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By Gauss's law, the total electric flux through a closed surface enclosing a net charge $Q_{enc}$ is

A$Q_{enc}\,\varepsilon_0$
B$\dfrac{Q_{enc}}{\varepsilon_0}$
C$\dfrac{\varepsilon_0}{Q_{enc}}$
D$\dfrac{Q_{enc}}{4\pi\varepsilon_0}$
Answer & Solution
Correct answer: B. $\dfrac{Q_{enc}}{\varepsilon_0}$
1. Gauss's law: $\displaystyle\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{enc}}{\varepsilon_0}$. 2. The flux depends only on the **enclosed** charge, not on the shape or size of the surface or where the charge sits inside. 3. Charges outside the surface contribute zero net flux. _Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', §1.6 (Gauss law) / Long Answer Q.11_
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