A uniformly charged conducting sphere of radius 1.3 m has a surface charge density of \( 70\,\mu\mathrm{C\,m^{-2}} \). What is the total electric flux leaving the surface of the sphere?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 1.7\times10^{8}\,\mathrm{N\,m^{2}\,C^{-1}} \)
By Gauss's law, the total flux leaving a closed surface is the enclosed charge divided by \( \varepsilon_0 \). The charge on the sphere is the surface charge density times the surface area.
\( \displaystyle q = \sigma \times 4\pi R^{2} = 70\times10^{-6} \times 4 \times 3.14 \times (1.3)^{2} = 1.49\times10^{-3}\,\mathrm{C} \)
\( \displaystyle \Phi = \frac{q}{\varepsilon_0} = \frac{1.49\times10^{-3}}{8.85\times10^{-12}} \approx 1.7\times10^{8}\,\mathrm{N\,m^{2}\,C^{-1}} \)
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 – …
