Suppose we have 2 media with electrical parameters ${\varepsilon _1},\,{\sigma _1}$, respectively ${\varepsilon _2},\,{\sigma _2}$, separated by the plane surface $\Sigma $; electrical charge surface density on $\Sigma $ is ${\rho _s} = 0$.We denote by $\overrightarrow {{E_1}} ,\,\overrightarrow {{E_2}} $ the electric field vectors in the two environments, and by $\overrightarrow {{J_1}} ,\,\overrightarrow {{J_2}} $ the corresponding current density, with ${J_1} = {\sigma _1} \cdot {E_1}$ and ${J_2} = {\sigma _2} \cdot {E_2}$. Let ${\alpha _1},\,{\alpha _2}$ be the angles between the normal n at the surface $\Sigma $ and the vectors $\overrightarrow {{E_1}} ,\,\overrightarrow {{E_2}} $.

By assuming the boundary conditions ${E_{t1}} = {E_{t2}} , {D_{n1}} = {D_{n2}}$, we obtain the the refractive condition of the electric field lines ${\varepsilon _1} \cdot tg\left( {{\alpha _2}} \right) = {\varepsilon _2} \cdot tg\left( {{\alpha _1}} \right)$.

By assuming the boundary conditions ${E_{t1}} = {E_{t2}} , {J_{n1}} = {J_{n2}}$, we obtain the the refractive condition of the electric field ${\sigma _1} \cdot tg\left( {{\alpha _2}} \right) = {\sigma _2} \cdot tg\left( {{\alpha _1}} \right)$; the boundary condition ${J_{n1}} = {J_{n2}}$ is obtained from the electric current continuity equation.

Because the values of ${\varepsilon _1},\,{\sigma _1}$ and ${\varepsilon _2},\,{\sigma _2}$ are arbitrary material parameters, this yield different values of the angle of refraction of the electric field for the two boundary conditions set. What is the explanation of this paradox?

  • $\begingroup$ Remark: all capital nicks look bad. You can change your nick on your profile page. Your acceptance here depends highly on the votes you get; and it is absolutely up to the voter, how do they vote. Btw, your first question looks quite interesting and high-level. $\endgroup$
    – peterh
    Jul 16, 2016 at 17:13
  • $\begingroup$ You need to treat the dielectric and conductive properties of the two materials at the same time and the fields have to satisfy Maxwell's equations. In case of current flow, that's especially true for the continuity equation because charge can not be created or destroyed (at least not by this physical process). If your first set of boundary conditions doesn't satisfy that, then it's wrong. $\endgroup$
    – CuriousOne
    Jul 16, 2016 at 18:51
  • $\begingroup$ @peterh: Thank you for the observations and assessment which you had made. I hope the new form of the profile is better ... $\endgroup$ Jul 17, 2016 at 6:12
  • $\begingroup$ @CuriousOne: I suspected that the boundary conditions must treat simultaneously dielectric and conductive properties of the two media. Unfortunately, the cases I have found in the literature refers to the ideal cases dielectric-dielectric, dielectric-conductor, but not to my case, conductor-conductor. What is the general form of the Maxwell equations, able to handle this case? $\endgroup$ Jul 17, 2016 at 6:12

1 Answer 1


The explanation of your paradox is that the boundary condition Dn1 = Dn2 does not hold in the case of current flowing across the boundary. There is a (free) sheet charge 𝜂 at the interface so that the electric displacement becomes discontinuous Dn2 - Dn2 = 𝜂. The correct normal electric field boundary condition is σ1⋅En1 = σ2⋅En2 as deduced from the normal current continuity Jn1 = Jn2. This causes a discontinuity of the normal dielectric displacement und thus the interface sheet charge 𝜂. The build-up of 𝜂 can be thought of to be caused by different normal interface current densities before the situation settles in steady state. Thus the above second refractive condition of the electric field lines has to be used at interfaces of conducting media.


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