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NCERT Exemplar · Q10

Q.If there were only one type of charge in the universe, then

(a) ∮SE⃗⋅dS⃗≠0\oint_S \vec{E}\cdot d\vec{S} \neq 0 on any surface.
(b) ∮SE⃗⋅dS⃗=0\oint_S \vec{E}\cdot d\vec{S} = 0 if the charge is outside the surface.
(c) ∮SE⃗⋅dS⃗\oint_S \vec{E}\cdot d\vec{S} could not be defined.
(d) ∮SE⃗⋅dS⃗=qε0\oint_S \vec{E}\cdot d\vec{S} = \dfrac{q}{\varepsilon_0} if charges of magnitude qq were inside the surface.
Rajasthan RbseMCQ· 1mImportance★★★★★
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Gauss's law ∮SE⃗⋅dS⃗=qenc/ε0\oint_S \vec{E}\cdot d\vec{S}=q_{enc}/\varepsilon_0 ties the net flux only to the enclosed charge, and holds regardless of how many kinds of charge exist. A charge outside gives zero net flux; a charge qq inside gives q/ε0q/\varepsilon_0. Correct: (b) and (d).

Concept understanding

Gauss's law is a geometric consequence of the inverse-square Coulomb field. For any closed surface,

∮SE⃗⋅dS⃗=qencε0.\oint_S \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{\varepsilon_0}.

Nothing in its derivation requires the existence of both positive and negative charge — it depends only on the 1/r21/r^2 field and the solid-angle geometry of a closed surface.

Testing each option

  • (a) claims the flux is never zero. This is false: the field lines from a charge lying outside the surface enter one side and leave the other, so their contributions cancel and the net flux vanishes.
  • (b) If the charge is outside the surface, qenc=0q_{enc}=0, so ∮SE⃗⋅dS⃗=0\oint_S \vec{E}\cdot d\vec{S}=0. True. …

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