The separation between the plates of a charged parallel plate capacitor is increased. Which of the following quantities will change ?

In case of isolated capacitor, charge on it remains same, while capacitance decreases with increase in separation between the plates, As ,$q=VC$ and so $V$will increase. The electric field between the plates will not charge and so energy density, $\frac{1}{2}{\in }_0{E}^2$remains the change. Also U = $\frac{Q^2}{2C}. Since C decreases , U increases .$