When is the free charge density zero at the boundary of dielectrics It is known that across the interface of two different dielectrics, the electric displacement field must satisfy
$$(\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{\hat{n}}=\sigma$$
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?
 A: 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. 
A: 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:



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.


*

*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!

*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!
