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Gauss's law in integral form states that the electric flux through a CLOSED surface equals
Athe total enclosed charge divided by $\varepsilon_0$
Bthe total charge in the universe divided by $4\pi\varepsilon_0$
Cthe electric field at the centre of the surface
Dthe volume enclosed by the surface, times $\varepsilon_0$
Answer & Solution
Correct answer: A. the total enclosed charge divided by $\varepsilon_0$
1. NCERT §1.11 states Gauss's law: $\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_\text{enc}}{\varepsilon_0}$.
2. The LEFT side is the electric flux through the closed (Gaussian) surface.
3. The RIGHT side depends only on the charge ENCLOSED by the surface — external charges contribute zero net flux.
4. Useful exactly when the geometry has symmetry (spherical, cylindrical, planar) — the symmetry lets you pull $E$ outside the surface integral.
5. Option B confuses the constant. Option C confuses a point value with a flux. Option D has wrong dimensions.
_Source: NCERT Class 12 Physics Part 1, Ch 1, §1.11 (Gauss's Law), p. 29–30._
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