Gauss law when dealing with materials with conductivy Suppose we have a parallel plate capacitor filled with two dielectrics materials, one with conductivity $\sigma_1$ and permittivity $\epsilon_1$ and the other one with conductivity $\sigma_2$ and permittivity $\epsilon_2$. Each dielectric has thickness equal to half of the distance that separates the plates. The capacitor is connected to a battery of potential V. I am asked to find out the electric field between the plates. 
Applying Gauss law, I find that the electric displacement vector inside the capacitor is equal to the superficial charge density, $\sigma$. 
From here, I can calculate $\sigma$, supposing we are dealing with linear dielectrics:
$V = \int_0^\frac{d}{2} \frac{D}{\epsilon_1} dl + \int_\frac{d}{2}^d \frac{D}{\epsilon_2} dl = \frac{\sigma d \left( \epsilon_1 + \epsilon_2 \right)}{2\epsilon_1\epsilon_2} \iff \sigma = \frac{2V\epsilon_1\epsilon_2}{d(\epsilon_1+\epsilon_2)}$
From here I conclude that:
$E_1 = \frac{\sigma}{\epsilon_1} = \frac{2V\epsilon_2}{d(\epsilon_1+\epsilon_2)}$
$E_2 = \frac{\sigma}{\epsilon_2} = \frac{2V\epsilon_1}{d(\epsilon_1+\epsilon_2)}$
Problem is that acoording to my professor the solution to this part of the exercise is:
$E_1 = \frac{2V\sigma_2}{d(\sigma_1+\sigma_2)}$
$E_2 = \frac{2V\sigma_1}{d(\sigma_1+\sigma_2)}$
Which he obtains by imposing boundary conditions and calculating the current densities. 
My question is: why is my procedure wrong? What have I assumed that is not correct?
 A: Your technique is not wrong... if $\sigma_1=\sigma_2=0$. See, if the materials you are working with can conduct current, the parts of the system with free charges will not just be the plates, but any part of the (partially) conducting media can also have free charges brought by the current that flow in these materials. This is your mistaken assumption.
In this problem, the interface between $\epsilon_1,\sigma_1$ and $\epsilon_2,\sigma_2$ can have a free charge density $\sigma'$ since the system in equilibrium can have brought them from either plate.
Assuming this, your potential equation now reads
$$V = \int_0^\frac{d}{2} \frac{D_-}{\epsilon_1} dl + \int_\frac{d}{2}^d \frac{D_+}{\epsilon_2} dl,\ \text{ with }\ D_\pm=(\sigma+\sigma'/2).$$
$$\implies V = \frac{\sigma d \left( \epsilon_1 + \epsilon_2 \right)}{2\epsilon_1\epsilon_2} + \frac{\sigma' d \left( \epsilon_1 - \epsilon_2 \right)}{2\epsilon_1\epsilon_2}.$$
The idea is that this new interface charge density is a free parameter. Its value depends on the current equilibrium state since this is our other boundary condition, or in other words, the system's equilibrium state can only be fully described by both charge and current boundary conditions. Your professor simplifies this by first finding the current equations and seeing that they do not depend on the charge boundary conditions (thus fully explain the, say, voltage and electric fields fully).
(Edit:) A nice way to illustrate this is to assign $\sigma=0$ or $\sigma=\infty$ to one of the parts. For example, in the case where $\sigma_1=0$, there cannot be any current flowing in the first region. This necessarily means that $E_2=0$, since otherwise the current coming from the 2nd plate would always want to flow into the first region. For another example, if $\sigma_2\to\infty$, the second part, again, cannot have any electric fields inside of it since, well, there are no electric field inside (perfect) conductors. These happen regardless of dielectric properties.
