Infinite plane sheets of charge are placed parallel to each other. List-I gives the surface charge densities of the sheets and the region considered. Match List-I with the magnitude of the electric field in that region in List-II.
| List-I | List-II |
|---|---|
| (A) A single sheet of density \(\sigma\); any point near it | (I) \(\sigma/\varepsilon_0\) |
| (B) Two sheets of densities \(+\sigma\) and \(+\sigma\); a point outside both sheets | (II) \(\sigma/2\varepsilon_0\) |
| (C) Two sheets of densities \(+\sigma\) and \(-\sigma\); a point outside both sheets | (III) Zero |
| (D) Two sheets of densities \(+2\sigma\) and \(-\sigma\); a point between the sheets | (IV) \(3\sigma/2\varepsilon_0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Each sheet of density \( \sigma \) gives a field of magnitude \( \sigma/2\varepsilon_0 \) that does not depend on distance. It points away from a positive sheet and towards a negative sheet. The fields of two sheets add as vectors.
(A) A single sheet of density \(\sigma\) — This matches (II) \( \sigma/2\varepsilon_0 \). This is the field of one sheet at any distance from it.
(B) Sheets \(+\sigma\) and \(+\sigma\), point outside — This matches (I) \( \sigma/\varepsilon_0 \). Outside both sheets, the two fields point the same way, away from both sheets. They add to \( 2\times\sigma/2\varepsilon_0 \).
(C) Sheets \(+\sigma\) and \(-\sigma\), point outside — This matches (III) zero. Outside, the field of the positive sheet points away from it and the field of the negative sheet points towards it. These are equal and opposite.
(D) Sheets \(+2\sigma\) and \(-\sigma\), point between — This matches (IV) \( 3\sigma/2\varepsilon_0 \). Between the sheets, both fields point from the positive sheet towards the negative sheet. They add to \( \sigma/\varepsilon_0+\sigma/2\varepsilon_0 \).
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,”
\( 2EA=\sigma A/\varepsilon_0 \quad \text{or,} \quad \)\( E=\sigma/2\varepsilon_0 \)
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.”