A \( 16 \, \mu\text{F} \) capacitor charged to \( 10 \, \text{V} \) is connected to an uncharged \( 16 \, \mu\text{F} \)
**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial charge: Q = 16 × 10⁻⁶ × 10 = 1.6 × 10⁻⁴ C . Total capacitance: 16 + 16 = 32 μF . Final voltage: V = (Q/C) = (1.6 × 10⁻⁴/32 × 10⁻⁶) = 5 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,
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications
