homework and exercises - Electric field for a partially inserted dielectric slab in a parallel plate capacitor

I’m working on a problem that I think I solved incorrectly yet I don’t know where my mistake is. Suppose we partially insert a dielectric slab (with constant $\kappa$) inside a parallel plate capacitor (in a vacuum) where the area of both plates is $A$. Let $A_D$ be the area that the dielectric occupies and let $A_v$ be the area the vacuum occupies $\Rightarrow A = A_v + A_D$.

Now using Gauss Law in both cases we get that $$D_v = \frac{q}{(A - A_D)}, \ D_D = \frac{q}{A_D}$$

Then $D_v = \varepsilon_0 E_v$ and $D_D = \kappa \varepsilon_0 E_D$ so we know the electric field for both cases. As we can see if $A_D = 0$ and $A_D = A$ we get what we should for $E_v$ and $E_D$ respectively.

Yet I think my answer is wrong, we should expect the electric field to be zero for $E_v$ and $E_D$ when $A_D = A$ and $A_D = 0$ respectively because in those cases either the dielectric is completely inside the capacitor or it is completely outside the capacitor, however this doesn’t happen and they both diverge to infinity in their respective cases. What am I missing here?

Answer

Sometimes, it is good to fall back on the basics. The plates are equipotential surfaces. Therefore, the electric field between them is still uniform and identical for both regions. However, what's different is the electric displacement and the surface charge densities of the two regions. The charges are no longer uniformly distributed on the plates. We can use Gauss' Law to find the electric displacements and verify that they result in the same electric field for both regions. $$\oint \mathbf{D} \cdot d \mathbf{a} = Q_f$$