SI unit of oint_s E.dS is -

The correct answer is Option (3) → Nm²C⁻¹

The surface integral of the electric field over a closed surface is the electric flux. Its unit is the unit of field multiplied by the unit of area.

    \( \displaystyle [E] = \mathrm{N\,C^{-1}}, \quad [dS] = \mathrm{m^{2}} \)

    \( \displaystyle \left[\oint_s E.dS\right] = \mathrm{N\,C^{-1}} \times \mathrm{m^{2}} = \mathrm{N\,m^{2}\,C^{-1}} \)

The same unit follows from Gauss’s law, \( \phi = q/\varepsilon_0 \): \( \mathrm{C} \div (\mathrm{C^{2}\,N^{-1}\,m^{-2}}) = \mathrm{N\,m^{2}\,C^{-1}} \).

NCERT (2026-27):
Electric Flux – “… Electric flux \( \Delta\phi \) through an area element \( \Delta S \) is defined by

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

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

Electric Flux – “… The unit of electric flux is N C^–1 m².”