# Can electric displacement field be zero if electric field is not?

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 $$\chi > 0$$.
• 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. – Tendero Mar 22 at 17:26
• 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 $\vec{D} = 0$. – lr1985 Mar 22 at 18:08
• 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? – Tendero Mar 22 at 20:41
• @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. – GiorgioP Mar 22 at 23:36