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:

By Gauss's law, the electric flux through any closed surface equals the total charge enclosed divided by $$\varepsilon_0$$:

$$\Phi = \frac{Q_{enclosed}}{\varepsilon_0}$$

The positive plate of the capacitor carries charge $$Q = CV$$.

Since the closed surface encloses only the positive plate:

$$\Phi = \frac{CV}{\varepsilon_0}$$

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