Units of Flux Density: Electric Field vs Displacement Vector

[gralla55]

While I do understand the use for the displacement vector D = (e0)E + P, I don't quite understand why they label it as "flux density". The units for the displacement vector are C/m^2. The units for electric flux are Vm, so wouldn't electric flux density become Vm/m^2 = V/m which is just equal to the electric field?

edit: C/m^2 just looks like surface charge density...

 

[vanhees71]

For this reason, I hate the SI. It's totally unintuitive. Anyway, let's work through the dimensional analysis. According to Gauß's law, which reads in differential form
$$\vec{\nabla} \cdot \vec{D}=\rho,$$
where $\rho$ is the charge density (i.e., charge per unit volume), which has the SI unit $\mathrm{C}/\mathrm{m}^3$ (Coulomb per cubic metre). The Nabla operator is a differential operator wrt. to space coordinates and thus has the unit $1/\mathrm{m}$.

This implies that $\vec{D}$ has the unit $\mathrm{C}/\mathrm{m}^2$.