Question 11 of 180

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:

A$\dfrac{CV}{{\epsilon }_{0}}$BZeroC$\dfrac{2CV}{{\epsilon }_{0}}$D$\dfrac{CV}{2{\epsilon }_{0}}$

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Answer: (A) $\dfrac{CV}{{\epsilon }_{0}}$

The charge enclosed is given by ${q}_{inc}=CV$
Gauss' law states that the flux through any closed surface is equal to the charge enclosed by the surface.
Using Gauss' law the flux is, $\phi =\dfrac{{q}_{\text{inc}}}{{\epsilon }_{0}}=\left(\dfrac{CV}{{\epsilon }_{0}}\right)$