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The electric polarization $P$ is electric dipole moment per unit volume. A polarized body acts as the bound volume and surface charge densityIndeed, summing the scalar potential of its dipoles and integrating by parts givesThis is exactly the potential of ${\rho }_{\mathrm{b}}$ and ${\sigma }_{\mathrm{b}}$.

For the central free charge, Gauss's law for the electric displacement field givesThus on the sphere,

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume. It producesThe electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ is useful becauseso it separates free charge from the bound charge represented by $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{D}=ϵ\mathbf{E}$.

Spherical symmetry and Gauss's law for $\mathbf{D}$ give, on both sides of the interface,ThusWith $\mathbf{n}={\mathbf{e}}_r$ pointing from medium one to medium two, the net bound sheet charge at $r=R$ is the polarization jumpIf one assigns surface charge separately to the inner body and outer medium, their contributions are ${\mathbf{P}}_1\cdot \mathbf{n}$ and $-{\mathbf{P}}_2\cdot \mathbf{n}$; their sum is the displayed physical interface charge.