A charge q is placed at the centre of a sphere of radius r. The electric flux crossing the sphere is \( \phi \). If both the charge and radius of the sphere are doubled, the electric flux crossing the sphere will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\phi \)
By Gauss's law, the flux through a closed surface depends only on the charge enclosed, not on the size or shape of the surface.
\( \displaystyle \phi = \frac{q}{\varepsilon_0} \)
Doubling the radius changes nothing. Doubling the charge doubles the flux.
\( \displaystyle \phi' = \frac{2q}{\varepsilon_0} = 2\phi \)
NCERT (2026-27):
Gauss’s Law – “… We state Gauss’s law without proof:
\( \displaystyle \text{Electric flux through a closed surface } S = q/\varepsilon_0 \qquad (1.31) \)
q = total charge enclosed by S.”
Gauss’s Law – “… (i) Gauss’s law is true for any closed surface, no matter what its shape or size.”