Given: a capacitor of capacitance C charged to a potential V

Gauss’s law:

Electric flux ϕ) through a closed surface S is given by

Where,

Q is the charge enclosed by S

ε_o is the permittivity of the free space

According to the gauss law

Electric flux,

Where

Q is the total charge enclosed in the gaussian surface

ε_o is the absolute permittivity of the vacuum

Since, the two plates of capacitor contains equal and opposite charges

∴ Total charge enclosed by the surface ⇒ Q-Q=0

Putting the values of total charge in gauss law, we get

∴ the electric flux through the closed surface enclosing the capacitor=0.