The electric field intensity due to an infinite thin plane sheet of surface charge density σ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\sigma}{2\varepsilon_0} \)
Use Gauss's law with a cylindrical (pill-box) Gaussian surface of end area A that cuts the sheet. By symmetry the field is perpendicular to the sheet and has the same magnitude E on both sides, so the flux comes out through both flat ends, \( 2EA \), and none through the curved side. The charge enclosed is \( \sigma A \).
\( \displaystyle 2EA = \frac{\sigma A}{\varepsilon_0} \Rightarrow E = \frac{\sigma}{2\varepsilon_0} \)
The field is directed away from the sheet for \( \sigma \gt 0 \), and it does not depend on the distance from the sheet.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Summary – “… (ii) Infinite thin plane sheet of uniform surface charge density σ
\( \displaystyle \mathbf{E} = \frac{\sigma}{2\varepsilon_0}\,\hat{\mathbf{n}} \)
where \( \hat{\mathbf{n}} \) is a unit vector normal to the plane, outward on either side.”