At a point on the surface of a charged conductor, the surface charge density is \(8.85\times10^{-7}\ \mathrm{C\,m^{-2}}\). A small pill box (a short closed cylinder) of cross-section \(2\ \mathrm{cm^{2}}\) is drawn across the surface at this point. One flat face is just inside the conductor and the other is just outside. The electric flux through the pill box is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(20\ \mathrm{N\,m^{2}\,C^{-1}}\)
By Gauss’s law, the flux through a closed surface is the enclosed charge divided by \( \varepsilon_0 \). The pill box encloses the surface charge \( \sigma\,\delta S \) only. The field inside the conductor is zero, so the face inside gives no flux. The field outside is normal to the surface, so only the outer face receives flux.
Here \( \delta S=2\ \mathrm{cm^{2}}=2\times10^{-4}\ \mathrm{m^{2}} \).
\( \begin{aligned} \phi &= \frac{\sigma\,\delta S}{\varepsilon_0} \\ &= \frac{8.85\times10^{-7}\times2\times10^{-4}}{8.85\times10^{-12}} \\ &= 20\ \mathrm{N\,m^{2}\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Electrostatics of Conductors – “Just inside the surface, the electrostatic field is zero; just outside, the field is normal to the surface with magnitude E. Thus, the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. … The charge enclosed by the pill box is σδS.”