In the given circuit, the two-way switch S is initially kept at position 1 for a long time, fully charging the capacitor C₁ = 4 F using a 50 V battery. The switch is then shifted to position 2, disconnecting the battery and connecting C₁ to an initially uncharged capacitor C₂ = 6 F . The amount of electrostatic energy dissipated as heat during the redistribution of charge is:

Step-by-step solution

When the switch is at position 1, capacitor C₁ is charged to potential V₁ = 50 V . When the switch is shifted to position 2, C₁ is connected across the initially uncharged capacitor C₂ ( V₂ = 0 ). The electrostatic energy dissipated as heat during the redistribution of charge is given by the formula: U = 1 2 C₁ C₂ C₁ + C₂ (V₁ - V₂)^2 Substituting the given values C₁ = 4 F , C₂ = 6 F , V₁ = 50 V , and V₂ = 0 V : U = 1 2 (4 10⁻⁶) (6 10⁻⁶) (4 + 6) 10⁻⁶ (50 - 0)^2 U = 1 2 24 10⁻¹² 10 10⁻⁶ 2500 U = 1 2 2.4 10⁻⁶ 2500 U =