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(A) For an isolated parallel plate capacitor,the charge $Q$ on the plates remains constant.The electric field $E$ produced by one plate at the locatio

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Independent of the distance between the plates

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(A) For an isolated parallel plate capacitor,the charge $Q$ on the plates remains constant.The electric field $E$ produced by one plate at the location of the other plate is given by $E = \frac{\sigma}{2\varepsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.The electrostatic force $F$ experienced by one plate due to the electric field of the other is $F = Q 	imes E$.Substituting the expression for $E$,we get $F = Q 	imes \frac{Q}{2A\varepsilon_0} = \frac{Q^2}{2A\varepsilon_0}$.Since $Q$,$A$,and $\varepsilon_0$ are constants,the force $F$ is independent of the distance $d$ between the plates.

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

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