A cube of side a, shown in the figure given below is placed in an electric field Ex = 100x0.5 V/m along x-axis. The left face of the cube is at a distance of 'a' from the origin. The electric flux through the cube is (Assume that a = 1 m)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 41.4 Vm
The electric flux through a surface is \( \phi = \oint \vec{E}\cdot d\vec{A} \), and the flux through each face is counted with its outward normal. The field has only an x-component, so the four faces parallel to the x-axis contribute nothing; only the left and right faces matter.
The left face is at x = a = 1 m, with its outward normal along −x; the right face is at x = 2a = 2 m, with its outward normal along +x. Each face has area a2 = 1 m2.
\( E_L = 100 \times (1)^{0.5}\,\mathrm{V\,m^{-1}} = 100\,\mathrm{V\,m^{-1}} \)
\( \displaystyle E_R = 100 \times (2)^{0.5}\,\mathrm{V\,m^{-1}} = 100\sqrt{2}\,\mathrm{V\,m^{-1}} \approx 141.4\,\mathrm{V\,m^{-1}} \)
\( \phi_L = -E_L a^{2} = -100\,\mathrm{V\,m} \)
\( \phi_R = +E_R a^{2} = +141.4\,\mathrm{V\,m} \)
\( \phi = \phi_L + \phi_R = 141.4\,\mathrm{V\,m} - 100\,\mathrm{V\,m} = 41.4\,\mathrm{V\,m} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Gauss’s Law – …
