If the electric field is given by \( (6\hat{i} + 3\hat{j} + 4\hat{k})\,\mathrm{V/m} \). The electric flux through a surface of area \( 20\,\mathrm{m^{2}} \) lying in Y-Z plane would be (in V-m).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 120
The electric flux through a plane surface is \( \phi = \vec{E}\cdot\vec{A} \), where the area vector \( \vec{A} \) is perpendicular to the surface. A surface lying in the Y-Z plane has its normal along the x-axis, so \( \vec{A} = 20\,\hat{i}\,\mathrm{m^{2}} \).
\( \phi = (6\hat{i} + 3\hat{j} + 4\hat{k})\cdot(20\hat{i}) = 6\times20 = 120\,\mathrm{V\,m} \)
Only the component of the field along x (perpendicular to the surface) contributes; the y and z components lie in the plane of the surface and give no flux.
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 – …
