An uncharged metal slab is placed in a uniform external electric field of \(2\times10^{5}\ \mathrm{V\,m^{-1}}\), with its faces normal to the field. In the static situation, the magnitude of the surface charge density induced on each face of the slab is (Take \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{C^{2}\,N^{-1}\,m^{-2}}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.77\times10^{-6}\ \mathrm{C\,m^{-2}}\)
The free electrons of the metal move until the field inside the slab is zero. The charges \( +\sigma \) and \( -\sigma \) left on the two faces act like two charged sheets. Between them they produce a field \( \sigma/\varepsilon_0 \), opposite to the external field. For the net field inside to be zero, this field must equal \( E_0 \).
\( \begin{aligned} \frac{\sigma}{\varepsilon_0} &= E_0 \\ \sigma &= \varepsilon_0E_0 \\ &= 8.85\times10^{-12}\times2\times10^{5} \\ &= 1.77\times10^{-6}\ \mathrm{C\,m^{-2}} \end{aligned} \)
The slab as a whole stays neutral, but each face carries this charge density.
NCERT (2026-27):
Dielectrics and Polarisation – “The free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor. This happens until, in the static situation, the two fields cancel each other and the net electrostatic field in the conductor is zero.”
The Parallel Plate Capacitor – “In the inner region between the plates 1 and 2, the electric fields due to the two charged plates add up, giving”
\( E=\frac{\sigma}{2\varepsilon_0}+\frac{\sigma}{2\varepsilon_0}=\frac{\sigma}{\varepsilon_0} \)\( \qquad (2.41) \)