The electrostatic force $F$ between two point charges $q_1$ and $q_2$ changes as the distance $d$ between the charges changes as seen from Coulomb's law: $$F=\frac{q_1q_2}{4\pi \varepsilon_0 d^2}.$$
However, the electrostatic force between two parallel plates with constant charge is constant regardless of distance. How can this be?
For reference, the electrostatic force between two parallel plates of area $A$ separated by distance $d$ and holding charge $Q$ is $$ \begin{align} F &=\frac{QE}{2}=\frac{QV}{2d} \end{align},$$ where $E$ is the electric field and $V$ is the voltage difference between the plates [Ref1, Ref2]. For a parallel plate capacitor (as these two plates are), $$V=\frac{Q}{C} \text{ and } C= \frac{\varepsilon A}{d},$$ where $C$ is capacitance and $\varepsilon$ is permittivity so $$F=\frac{Q^2}{2\varepsilon A}.$$