Practice question

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (7 × 10⁻⁹/4²) = 9 × 10⁹ × (7 × 10⁻⁹/16) = 3.9375 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common