Maxwell's equations not satisfied simultaneously? In an exercise in Electromagnetism, it is stated that $\vec{E}$ is uniform between the plates of a capacitor, and is given by
$$\vec{E} = \frac{Q_0}{\epsilon_0 S} \cos(\omega t)  \cdot \hat{z}$$
Supposing that the media between the plates is non-conductive, $\sigma = 0$, then $\vec{J} = \sigma \cdot \vec{E} = \vec{0}$.
From here we have the following 2 conclusions from the differential form of Maxwell's equations:

*

*$$\frac{\partial \vec{B}}{\partial t} = \vec{0}$$ from Faraday's law ,since curl $\vec{E} = \vec{0}.$

*$$ \text{rot}\vec{B} = -\omega \frac{\mu_0 Q_0}{S} \sin(\omega t) \cdot \hat{z},$$ from rot$\vec{B} = \mu_0 \vec{J} + \frac{1}{c^2} \frac{\partial \vec{E}}{\partial t}$ since $\vec{J} = \vec{0}$.

Is it possible to interchange the partial derivative with respect to time with the rot operator? If so, why? Therefore can both 1 and 2 be true simultaneously? Because then, if 2 is differentiated with respect to time, 1 cannot be true.
Thanks in advance.
 A: Yes, it is possible to interchange the time derivative with the curl (or rot) operator. This is just a consequence of Schwarz's theorem and the fact that the expression you gave is twice continuously differentiable.
As for your paradox, the issue is with the hypothesis that $\vec{J} = \vec{0}$. Notice that for one to get the oscillating field in the middle of the capacitor, the charge in the plates of the capacitor must be changing, hence $\frac{\partial \rho}{\partial t} \neq 0$. However, conservation of charge demands $\frac{\partial \rho}{\partial t} + \mathrm{div} \vec{J} = 0$. Therefore, we conclude that $\mathrm{div} \vec{J} \neq 0$, which would be absurd if $\vec{J} = \vec{0}$ everywhere.
If the medium between the capacitor is non-conductive, there will be no current in there, but there must be current somewhere else for the charge in the plates of the capacitor to be changing. If there is no current, then the problem you proposed does not exist for $\omega \neq 0$, for it disrespects charge conservation.
Edit: as mentioned in the comments, the electric field will be more complicated than the one given in the question. The time derivatives of the charge and current densities will contribute to the electric and magnetic fields and give additional terms, as one can see from Jefimenko's equations.
