It is known that across the interface of two different dielectrics, the electric displacement field must satisfy


where $\sigma$ is free surface density charge in the boundary.

My question is: if both materials are dielectrics (i.e. thay have no free charge), how could $\sigma$ (which is free charge indeed) appear at the boundary?

  • $\begingroup$ IF there is a free charge $\sigma$ then the displacement is discontinuous. That equation does not imply that there MUST be free charges all the time at the boundary. If there isnt any, then the normal component of electric displacement is continuous $\endgroup$ – Prasad Mani Oct 9 '16 at 15:57
  • $\begingroup$ @PrasadMani Yes, I wanted to know in which cases could that charge appear. Any example would be appreciated. $\endgroup$ – Tendero Oct 9 '16 at 15:59

In dielectrics with different permittivities but no conductivity, there will be no free charge at the interface upon application of en electric field. However, if the dielectrics also possess different conductivities, which leads to a current flowing across the interface, in general, a free interface charge will accumulate at the interface so that the stationary normal electric currents (produced by the normal electric fields together with the conductivities) fulfill the current continuity condition. If the conductivities are equal, there will be no interface charge generation.

  • $\begingroup$ Thanks for the answer. I get that from the math, but I don't understand it qualitatively. Where does the free charge come from if neither of the dielectrics have any? And what about charge conservation? There is charge accumulating there... isn't that a problem from the conservation standpoint? $\endgroup$ – Tendero Oct 9 '16 at 16:06
  • 1
    $\begingroup$ The free charge is supplied by the current that come from the metal electrodes. No there isn't any problem with charge conservation the interface charge accumulating at the interface comes from the electrodes and, according to Gauss law, equal and opposite charges are induced on the metal electrodes so that the total charge (including the electrodes) is zero. $\endgroup$ – freecharly Oct 9 '16 at 16:09

What you considered is right. Actually, For electrostatic fields, between 2 dielectrics, no free charge will be lying there at the boundary, so the free charge density $\sigma_0$ (or $\sigma_f$, some common denotations) is 0, i.e. $$ (\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{\hat{n}} = \sigma_0 = 0 $$ which is right what we used as one of the boundary conditions.

Important notes:

  • It works for both dielectrics and conductors. The perpendicular continuity of $\mathbf{D}$ is for dielectrics. So when it comes to conductors, don't forget the free charge. In fact, in many problems given a free charge density, you can use the formula to obtain the $\mathbf{D}$ conveniently.
  • When currents exist, there could be additional free charge at the boundary between dielectrics (to satisfy the continuity of currents), which means $\sigma_0$ is not necessarily 0 even there are only dielectrics, as @freecharly proposed. To clarify in detail, I'd like to give an example here:

A parallel-plate capacitor with voltage applied

Now consider a parallel-plate capacitor with voltage U applied on the plates. Parameters like thickness d, dielectric constant $\epsilon$, conductivity $\sigma$ (not to confused with charge density!), are given. Use the boundary condition we've been talking about, we will get the free charge density $$ \sigma_0 = D_2 - D_1 = \epsilon_2 E_2 - \epsilon_1 E_1 $$ Using Ohm's law and by some simple calculation, we eventually get $$ \sigma_0 = \frac{\epsilon_2 \sigma_1 - \epsilon_1 \sigma_2}{\sigma_1 d_2 + \sigma_2 d_1}U $$

Now let's see if it works just as we analyzed.

  1. The free charge at the boundary is proportionally linked to voltage U, implying that there should be no free charge without applied U (no currents then), as expected!
  2. While U being applied, it's possible to let the $\sigma_0$ remain 0 as long as the numerator is 0, i.e. $$ \epsilon_2 \sigma_1 = \epsilon_1 \sigma_2 \tag{$*$} $$

So I'd have to point out that to let conductivities be equal is not enough for zero interface free charge, but should've been ($*$). In fact, this result can be directly derived by currents continuity at a boundary and Ohm's law, without considering a specific example above. But an example is intuitive. :)

I hope the example and its discussion answered right what you asked about. That's a common question for learners, including myself!


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.