If ointE· dS=0 over a closed surface, then:(i) the electric field ins…

The correct answer is Option (1) → Only (iii) is correct

By Gauss’s law, the flux through a closed surface is \( q/\varepsilon_0 \), where \( q \) is the net charge enclosed. So zero flux means only that the net enclosed charge is zero. It does not fix the field itself.

(i) The electric field inside the surface and on it is zero — False. A closed surface placed in a uniform external field has zero flux, yet the field inside and on it is not zero.

(ii) The electric field inside the surface is necessarily uniform — False. A charge \( +q \) and a charge \( -q \) enclosed together give zero net charge and zero flux, but their field inside is far from uniform.

(iii) The number of flux lines entering the surface must be equal to the number of flux lines leaving it — True. Flux is proportional to the number of field lines crossing a surface. Lines going out count as positive and lines coming in count as negative. So zero net flux means as many lines enter as leave.

(iv) The electric field at every point of the surface must be normal to the surface — False. For a closed cylinder with its axis along a uniform field, the flux is zero. But on the curved surface, the field is parallel to the surface. So the field is not normal to the surface at every point.

NCERT (2026-27):
Gauss’s Law – “… Here the electric field is uniform and we are considering a closed cylindrical surface, with its axis parallel to the uniform field E. … Thus, the total flux is zero, as expected by Gauss’s law.”

Gauss’s Law – “… Now the normal to the surface 3 at every point is perpendicular to E, so by definition of flux, φ₃ = 0.”

Electric Flux – “We now come to the definition of electric flux. Electric flux Δφ through an area element ΔS is defined by

    \( \Delta\phi=\mathbf{E}\cdot\Delta\mathbf{S}=E\,\Delta S\cos\theta \qquad (1.11) \)

which, as seen before, is proportional to the number of field lines cutting the area element.”