Case: Read the passage and answer the question.
Conductors contain mobile charge carriers. In metallic conductors, these charge carriers are electrons. In a metal, the outer (valence) electrons part away from their atoms and are free to move. These electrons are free within the metal but not free to leave the metal. The free electrons form a kind of ‘gas’; they collide with each other and with the ions and move randomly in different directions. In an external electric field, they drift against the direction of the field. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions.
Just outside a point P on the surface of a charged metal body, the electric field is \(4\times10^{5}\ \mathrm{N\,C^{-1}}\). The field points into the surface. What is the surface charge density at P? (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) → \(-3.54\times10^{-6}\ \mathrm{C\,m^{-2}}\)
Just outside a charged conductor, the field is normal to the surface and has magnitude \( E=\sigma/\varepsilon_0 \). It points outward when \( \sigma \) is positive and inward when \( \sigma \) is negative. Here the field points into the surface, so the charge at P is negative.
Given:
\( E=4\times10^{5}\ \mathrm{N\,C^{-1}} \)
\( \varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}} \)
\( \begin{aligned} |\sigma| &= \varepsilon_0E \\ &= 8.85\times10^{-12}\times4\times10^{5} \\ &= 3.54\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
The field is inward, so \( \sigma=-3.54\times10^{-6}\ \mathrm{C\,m^{-2}} \).
NCERT (2026-27):
Electrostatics of Conductors – “5. Electric field at the surface of a charged conductor”
\( \displaystyle \vec{E}=\frac{\sigma}{\varepsilon_0}\,\hat{n} \)\( \qquad (2.35) \)
“where σ is the surface charge density and n̂ is a unit vector normal to the surface in the outward direction.”
Electrostatics of Conductors – “For σ > 0, electric field is normal to the surface outward; for σ < 0, electric field is normal to the surface inward.”