 17:26
  • $\begingroup$ Since $\chi$ is always positive, $\vec{P}$ and $\vec{E}$ have always the same direction. As a result, they cannot cancel each other to yield $\vec{D} = 0$. $\endgroup$ – lr1985 Mar 22 at 18:08
  • $\begingroup$ But in a polarized sphere, there are no free charges. Thus, by Gauss' Law, $\mathbb{D}$ must equal $0$. But at the same time, $\mathbf{P}$ cannot be zero because there are polarization charges in the sphere. Where am I making the mistake in the reasoning? $\endgroup$ – Tendero Mar 22 at 20:41
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    $\begingroup$ @Tendero In the case of a polarized sphere, Gauss' Law tells you that the flux of ${\bf D}$ is zero. But from there you cannot deduce ${\bf D}=0$ because the field is not spherically symmetric. $\endgroup$ – GiorgioP Mar 22 at 23:36
  • $\begingroup$ @GiorgioP You are absolutely right, that's what I was missing. Thanks $\endgroup$ – Tendero Mar 23 at 0:31

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