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. Outside shell: E = (k q/r²) . E = 9 × 10⁹ × (12 × 10⁻⁶/(0.22)²) = 9 × 10⁹ × (12 × 10⁻⁶/0.0484) = 2.23 × 10⁶ N/C . Substituting values gives 2.23 × 10⁶ N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.