There is an electric field between the plates because there is a potential difference between the capacitor plates. Namely, one face of the capacitor has charge $+Q$, and the other face has charge $-Q$. The voltage will be given by
$$ V(x) = \frac{Q}{C} = -E \ x . $$
The way to show this is to first use Gauss's law with the "pillbox". Each plate gives a contribution of
$$ E_{plate} = \frac{\sigma}{2\epsilon_0} $$
so that when you add them you get that
$$ E_{total} = \frac{\sigma}{\epsilon_0} $$
where $\sigma = \frac{Q}{A}$ where $Q$ is the charge induces on the capacitor and $A$ is the area of the plate.
To relate this to your question of $\nabla \cdot\mathbf{E}$, note that this is just the differential form of Gauss's law. Namely we have that
$$\iint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{1}{\epsilon_0} \iiint_V \nabla \cdot E \ dV = \frac{1}{\epsilon_0}\iiint_V \rho(\mathbf{r}) \ dV \equiv \frac{Q_{nec}}{\epsilon_0} $$
I could go through the derivation of the electric field of a parallel plate capacitor, but these are easier to find than a parking space in Arkansas.