Given, parallel plate is charged and then isolated. Therefore,Q = constant.
Now, capacitance is given as C
=⇒C∝ So, if the separation between plates is increased then, C decreases.
Also, we know
Q=CV⇒V=⇒V∝ So, if C decreases then potential increases.
Thus, on increasing the plate separation charge remains constant, potential increases and capacitance decreases.