Coulomb's law for Electric Displacement vector

I am quite a newbie at Electromagnetism and I have been reading Intro to Electrodynamics by Griffiths. I have a doubt and I would highly appreciate any help :)

The author states that for the Electric Displacement vector $D = \epsilon_0E+P$ obeys its own Gauss Law: $\nabla \cdot D= \rho_{\text{free charge}}$ but that there is not any Coulomb's Law for D since $\nabla \times D \stackrel{\nabla \times E = 0}{=} \nabla \times P \neq 0\text{ (generally)}$ (so D it is not like E).

However, when we find ourselves inside a linear isotropic homogeneous dielectric ($P = \text{constant}$) we know that $\nabla \times P = 0 = \nabla \times D$ (the same as E). Later he concludes that

$D = \epsilon_0 E_{\text{vacuum}}$ where $E_{\text{vacuum}} =$ the electric field without the dielectric (generated by $\rho_{\text{free charge}}$) and I can't imagine how he reached there (that's my question).

Thus we DO have a Coulomb's Law for D in this particular case using the free charge to get the field.

I understand that $\nabla \times D = 0 = \epsilon_0\nabla \times E$ and $\nabla \cdot D = \rho_{\text{free charge}}\stackrel{\text{Gauss Law}}{=} \epsilon_0\nabla \cdot E_{\text{vacuum}}$ but as far as I am concerned that is not sufficient to state that $D=\epsilon_0E_{\text{vacuum}}$ (it could be proved if D was parallel to E though). How can I prove this equality?

Thanks in advance, some relevant images:

This is all about the structure of the equations. The electrostatic field $\mathbf{E}$ satisfies the two equations

$$\nabla\cdot \mathbf{E}=\dfrac{\rho}{\epsilon_0},\quad \nabla\times \mathbf{E}=0,$$

together with boundary conditions on crossing charged surfaces, for example.

What the author means is the following. We know that $\mathbf{D}$ satisfies the equations

$$\nabla\cdot \mathbf{D}=\rho_{\text{free}},\quad\nabla\times\mathbf{D}=\nabla\times \mathbf{P}.$$

Now when $\nabla\times \mathbf{P}=0$, for example, when $\mathbf{P}$ is constant, we have that

$$\nabla \cdot \mathbf{D}=\rho_{\text{free}},\quad \nabla \times \mathbf{D}=0.$$

As a mathematical problem of solving a differential equation this is the same as the electrostatic problem I said above, with the identification that the charge density should be $\rho = \epsilon_0\rho_{\text{free}}$.

To see this, divide by $\epsilon_0$ the first equation and notice that

$$\nabla\cdot \left(\dfrac{1}{\epsilon_0}\mathbf{D}\right)=\dfrac{\rho_\text{free}}{\epsilon_0},\quad \nabla \times \mathbf{D}=0.$$

Thus since these are the same equations for the electrostatic field in vacuum generated by a charge density $\rho_{\text{free}}$, which we shall call $\mathbf{E}_{\text{vac}}$ you have that

$$\dfrac{1}{\epsilon_0}\mathbf{D}=\mathbf{E}_{\text{vac}}$$

which gives your equation $\mathbf{D}=\epsilon_0 \mathbf{E}_{\text{vac}}$.

• Thanks I think I've got it. As I see it, the key is that they share the same "boundary conditions" over a surface, thus, the electrostatics uniqueness theorem states they must be the same. – J. Doe Nov 3 '17 at 7:38

Ah homogeneous linear dielectric is defined by the property

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

so that

$\textbf{D} = \epsilon_0 \textbf{E} + \textbf{P} = \epsilon_0 \textbf{E} + \epsilon_0 \chi \textbf{E} = \epsilon_0 (1 + \chi) \textbf{E} \overset{def}{=} \epsilon_0 \textbf{E}_{vac}$

I think it is just a definition.