A charge \( Q \) is placed at the centre of a cube of side \( a \). The flux linked with each face of the cube is \( \phi \). If the side of the cube is doubled, then the outward flux through each face of the cube will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \phi \)
By Gauss’s law, the total flux through any closed surface depends only on the charge enclosed, not on the size or shape of the surface.
\( \phi_{\mathrm{total}} = Q/\varepsilon_{0} \)
With the charge at the centre, the six faces of the cube are placed symmetrically about it, so each face carries one-sixth of the total flux:
\( \displaystyle \phi = \frac{Q}{6\varepsilon_{0}} \)
This contains no \( a \). Doubling the side still encloses the same charge \( Q \) with the same symmetry, so the flux through each face stays \( \phi \).