An infinite plane sheet carries a uniform surface charge density \(\sigma\). Consider the following statements about the electric field near the sheet:
(A) The field is directed normal to the sheet.
(B) The magnitude of the field is \(\sigma/2\varepsilon_0\).
(C) The magnitude of the field decreases in inverse proportion to the distance from the sheet.
(D) If the sheet is negatively charged, the field points towards the sheet.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
For an infinite plane sheet, symmetry makes the field normal (at right angles) to the sheet. Gauss’s law over a small cylinder across the sheet gives \( 2EA=\sigma A/\varepsilon_0 \).
(A) The field is directed normal to the sheet. — True. By symmetry the field cannot have a component along the sheet.
(B) The magnitude of the field is \( \sigma/2\varepsilon_0 \). — True. From \( 2EA=\sigma A/\varepsilon_0 \), \( E=\sigma/2\varepsilon_0 \).
(C) The magnitude of the field decreases in inverse proportion to the distance from the sheet. — False. The result \( \sigma/2\varepsilon_0 \) has no distance in it. So the field of an infinite sheet is the same at every distance. A field that varies as \( 1/r \) is found near a long straight wire.
(D) If the sheet is negatively charged, the field points towards the sheet. — True. The field points away from a positive sheet and towards a negative one.
NCERT (2026-27):
Field due to a uniformly charged infinite plane sheet – “2EA = σA/ε0 or, E = σ/2ε0 Vectorically, … where n̂ is a unit vector normal to the plane and going away from it.”
Field due to a uniformly charged infinite plane sheet – “E is directed away from the plate if σ is positive and toward the plate if σ is negative. Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also.”