A point charge \( 17.7\,\mu\mathrm{C} \) is placed at a perpendicular distance of 5 cm from the centre of a square. The amount of electric flux going through the square will be
(given \( \epsilon_0 = 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.33\times10^{5}\,\mathrm{N\,m^{2}/C} \)
By Gauss's law the total flux of a charge \( q \) through a closed surface is \( q/\epsilon_0 \). Take the square as one face of a cube of side 10 cm, with the charge at the centre of the cube (the charge is 5 cm from the face, which is half the side).
The six faces of the cube are identical, so by symmetry each face carries one sixth of the total flux.
\( \displaystyle \phi = \frac{1}{6}\,\frac{q}{\epsilon_0} \)
\( \displaystyle \phi = \frac{17.7\times10^{-6}}{6\times 8.85\times10^{-12}} = 3.33\times10^{5}\,\mathrm{N\,m^{2}/C} \)
Note: The paper does not give the side of the square. The key follows if the square is a face of a cube with the charge at the centre of the cube.
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 – …
