Interface conditions for electromagnetic fields

${\displaystyle \mathbf {n} _{12}\times (\mathbf {E} _{2}-\mathbf {E} _{1})=\mathbf {0} }$

where:
${\displaystyle \mathbf {n} _{12}}$ is normal vector from medium 1 to medium 2.

Therefore, the tangential component of E is continuous across the interface.

Outline of proof from Faraday's law

We begin with the integral form of Faraday's law: ${\displaystyle \oint _{\partial \Sigma }\mathbf {E} \cdot d{\boldsymbol {\ell }}=-\int _{\Sigma }{\frac {\partial \mathbf {B} }{\partial t}}\cdot d\mathbf {A} }$ Choose ${\displaystyle \Sigma }$ as a small square across the interface. Then, have the sides perpendicular to the interface shrink to infinitesimal length. The area of integration now looks like a line, which has zero area. In other words: ${\displaystyle \lim _{linelength\to 0}\mathbf {A} =0}$ Since ${\displaystyle \partial \mathbf {B} /\partial t}$ remains finite in this limit, the whole right hand side goes to zero. All that is left is: ${\displaystyle \oint _{\partial \Sigma }\mathbf {E} \cdot d{\boldsymbol {\ell }}=0}$ Two of our sides are infinitesimally small, leaving only ${\displaystyle \int _{medium1}\mathbf {E} \cdot d{\boldsymbol {\ell }}+\int _{medium2}\mathbf {E} \cdot d{\boldsymbol {\ell }}=0}$ Assuming we made our square small enough that E is roughly constant, its magnitude can be pulled out of the integral. As the remaining sides to our original loop, the ${\displaystyle d{\boldsymbol {\ell }}}$ in each region run in opposite directions, so we define one of them as the tangent unit vector ${\displaystyle {\boldsymbol {t}}}$ and the other as ${\displaystyle -{\boldsymbol {t}}}$ ${\displaystyle (\mathbf {E} _{2}\cdot {\boldsymbol {t}})l-(\mathbf {E} _{1}\cdot {\boldsymbol {t}})l=\mathbf {0} }$ After dividing by l, and rearranging, ${\displaystyle (\mathbf {E} _{2}-\mathbf {E} _{1})\cdot {\boldsymbol {t}}=\mathbf {0} }$ This argument works for any tangential direction. The difference in electric field dotted into any tangential vector is zero, meaning only the components of ${\displaystyle \mathbf {E} }$ parallel to the normal vector can change between mediums. Thus, the difference in electric field vector is parallel to the normal vector. Two parallel vectors always have a cross product of zero. ${\displaystyle \mathbf {n} _{12}\times (\mathbf {E} _{2}-\mathbf {E} _{1})=\mathbf {0} }$

${\displaystyle (\mathbf {D} _{2}-\mathbf {D} _{1})\cdot \mathbf {n} _{12}=\sigma _{s}}$

${\displaystyle \mathbf {n} _{12}}$ is the unit normal vector from medium 1 to medium 2.
${\displaystyle \sigma _{s}}$ is the surface charge density between the media (unbounded charges only, not coming from polarization of the materials).

This can be deduced by using Gauss's law and similar reasoning as above.

Therefore, the normal component of D has a step of surface charge on the interface surface. If there is no surface charge on the interface, the normal component of D is continuous.

${\displaystyle (\mathbf {B} _{2}-\mathbf {B} _{1})\cdot \mathbf {n} _{12}=0}$

where:
${\displaystyle \mathbf {n} _{12}}$ is normal vector from medium 1 to medium 2.

Therefore, the normal component of B is continuous across the interface (the same in both media). (The tangential components are n the ratio of the permeabilities)

${\displaystyle \mathbf {n} _{12}\times (\mathbf {H} _{2}-\mathbf {H} _{1})=\mathbf {j} _{s}}$

where:
${\displaystyle \mathbf {n} _{12}}$ is the unit normal vector from medium 1 to medium 2.
${\displaystyle \mathbf {j} _{s}}$ is the surface current density between the two media (unbounded current only, not coming from polarisation of the materials).

Therefore, the tangential component of H is continuous across the interface if there's no surface current present The normal components of H in the two media are in the ratio of the permeabilities.

There are no charges nor surface currents at the interface, and so the tangential component of H and the normal component of D are both continuous.

There are charges and surface currents at the interface, and so the tangential component of H and the normal component of D are not continuous.