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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:
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Verified Answer
The correct answer is:
$\frac{\mathrm{CV}}{\varepsilon_0}$
From the Gauss's law
$\begin{aligned}
& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{Q}}{\epsilon_0} \quad \because \mathrm{Q}=\mathrm{CV} \\
& \therefore \phi=\frac{\mathrm{CV}}{\epsilon_0}
\end{aligned}$
$\begin{aligned}
& \phi=\frac{\mathrm{q}_{\text {in }}}{\epsilon_0}=\frac{\mathrm{Q}}{\epsilon_0} \quad \because \mathrm{Q}=\mathrm{CV} \\
& \therefore \phi=\frac{\mathrm{CV}}{\epsilon_0}
\end{aligned}$
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