# Does the electric displacement field account for free charges?

I'm trying to understand the electric displacement field $$\mathbf{D}$$ for a lossy material with conductivity $$\sigma$$ and permittivity $$\epsilon$$.

Most textbooks I've looked at only consider perfect dielectrics when discussing $$\mathbf{D}$$. However, the first two sentences of the Wikipedia article on $$\mathbf{D}$$ state

In physics, the electric displacement field (denoted by D) or electric induction is a vector field that appears in Maxwell's equations. It accounts for the effects of free and bound charge within materials.

But then in the next section (as in most textbooks), $$\mathbf{D}$$ is defined as \begin{align} \mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P}.\tag{1}\label{eq} \end{align} Now clearly $$\mathbf{P}$$ accounts for only bound charges. So what accounts for free charges? If the Wikipedia quote is indeed true, shouldn't $$\sigma$$ appear somewhere in (\ref{eq})?

## 1 Answer

Displacement field $$\mathbf D$$ is defined by $$\mathbf D = \epsilon_0 \mathbf E + \mathbf P.$$

Statement "accounts for the effects of free and bound charge" is a weird, probably incorrect statement, because just giving values of components of $$D_x, D_y, D_z$$ in some region of space does not give us enough information on the bound charge. It does give us information on free charge, because its density obeys the equation

$$\nabla \cdot \mathbf D = \rho_f$$ or in other way $$\frac{\partial D_x}{\partial x} + \frac{\partial D_y}{\partial y} + \frac{\partial D_z}{\partial z} = \rho_f.$$

A correct statement would be that value of $$\mathbf D$$ at some point of space is influenced both by free and bound charge. This is because (in electrostatics) $$\mathbf E$$ is function of total charge density in all space, and $$\mathbf P$$ is function of how bound charge density is shifted from its neutral position (and in the simple cases, can be determined from bound charge distribution and boundary conditions for $$\mathbf P$$).