The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is: -
(1) independent of the distance between the plates.
(2) linearly proportional to the distance between the plates
(3) proportional to the square root of the distance between the plates.
(4) inversely proportional to the distance between the plates.

Hint: We are given a parallel plate capacitor. We know that the two plates are oppositely charged, the charges are equal in magnitude but opposite in polarity. and when connected to a source of emf, it conducts electricity but after it gets fully charged it stops, so there must be presence of an electric field.

Complete step by step answer:
Let q be the charge on each plate, the electric force is given as \[F=qE\]. Also, we know the electric fixed between two charged plates is given as \[E=\dfrac{\sigma }{2{{\varepsilon }_{0}}}\]
So, force becomes, \[F=\dfrac{q\sigma }{2{{\varepsilon }_{0}}}\]. Now we know \[\sigma \]is surface charge density, expanding it we get, \[F=\dfrac{{{q}^{2}}}{2A{{\varepsilon }_{0}}}\]
Thus, from the above equation, Electrostatic force is independent of distance between plates.

So, the correct answer is “Option 1”.

Additional Information:
Apart from parallel plate capacitors, there exist a class of capacitors which are cylindrical in shape. A cylindrical capacitor consists of a hollow or a solid cylindrical conductor surrounded by another concentric hollow spherical cylinder. Here the length is the same for both but the radius is different. The charges are equal in magnitude but opposite in polarity.

Note:
While using the surface charge density the total charge is divided by the total area. We know for a conductor any excess charge resides on the surface of the conductor and there is no charge in the vicinity of the conductor. That's why the electric field inside a conductor is zero.