A point charge of \( 2\,\mu\mathrm{C} \) is at the centre of a cubic Gaussian surface of side \( 9\,\mathrm{cm} \). The net electric flux through the surface is : [Given \( \varepsilon_{0} = 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \)]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.26\times10^{5}\,\mathrm{N\,m^{2}\,C^{-1}} \)
By Gauss’s law, the total flux through a closed surface depends only on the charge enclosed, not on the shape or size of the surface. The side of the cube (9 cm) is not needed.
\( \displaystyle \phi = \frac{q}{\varepsilon_0} \)
\( \displaystyle \phi = \frac{2\times10^{-6}\,\mathrm{C}}{8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}} \)
\( \phi \approx 2.26\times10^{5}\,\mathrm{N\,m^{2}\,C^{-1}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
