A charge of \(3.54\times10^{-10}\ \mathrm{C}\) is spread evenly over a small area of \(2\ \mathrm{cm^{2}}\) on the surface of a charged metal body. The magnitude of the electric field just outside this area is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2.0\times10^{5}\ \mathrm{N\,C^{-1}}\)
Just outside a charged conductor, the field is normal to the surface. Its magnitude depends only on the surface charge density \( \sigma \) at that spot, \( E=\sigma/\varepsilon_0 \). So we first find \( \sigma \) from the charge and the area.
Here \( A=2\ \mathrm{cm^{2}}=2\times10^{-4}\ \mathrm{m^{2}} \).
\( \begin{aligned} \sigma &= \frac{q}{A} \\ &= \frac{3.54\times10^{-10}}{2\times10^{-4}} \\ &= 1.77\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
\( \begin{aligned} E &= \frac{\sigma}{\varepsilon_0} \\ &= \frac{1.77\times10^{-6}}{8.85\times10^{-12}} \\ &= 2.0\times10^{5}\ \mathrm{N\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Summary – “8. Electrostatics field E is zero in the interior of a conductor; just outside the surface of a charged conductor, E is normal to the surface given by”
\( \vec E=\frac{\sigma}{\varepsilon_0}\,\hat n \)