A uniform electric field of magnitude E = 100 NC−1 exists in the space in x-direction. The flux of this field through a plane square area of edge 10 cm placed in the y-z plane would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1.0 Nm2C−1
The electric flux through a flat surface in a uniform field is
\( \displaystyle \phi = EA\cos\theta \)
where \( \theta \) is the angle between the field and the normal to the surface. A square in the y-z plane has its normal along the x-axis, which is the direction of the field, so \( \theta = 0^\circ \).
\( E = 100\,\mathrm{N\,C^{-1}} \)
\( A = (10\,\mathrm{cm})^{2} = (0.1\,\mathrm{m})^{2} = 1.0\times10^{-2}\,\mathrm{m^{2}} \)
\( \displaystyle \phi = 100\,\mathrm{N\,C^{-1}} \times 1.0\times10^{-2}\,\mathrm{m^{2}} \times \cos 0^\circ \)
\( \displaystyle \phi = 1.0\,\mathrm{N\,m^{2}\,C^{-1}} \)
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 – …
