Gauss's Law

Gauss's Law applied to an infinite plane sheet of charge yields the electric field by choosing a cylindrical Gaussian pillbox with faces parallel to the sheet. Since the field is symmetric and perpendicular to the sheet on both sides, the flux through the two flat faces gives (2EA = \frac{\sigma A}{\varepsilon_0}), leading to (E = \frac{\sigma}{2\varepsilon_0}). Critically, this result is independent of the distance from the sheet — the field is uniform throughout space on each side. With (\sigma = 4 \times 10^{-6}) C/m² and (\varepsilon_0 = 8.85 \times 10^{-12}) C²/N·m²: (E = \frac{4 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = \frac{4 \times 10^{-6}}{1.77 \times 10^{-11}} \approx 2.26 \times 10^5) N/C. Option (4.52 \times 10^5) N/C is incorrect because it uses (E = \sigma/\varepsilon_0) (missing the factor of 2). Option (1.13 \times 10^5) N/C is incorrect because it introduces an extra factor of 2 in the denominator. Option (9.04 \times 10^5) N/C is incorrect because it is four times the correct answer. Notably, the distance of 10 cm is irrelevant — the field from an infinite sheet is uniform. Plausibility check: (10^5) N/C scale for microcoulomb/m² surface density is consistent with standard JEE problems.