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 ( r > R ): E = (k q/r²) . k = 9 × 10⁹ N·m²/C² , q = 16 × 10⁻⁶ C , r = 0.3 m . E = 9 × 10⁹ × (16 × 10⁻⁶/(0.3)²) = 9 × 10⁹ × (16 × 10⁻⁶/0.09) = 1.6 × 10⁶ N/C . Substituting values gives 1.6 × 10⁶ N/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.