A capacitor of capacitance C is connected to a cell of emf V and when fully charged, it is disconnected. Now the separation between the plates is doubled. The change in flux of electric field through a closed surface enclosing the capacitor is

Zero

Flux = $\frac{q_{incl}}{\varepsilon_0}$

The two plates of the capacitor have equal and opposite charges.

Hence, total charge enclosed by the given surface = 0

∴ Flux is zero is both conditions.

Hence change in flux = 0.