According to Gauss's theorem in electrostatics, total electric flux...

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According to Gauss's
theorem in electrostatics, total electric flux over a closed surface \( S \) in vacuum is \( 1 / \epsilon_{0} \) times the total charge \( (Q) \) contained inside \( S \)
i.e., \( \quad \phi_{E}=\oint_{s s} \vec{E} \cdot \overrightarrow{d s}=\frac{Q}{\epsilon_{0}} \)
The charges enclosed may be distributed any way. If the medium surrounding the charge has a dielectric constant \( K \). then
\[
\phi_{E}=\frac{Q}{\epsilon}=\frac{Q}{K \in_{0}}
\]
Charges situated outside the surface make no contribution to electric flux.

A cube of side 1 metre encloses a charge of 1 coulomb in vacuum. What is the electric flux from any one surface of the cube ?
(a) \( \frac{1}{\epsilon_{0}} \)
(b) \( \frac{1}{6 \epsilon_{0}} \)
(c) \( \frac{6}{\epsilon_{0}} \)
(d) \( \frac{\epsilon_{0}}{6} \)
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