Dielectrics and Polarisation

The Field Inside Matter

Put a dielectric in a field and its molecules acquire aligned dipole moments. The polarisation $\mathbf{P}$ is the dipole moment per unit volume, and it is equivalent to a set of bound charge densities:

$$\sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}}, \qquad \rho_b = -\nabla \cdot \mathbf{P}$$

These are real charges, they are simply not free to move through the material.

Because bound charge is awkward to track, electrostatics introduces the displacement field

$$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}$$

whose divergence counts only the free charge: $\nabla \cdot \mathbf{D} = \rho_f$. That is Gauss's law in a form you can actually apply, since free charge is what you put there deliberately.

For a linear isotropic medium the response is proportional to the field:

$$\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}, \qquad \mathbf{D} = \varepsilon_0 \varepsilon_r \mathbf{E}$$

with $\varepsilon_r = 1 + \chi_e$ the relative permittivity. Since $\chi_e > 0$ for every ordinary dielectric, $\varepsilon_r > 1$ always.

The practical consequence is that a dielectric weakens the field inside it. Insert one into a charged capacitor at fixed charge and the field drops by $\varepsilon_r$, so the voltage drops and the capacitance rises by the same factor. The bound surface charge sits antiparallel to the free charge on the plates and partly cancels it.

Common pitfall: treating $\mathbf{D}$ as "the real field". $\mathbf{E}$ is what exerts force on a charge; $\mathbf{D}$ is a bookkeeping device that hides bound charge so Gauss's law stays usable. Neither is more fundamental, they answer different questions.