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

Explanation:

The electric force 'F' experienced by the charge 'q' is equal to the product of the magnitude of charge 'q' and the electric field 'E'.It is written as;

and electric field in isolated plate capacitor C is equal to;

Here,  the surface charge density,  is the permittivity of free space.

As we know that force experienced in the charge 'q' is,

  ----(1)

where electric field,  

Now, putting the value of an electric field in equation (1) we get;

 ----(2)

Now, surface charge density  is written as;

Where Q is the charge and A is the area. Putting this value in equation (2) we get;

 ----(3)

In equation (3) we can see that the F is Independent of the distance between plates. 

Hence, option 1) is the correct answer.