A charge q is enclosed by a spherical Gaussian surface of radius R. If the radius of the sphere is doubled, how will the outward electric flux change?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Remains the same
Gauss’s law says that the net flux through a closed surface depends only on the charge enclosed, not on the size or shape of the surface.
\( \displaystyle \phi_E = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} \)
The charge q stays inside the sphere when the radius is doubled, so \( \phi_E \) is unchanged. Directly: the field at the surface falls to one-fourth (\( E\propto 1/r^{2} \)), but the surface area becomes four times (\( A\propto r^{2} \)), and the product \( EA \) is the same.
NCERT (2026-27):
Gauss’s Law – “… We state Gauss’s law without proof: Electric flux through a closed surface S
\( \displaystyle = q/\varepsilon_0 \qquad (1.31) \)
q = total charge enclosed by S. … (i) Gauss’s law is true for any closed surface, no matter what its shape or size.”