Practice question

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (3 × 10⁻⁸/0.03) = 9 × 10⁹ × 10⁻⁶ = 9000 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 potential V = Q_total/C_total, result 9000 V follows, reflecting potential-capacitance relations.