You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you increase the plate separation, do the following quantities increase decrease or stay the same?
i)C
ii)Q
iii) E between the plates
iv)delta V

Explanation:

The capacity C will decrease because it depends on the plate separation . Capacity is inversely proportional to the plate separation.

The charge will remain the unchanged  because , the source of charge is battery which has been removed.

Potential difference V = Charge / capacity

As capacity decreases , V will increase , since charge remains fixed.

electric field

E =  V / d ( d is plate separation )

As V increases and d also increases , E remains the same or unchanged.