The electric field in a region is given by \( \vec{E} = \displaystyle\frac{3}{5}E_{o}\hat{\imath} + \displaystyle\frac{4}{5}E_{o}\hat{\jmath} \) where Eo = 2 × 103 N/C. What is the electric flux of this field through a rectangular surface of area 0.02 m2 parallel to y − z plane?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 24 Nm2/C
The electric flux through a flat surface is \( \phi = \vec{E}\cdot\vec{A} \), so only the field component along the area vector (normal to the surface) contributes. A surface parallel to the y–z plane has its normal along the x-axis, so only the \( \hat{\imath} \) component of the field counts.
\( \vec{A} = 0.02\,\hat{\imath}\ \mathrm{m^{2}} \)
\( \displaystyle \phi = E_xA = \frac{3}{5}E_oA \)
\( \displaystyle \phi = \frac{3}{5} \times (2\times10^{3}\,\mathrm{N/C}) \times (0.02\,\mathrm{m^{2}}) \)
\( \phi = 24\,\mathrm{N\,m^{2}/C} \)
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 – …
