For an infinite plane sheet carrying uniform surface charge density $\sigma$, the magnitude of the electric field intensity at a perpendicular distance $r$ from the sheet is: 
A$\dfrac{\sigma r}{\varepsilon_0}$
B$\dfrac{\sigma}{\varepsilon_0}$
C$\dfrac{\sigma}{4\pi\varepsilon_0}$
D$\dfrac{\sigma}{2\varepsilon_0}$
Answer & Solution
Correct answer: D. $\dfrac{\sigma}{2\varepsilon_0}$
Consider a Gaussian cylinder of cross-section A piercing the sheet symmetrically. Flux exits both flat ends: $\phi = 2EA$. Enclosed charge $= \sigma A$, so $2EA = \sigma A/\varepsilon_0$ ⇒ $E = \sigma/(2\varepsilon_0)$, independent of $r$. (Inside a parallel-plate capacitor, two such sheets add: $E = \sigma/\varepsilon_0$.)
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