Quick question about perpendicular electric field discontinuity In this boundary condition:
$$E_{\rm above}^\perp-E_{\rm below}^\perp=\frac{\sigma}{\varepsilon_0}$$
When dielectrics are involved, does $\sigma$ refer to free charge density or total (free + bounded) charge density? I can't find this information anywhere. Maybe $\sigma$ could be total charge density but since the net bound charge is zero, it is equal to free charge density?
 A: The condition of the discontinuity of the normal component of the electric field can be derived as a consequence of Gauss' law that, in the case of macroscopic fields, is expressed in terms of a surface integral of the normal component of the field ${\bf D}$:
$$
\oint_{\Sigma}  D^\perp  d{ S}=q_{free}\tag{1}
$$
Where $\Sigma$ is a closed boundary surface of a volume containing the total free charge $q$ (not including bounded charges).
Application of equation $(1)$ to a small cylinder with sides orthogonal to the  surface where the surface density $\sigma$ is confined allows to get the discontinuity condition
$$D_{\rm above}^\perp-D_{\rm below}^\perp=\sigma$$
where $\sigma$ has to be the surface density of free charges. In the vacuum, it becomes
$$E_{\rm above}^\perp-E_{\rm below}^\perp=\frac{\sigma}{\varepsilon_0}$$. In general, it should be written as
$$\varepsilon_{\rm above}E_{\rm above}^\perp-\varepsilon_{\rm below}E_{\rm below}^\perp=\sigma$$.
A: $\sigma$ here refers to the surface charge density. See Interface conditions for electromagnetic fields.
