Paradox of two plates which are supposed to be equipotential Let's have large two conductor plate of area $A$  and are filled with two adjacent dielectrics of the same size $\epsilon_{1}$ and $\epsilon{2}$. The charge of the condensers are fixed by $Q$. Analyzing using Gauss Law, the displacement field is the same in the two media, so the electric field is different for each material. So if I measured the potential difference in the plates for each media, it's going to be different. But the conductor is an equipotential, so how can it be possible?

 A: The answer is: the potential won't be different.
The misunderstanding is that in order to think this way you are assuming the charge Q will be uniformely distributed across each plate. This is what will happen:

*

*Since the total charges are fixed, when inserting the dielectrics, the charge densities will be: $\sigma_1 = \dfrac{Q_1}{A/2}$ and $\sigma_2 = \dfrac{Q-Q_1}{A/2}$.

*The displacement fields inside the capacitor will be: $D_1 = \sigma_1$ and $D_2 = \sigma_2$.

*The electric field is uniform by hipothesis in the electrostatic regime, because otherwise charge would flow until electrostatic regime is reached: $E_1 = E_2 = \frac{\sigma_1}{\epsilon_1} = \frac{\sigma_2}{\epsilon_2}$

*There is a charge imbalance in both sides. Solving the system:
$$\frac{Q_1}{\epsilon_1} = \frac{Q-Q_1}{\epsilon_2} \rightarrow Q_1 = \dfrac{Q}{1+\epsilon_2/\epsilon_1}$$

*The charge-density imbalance is: $\sigma_1 = \dfrac{Q}{({1+\epsilon_2/\epsilon_1})(A/2)}$ and $\sigma_2 = \dfrac{Q}{({1+\epsilon_1/\epsilon_2})(A/2)}$.

Note added:
If you compute the polarization charge in each dielectric, you will see the total charge density (metal + polarization) remain uniform.
