A point charge −5 μC is at a distance 5 cm directly above the centre of a square of side 10 cm. The magnitude of electric flux through this square is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{5\times10^{-6}}{6\varepsilon_0}\,\mathrm{N\,C^{-1}\,m^{2}} \)
The charge is 5 cm above the centre of a 10 cm square, so the square can be taken as one face of a cube of side 10 cm with the charge at its centre. By Gauss’s law the total flux through the closed cube is \( q/\varepsilon_0 \), and by symmetry the six faces share it equally.
\( \displaystyle \phi_{\mathrm{cube}} = \frac{q}{\varepsilon_0} \)
\( \displaystyle |\phi_{\mathrm{square}}| = \frac{1}{6}\cdot\frac{|q|}{\varepsilon_0} = \frac{5\times10^{-6}}{6\varepsilon_0}\,\mathrm{N\,C^{-1}\,m^{2}} \)
The negative sign of the charge only means the flux is inward; the question asks for its magnitude.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
