= Solution
An <electrical insulator> has no freely conducting bulk charge in the ideal static description, but an applied <electric field> can displace bound positive and negative charge relative to one another. The resulting <electric dipole moments> per unit <volume> define the <polarization density> $\mathbf P$. Distortion of electronic charge clouds, relative displacement of <ions>, and partial alignment of permanent molecular <electric dipoles> can all contribute. Thermal agitation opposes the last mechanism.
For a weak field in a linear <isotropic> <dielectric>, the constitutive law is
$$
\mathbf P=\epsilon_0\chi_e\mathbf E,\qquad
\mathbf D=\epsilon_0\mathbf E+\mathbf P=\epsilon\mathbf E,\qquad
\epsilon=\epsilon_0(1+\chi_e)=\epsilon_0\epsilon_r.
$$
Here $\chi_e$ is the <electric susceptibility>, $\epsilon$ is the <permittivity>, and $\mathbf D$ is the <electric displacement field>. In an <anisotropic> material, $\chi_e$ is a <tensor> and $\mathbf P$ need not be parallel to $\mathbf E$. These response laws describe the macroscopic field, not necessarily the microscopic field acting on one <molecule>.
Spatial variation and surface termination of the <polarization density> give <bound charge and bound current> through the bound charge densities
$$
\rho_b=-\nabla\cdot\mathbf P,\qquad \sigma_b=\mathbf P\cdot\hat{\mathbf n},
$$
where $\hat{\mathbf n}$ is outward from the material. Thus uniform <polarization density> has no bulk bound charge, but generally has opposite bound charges on opposing surfaces. <Gauss's law> can be rewritten as
$$
\nabla\cdot\mathbf D=\rho_{\rm free}.
$$
With no <free surface> charge, the normal component of $\mathbf D$ is continuous across an interface; the tangential <electric field> is continuous in <electrostatics>.
For example, fill a <parallel-plate capacitor> of plate area $A$ and gap $d$ with a uniform <dielectric>, neglecting fringing. At fixed free plate charge $Q$, $D=Q/A$, so
$$
E=\frac{Q}{\epsilon A},\qquad V=Ed,\qquad
\boxed{C=\frac{\epsilon A}{d}=\epsilon_rC_0.}
$$
The induced bound surface charges oppose the original field, reducing the <electric field> and potential difference at fixed $Q$. At fixed applied potential difference, the field remains $V/d$ and the battery instead supplies more free charge. The <electric displacement field> separates these free-charge and bound-charge roles.
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