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The electric field due to an INFINITE plane sheet of uniform surface charge density $\sigma$ (in vacuum) is
A$\dfrac{\sigma}{2\varepsilon_0}$, perpendicular to the sheet, independent of distance
B$\dfrac{\sigma}{4\pi\varepsilon_0\,r^2}$
C$\dfrac{\sigma}{\varepsilon_0\,r}$, decreasing with distance
DZero everywhere by symmetry
Answer & Solution
Correct answer: A. $\dfrac{\sigma}{2\varepsilon_0}$, perpendicular to the sheet, independent of distance
1. Use Gauss's law with a PILLBOX Gaussian surface of cross-sectional area $A$, perpendicular to the sheet, half on each side.
2. By symmetry, $\vec{E}$ is perpendicular to the sheet, same magnitude on both sides, pointing AWAY (for $\sigma > 0$).
3. Flux through the two pillbox faces: $\Phi = 2EA$ (one face each side). Curved side of pillbox: zero flux because $\vec{E}$ is parallel to it.
4. Enclosed charge: $Q_\text{enc} = \sigma A$.
5. Gauss's law: $2EA = \sigma A / \varepsilon_0$, so $E = \dfrac{\sigma}{2\varepsilon_0}$.
6. KEY FEATURE: the field is UNIFORM and INDEPENDENT of distance from the sheet — striking but true for an infinite plane.
7. Option B is for a point charge. Option C invents distance dependence. Option D is wrong — field is non-zero on both sides.
_Source: NCERT Class 12 Physics Part 1, Ch 1, §1.14 (Gauss's law — infinite plane sheet), p. 31–32._
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