Displacment field for linear dielectric

Consider the following statement:

In the case of a linear dielectric of arbitrary shape, when no free charges are present, in an external applied electric field $\vec E_0$, then $\vec D=0$.

Is this statement always true (considering that $\nabla \times \vec D= \nabla \times \vec P$) and can it be proved? If it is false can you please provide a counter example.

Can a similar statement be made for the auxiliary field $H$ in terms of the magnetization $M$?

• Not sure I understand you correctly but is it trivially not true for homogeneous media, where $\vec{D}\propto \vec{P} \propto \vec{E}$, see [wiki] (en.wikipedia.org/wiki/Dielectric). $\vec{D}=0$ implies everything is zero? – ophelia Dec 22 '15 at 16:25
• It would be helpful to know where does the quoted statement come from. It seems to be never true to me also. – Mikael Kuisma Dec 23 '15 at 10:45

No, this is essentially never true. In a linear dielectric, $D=\varepsilon E$, so $D=0$ can only be true if $E=0$. In other words, the statement is saying that any linear dielectric always perfectly screens a uniform external electric field. That's not true. Conductors perfectly screen electric fields in their interior, but dielectrics don't (unless the permittivity goes to infinity).
A particularly simple example is the limit $\varepsilon \rightarrow \varepsilon_0$, in which case the dielectric acts like a vacuum and has no effect on the external field. So obviously D inside is nonzero.
• Also adding an interesting bit on why dielectrics cannot perfectly screen electric fields. In the absence of any free charges, bound charges in the dielectric's volume $\Omega$ are also null, $\rho_\text{free} = \rho_\text{bound} = 0$. Then from ${\vec \nabla} \cdot \vec{D} =0$, ${\vec \nabla} \times \vec{D} =0$ it follows that the potential within the dielectric must be harmonic, $\Delta V = 0$. Since free charges are absent on the boundary too, the potential must be continuous there, $V_{in} = V_\text{out}\Big|_{\partial\Omega}$, along with its normal derivatives, $\left(\frac{\partial V_\text{in}}{\partial n} - \frac{\partial V_\text{out}}{\partial n}\right)\Big|_{\partial\Omega} = \sigma_\text{free}\Big|_{\partial\Omega} = 0$. But if the displacement $\vec{D}$ were null in $\Omega$, so would the field $\vec{E}$, and the potential's normal derivative would have to vanish on the interior of the boundary. This would leave $\left(-\frac{\partial V_\text{out}}{\partial n}\right)\Big|_{\partial\Omega} = 0$ for a non-vanishing field outside the boundary, which cannot be.
The result is that there must be a non-vanishing bound/polarization charge density on the dielectric's boundary, $\sigma_\text{bound} = \vec{P}\cdot\vec{n}\neq 0$, and a non-vanishing displacement in the volume, $\vec{D} \neq 0$.