Q 12-01-083JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
A capacitor of capacitance $C$ is charged to a potential $V$. The flux of the electric field through a closed surface enclosing the positive plate of the capacitor is
Answer: (A) $\dfrac{CV}{\epsilon_0}$
The enclosed charge is $Q=CV$, so by Gauss's law $\phi=\dfrac{CV}{\epsilon_0}$.
Solution by Sreeraj P, M.Sc Physics