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A closed surface encloses a net charge of $4\ \mu$C. The total electric flux through the surface is:
A$4\ \mu$C, ignoring the $\varepsilon_0$ in the law
B$4.52\times 10^5$ N m$^2$/C, by $q/\varepsilon_0$
CZero, since flux through a closed surface is always zero
D$36\pi\times 10^9$ N m$^2$/C, ignoring the charge value
Answer & Solution
Correct answer: B. $4.52\times 10^5$ N m$^2$/C, by $q/\varepsilon_0$
$\Phi = q/\varepsilon_0 = 4\times 10^{-6}/8.85\times 10^{-12} \approx 4.52\times 10^5$ N m$^2$/C.
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