Two charges \( 5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 10 cm apart. What is the potential energy of t
**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. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (5 × 10⁻⁶ × (-5 × 10⁻⁶)/0.1) = 9 × 10⁹ × (-25 × 10⁻¹²/0.1) = -2.25 J . 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 -2.25 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics