The electrostatic force ( F ) between the metal plates of an isolated parallel plate capacitor ( C ) having a charge Q and area A is ( d = distance between the plates)

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The electric field produced by one plate of the capacitor at the location of the other plate is given by $E = \frac{σ}{2 ϵ _{0}} = \frac{Q}{2 A ϵ _{0}}$ .

The electrostatic force $F$ experienced by the other plate having charge $Q$ is $F = QE$ .

Substituting the value of $E$ , we get $F = Q (\frac{Q}{2 A ϵ _{0}}) = \frac{Q ^{2}}{2 A ϵ _{0}}$ .

Alternatively, the potential energy of the capacitor is $U = \frac{Q ^{2}}{2 C} = \frac{Q ^{2} d}{2 A ϵ _{0}}$ .

The magnitude of the force between the plates is $F = \frac{d U}{dd} = \frac{Q ^{2}}{2 A ϵ _{0}}$ .

This expression shows that the force $F$ depends only on the charge $Q$ , area $A$ , and permittivity $ϵ_{0}$ .

Thus, the electrostatic force is independent of the distance $d$ between the plates.

Answer: independent of $d$

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