The electric field at a point distance r away from a uniformly charged infinite plane sheet is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → constant
Applying Gauss’s law to a cylinder crossing the sheet, only the two flat faces carry flux, and their area does not depend on how far they are from the sheet.
\( \displaystyle E \times 2A = \frac{\sigma A}{\varepsilon_0} \)
\( \displaystyle E = \frac{\sigma}{2\varepsilon_0} \)
The field has no r in it, so it is the same at every distance.
Why the other options are incorrect
Option (1) proportional to r-1 — Incorrect. \( E \propto 1/r \) is the field of an infinite line charge.
Option (2) proportional to r-2 — Incorrect. \( E \propto 1/r^{2} \) is the field of a point charge or a charged sphere outside it.
Option (4) proportional to r-3/2 — Incorrect. No standard charge distribution gives \( r^{-3/2} \).
NCERT (2026-27):
Field Due to a Uniformly Charged Infinite Plane Sheet – “… Therefore the net flux through the Gaussian surface is 2 EA. The charge enclosed by the closed surface is σA. Therefore by Gauss’s law,
\( \displaystyle 2EA = \frac{\sigma A}{\varepsilon_0} \quad \text{or,} \quad E = \frac{\sigma}{2\varepsilon_0} \)”
Field Due to a Uniformly Charged Infinite Plane Sheet – “… Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also.”