An infinite plane sheet has a uniform surface charge density of \(1.77\times10^{-7}\ \mathrm{C\,m^{-2}}\). The magnitude of the electric field near the sheet is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(1.0\times10^{4}\ \mathrm{N\,C^{-1}}\)
Gauss’s law gives the field of an infinite plane sheet as \( E=\sigma/2\varepsilon_0 \). The factor 2 appears because the flux leaves the sheet through both sides.
\( \begin{aligned} E &= \frac{\sigma}{2\varepsilon_0} \\ &= \frac{1.77\times10^{-7}}{2\times8.85\times10^{-12}} \\ &= \frac{1.77\times10^{-7}}{1.77\times10^{-11}} \\ &= 1.0\times10^{4}\ \mathrm{N\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Field due to a uniformly charged infinite plane sheet – “2 EA = σA/ε0 or, E = σ/2ε0 …”