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Asked in GUJCET 2011 · Gauss's law and enclosed charge
Idea: Gauss's law, ∮⃗E· d⃗A = (q_enc)/(ε₀), holds for any closed surface, but it gives E only if E can be taken outside the integral.
That needs a closed surface matched to the symmetry of the charges: wherever the field crosses it, E is normal to it and has one fixed magnitude (on any other part, E runs along the surface and adds no flux).
Then the flux is just E times the area it crosses, EA = (q_enc)/(ε₀), which gives E directly.
Examples: a concentric sphere for a point charge or a charged shell, a coaxial cylinder for a line charge.
So the Gaussian surface is a symmetrical closed surface on which the field has a single fixed value.
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