The electric charges are distributed in a small volume. The electric flux through a spherical surface of radius \( 10\,\mathrm{cm} \) enclosing the charges is \( 20\,\mathrm{Vm} \). The electric flux through another concentric spherical surface of radius \( 20\,\mathrm{cm} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 20\,\mathrm{Vm} \)
By Gauss’s law, the total electric flux through a closed surface depends only on the charge enclosed:
\( \displaystyle \phi = \frac{q_{\mathrm{enclosed}}}{\varepsilon_0} \)
The larger sphere encloses the same charge, so the flux through it is the same, \( 20\,\mathrm{V\,m} \). Its radius does not matter: the field falls as \( 1/r^{2} \) while the area grows as \( r^{2} \), and their product stays constant.
Note: The same question was also asked in CUET (UG) 2023 on 20-06-2023 (Slot 3).
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
