Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 15 cm surrounding the total charge is 5 Vm. The flux over a concentric sphere of radius 10 cm surrounding the total charge will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.0 Vm
By Gauss’s law, the electric flux through a closed surface depends only on the charge enclosed, not on the size or shape of the surface.
\( \displaystyle \Phi = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} \)
Both spheres surround the whole charge, so the enclosed charge is the same and so is the flux:
\( \displaystyle \Phi_{10\,\mathrm{cm}} = \Phi_{15\,\mathrm{cm}} = 5.0\,\mathrm{V\,m} \)
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.”