Free, bound and total charge density

If in a medium we have:

$$\rho(\vec r)_{total} = \rho(\vec r)_{free} + \rho(\vec r)_{bound}$$.

Now we know:

$$-\nabla\vec P = \rho(\vec r)_{bound}$$

$$\nabla\vec D = \rho(\vec r)_{free}$$.

Since $$\vec D=\epsilon_0 \epsilon \vec E$$ we can say also that:

$$\epsilon_0 \epsilon \nabla\vec E = \rho(\vec r)_{free}$$.

If what I wrote is corrent, then what gives us the total charge density. Is there a field whose divergence gives us the total charge density, similarly to how the rotation of the magnetic field gives us the total current density in a material $$\nabla \times \vec B = \vec j_{total}$$.

If I understand you correctly the corresponding field should be $$\epsilon_0 \vec{E}$$. This is one of Maxwell's equations. Let me explain a little bit about the different fields because they might be the source of confusion.

An equation you did not mention is the relation between $$\epsilon$$ and $$\vec{P}$$. We can write $$\epsilon$$ via the electric suceptibility: $$\epsilon=1+\chi$$ where $$\chi$$ is defined to be a proportionality factor in the following: $$\vec{P}=\chi \epsilon_0 \vec{E}$$. This is of course an assumption which is not necessarily always true but let us assume it is for now.

If we use that we get $$\vec{D}=\epsilon_0 \vec{E}+\vec{P}$$. Taking the divergence of that one sees that $$\rho_{free}=\epsilon_0\vec{\nabla}\vec{E}-\rho_{bound}$$, leading to $$\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$$.

The equation $$\vec{D}=\epsilon_0 \vec{E}+\vec{P}$$ can as far as I know be seen as the definition of $$\vec{D}$$ and is always true, so what I just did was some kind of backwards argument. Usually $$\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$$ is seen as the fundamental equation and the rest gets derived from there.

The thing which is not always true generally is $$\vec{D}=\epsilon \epsilon_0 \vec{E}$$ (where $$\epsilon$$ is a constant number). In general $$\epsilon$$ can be a matrix, i.e. $$\vec{D}$$ can have a different direction than $$\vec{E}$$, or it can depend on $$\vec{E}$$ making the relation between $$\vec{D}$$ and $$\vec{E}$$ non-linear. But nonetheless $$\vec{D}=\epsilon_0 \vec{E}+\vec{P}$$ should always be true and $$\epsilon_0\vec{\nabla}\vec{E}=\rho_{total}$$ should also always be true as it is one of Maxwell's equations.

• One more question: in this equation $\vec D = \epsilon_0 \vec E + \vec P$, the electric field inside the dielectric is which of the three? Dec 1, 2021 at 18:41
• Maybe I am missing the point but I would say the electric field is $\vec{E}$ and nothing else. According to the Wiki $\vec{D}$ for example is the flux density. The question is what you want to do with it. For example if you want to use the electric field to compute the force on a particle (the Lorentz force), where one ordinarily sees the electric field, you might run into a problem because the Lorentz force is not always useful in materials. Dec 2, 2021 at 18:54