Note: In the following, I'll be using Lorentz-Heaviside units. For other unit systems, throw in factors of ε_0 or 4π as needed... The reason for the introducti

Note: In the following, I'll be using Lorentz-Heaviside units. For other unit systems, throw in factors of ε_0 or 4π as needed... The reason for the introduction of D is that even though a medium may have no net charge, it's still made from protons and electrons, meaning the charge density will be nonzero. We call this the density of 'bound' charges ρ_b. The medium is often modelled as consisting of tiny dipoles. If left to their own devices, the dipoles tend to align in such a way that the generated field largely cancels out. Now, let's add additional charges via a density of 'free' charges ρ_f, yielding a total charge density ρ = ρ_f + ρ_b. The bound charges will respond to the presence of the free charges, and the field they generate will no longer cancel out. We call this field -P, with P the electric polarization (the negative sign comes in through its definition as the dipole moment density of the dipole model). Defining the electric displacement field as D = E + P, Gauss's law div(E) = ρ can be decomposed into div(D) = ρ_f and div(P) = -ρ_b In many cases, the response of the medium will be approximately linear, and polarization will be proportional to the field generated by the free charges, P = η·D. Note that the definition of P was such that for positive η, the actual field will be opposing D. Equivalently, we can consider the response linear in the total field, P = χ·E if we introduce the electric susceptibility χ = η/(1-η). For the displacement field, we get D = E + P = (1 + χ)·E = ε·E with ε the permittivity. What's the point, you may ask? Simplification, as we can now just measure the linear response of the medium, instead of having to figure out the microscopic distribution of bound charges.