= Solution
An applied <electric field> displaces <bound charges> or reorients permanent dipoles, producing <electric polarization> $\mathbf P$, the dipole moment per unit volume. It gives <bound charges> $\rho_b=-\nabla\cdot\mathbf P$ and $\sigma_b=\mathbf P\cdot\mathbf n$. Define
$$
\mathbf D=\epsilon_0\mathbf E+\mathbf P,\qquad
\mathbf P=\epsilon_0\chi_e\mathbf E,\qquad
\epsilon=\epsilon_0(1+\chi_e)
$$
for a \b[linear isotropic <dielectric>]. Then $\nabla\cdot\mathbf D=\rho_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational <electric polarization> respond on different timescales; real <permittivity> is <frequency> dependent, and its dissipative part describes <dielectric loss>.
A uniform <dielectric> filling a <parallel-plate capacitor> raises its <capacitance> to $\epsilon A/d$. At fixed charge this reduces field <energy> $Q^2/(2C)$; at fixed <voltage> the stored <energy> $CV^2/2$ increases because the <voltage> source supplies charge and <energy>. Insulation and <energy> storage are useful applications, limited by loss and <dielectric breakdown>. Static interface conditions are continuous tangential $\mathbf E$ and a jump in normal $\mathbf D$ equal to free surface charge.
Back to article page
