A point charge of \(2\ \mu\mathrm{C}\) is at the centre of a sphere of radius 20 cm. The electric flux through a small part of the surface of the sphere, of area \(4\ \mathrm{cm^{2}}\), is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(180\ \mathrm{N\,m^{2}\,C^{-1}}\)
The field of the charge is radial. On the sphere it is normal to every small area and has the same size, \( E=q/4\pi\varepsilon_0r^{2} \). So the flux through a small area \( \Delta S \) of the sphere is \( \Delta\phi=E\,\Delta S \).
\( \begin{aligned} q &= 2\times10^{-6}\ \mathrm{C} \\ r &= 20\ \mathrm{cm}=0.2\ \mathrm{m} \\ \Delta S &= 4\ \mathrm{cm^{2}}=4\times10^{-4}\ \mathrm{m^{2}} \end{aligned} \)
\( \begin{aligned} E &= \frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}} \\ &= \frac{9\times10^{9}\times2\times10^{-6}}{(0.2)^{2}} \\ &= 4.5\times10^{5}\ \mathrm{N\,C^{-1}} \end{aligned} \)
\( \begin{aligned} \Delta\phi &= E\,\Delta S \\ &= 4.5\times10^{5}\times4\times10^{-4} \\ &= 180\ \mathrm{N\,m^{2}\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Gauss’s Law – “… Now, since the normal to a sphere at every point is along the radius vector at that point, the area element ΔS and rˆ have the same direction. Therefore,”
\( \displaystyle \Delta\phi=\frac{q}{4\pi\varepsilon_0r^{2}}\,\Delta S \)\( \qquad (1.29) \)