The electric charges are distributed in a small volume. The flux of the electric field through a sphere of radius 10 cm surrounding the total charge is 20 Vm. The flux over a concentric sphere of radius 20 cm will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 20 Vm
By Gauss's law, the electric flux through any closed surface depends only on the charge enclosed by it.
\( \displaystyle \Phi = \frac{q_{\mathrm{enc}}}{\varepsilon_0} \)
The sphere of radius 20 cm encloses the same total charge as the sphere of radius 10 cm. So the flux is the same, 20 Vm.
The field on the larger sphere is weaker (\( E \propto 1/r^{2} \)), but its area is larger (\( A \propto r^{2} \)) in the same ratio, so the product \( E \times A \) does not change.
Note: The same question was also asked in CUET (UG) 2026 on 19-05-2026 (Slot 1).
NCERT (2026-27):
Gauss’s Law – “… (i) Gauss’s law is true for any closed surface, no matter what its shape or size. (ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.”