Practice question

**Special distributions** like infinite line, plane, spherical shell demonstrate Gauss's law advantage. Outside shell, charge appears concentrated at centre; inside, q_enc = 0 implies vanishing field, key result for shielding. E = (sigma/2 ε₀) . E = (1.416 × 10⁻¹⁰/2 × 8.854 × 10⁻¹²) = 8 N/C . Substituting values gives 8.0 N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.