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: 

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: $E = \sigma/\varepsilon_0$.)