Let's assume the following for conceptual simplicity:
The plate is an insulator with uniform surface charge density $\sigma$.
The electric field $\mathbf E(\mathbf x)$ is uniform, so there exists some vector $\mathbf E_0$ for which $\mathbf E(\mathbf x) = \mathbf E_0$ for all $\mathbf x$.
In both cases, the force on the plate $P$ is given by
\mathbf F = \int_P dA \sigma \, \mathbf E(\mathbf x)
where $dA$ is a surface area element on the plate. Since the electric field was assumed constant, it comes out of the integral, and the remaining integral just gives the total charge $Q$ on the plate, so we get
\mathbf F = Q\mathbf E_0
The result is the same in both cases! However, if the electric field were not uniform, or if the charge density on the plate were not uniform, then in general the results would have been different.
Hope that helps!