Consider Laplace's equation in a rectangle with specified boundary conditions. This problem is solved when $\epsilon_1 = \epsilon_2$ in the following link. $$ \nabla \cdot \epsilon \nabla V=0$$
What if the $\epsilon_1$ and $\epsilon_2$ be different. We can still use the fact that at $x=a/2$, $V$ would be constant. However, its value is not $(V_1+V_2)/2$ since the symmetry is broken. Can we use the continuity of $dV/dx$ to find the value of V at $x=a/2$?