The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A is 

As we know, the total work in transferring a charge to a parallel plate capacitor is given as

                               $\mathrm{W}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}$                                                   … (i)

We can also write a relation for work done as, $\mathrm{W}=\mathrm{F}·\mathrm{d}$         … (ii)
From Eqs. (i) and (ii), we get

                               $\mathrm{W}=\frac{{\mathrm{Q}}^2}{2\mathrm{C}}=\mathrm{Fd}$

$\Rightarrow$                          $\mathrm{F}=\frac{{\mathrm{Q}}^2}{2\mathrm{Cd}}$                                                  … (iii)

As, the capacitance of a parallel plate capacitor is given as

                               $\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

Substituting the value of C in Eq. (iii), we get

                               $\mathrm{F}=\frac{{\mathrm{Q}}^2\mathrm{d}}{2{\mathrm{\epsilon }}_0\mathrm{Ad}}=\frac{{\mathrm{Q}}^2}{2{\mathrm{\epsilon }}_0\mathrm{A}}$

Thus, it means electrostatic force is independent of the distance between the plates.