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Suppose the universe contained only one type (sign) of charge. Then which statement is correct? (closed-surface flux integral of E.dS)

1. the closed-surface integral of E.dS is non-zero on any surface
2. the closed-surface integral of E.dS = 0 if the charge is outside the surface
3. the closed-surface integral of E.dS cannot be defined
4. the closed-surface integral of E.dS = q/e0 if charge of magnitude q lies inside the surface

Correct answer: the closed-surface integral of E.dS = 0 if the charge is outside the surface

Solution

Gauss's law is unaffected by the existence of only one sign of charge. Flux through a closed surface equals (enclosed charge)/e0. A charge entirely outside the surface contributes zero net flux. The option claiming the integral is non-zero on any surface is false (it is zero with no enclosed charge), and option (D) ignores sign/sense subtleties, but the always-true statement is that the flux is zero when the charge is outside.

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