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
Answer & explanation
Correct answer: option 1
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.