1
$\begingroup$

I'm studying from Reitz's Foundations of Electromagnetic Theory and trying to understand how the results obtained under electrostatic conditions are affected when there is a current flowing through material.

I understand that it isn't true anymore that the electric field inside a conductor is zero if there is a current going through the conductor. And I think (please correct me if I'm mistaken) that because of that, now the electric field doesn't need to be perpendicular to the surface of the conductor. So I guess this may affect the boundary conditions, but how are they affected?

Specifically, how do the electric displacement and the polarization behave when there is a current flowing from one material with some permitivity $\epsilon_1$ and conductivity $g_1$ to another one with permitivity $\epsilon_2$ and conductivity $g_2$? Is it still true that a free charge density is accumulated at the surface between both materials?

I mean, is this boundary condition: $$(\vec{D_1}-\vec{D_2}) \hat n =\sigma$$ still valid?

I'm quite confused, because I don't get what it would mean to have a surface charge when all the charge is actually flowing.

$\endgroup$
1
$\begingroup$

In the case of an interface of dielectrics with conductivity the application of an electric field leads to the build up of a free interface charge which modifies the normal fields so that current continuity holds at the interface, i.e., $$\vec{J_1}\hat n = g_1\vec{E_1}\hat n=g_2\vec{E_2}\hat n = \vec{J_2}\hat n$$ Your boundary condition for the interface charge is correct. See my answers to the similar questions here: Free charge in a dielectric and Surface charge density in conducting plate .

Note: The interface charge builds up in the first moment upon applying an electric field to the interface due to the different conductivities and at first equal fields. Very quickly a steady state situation arises where you have a stationary interface charge and equal current densities entering and leaving the interface.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.