A capacitor of capacitance 5.00 F is charged to 24.0 V , while a second capacitor of capacitance 6.0 F is charged to 12.0 V . The positive plate of the first capacitor is subsequently connected to the negative plate of the second, and vice versa. Determine the loss of electrostatic energy during this process.

Step-by-step solution

When the positive plate of the first capacitor is connected to the negative plate of the second, the common potential V is given by: V = C₁ V₁ - C₂ V₂ C₁ + C₂ The loss of electrostatic energy during this process is given by the formula: U = 1 2 C₁ C₂ C₁ + C₂ (V₁ + V₂)^2 Substituting the given values C₁ = 5.00 F , C₂ = 6.0 F , V₁ = 24.0 V , and V₂ = 12.0 V : U = 1 2 5.00 6.0 5.00 + 6.0 (24.0 + 12.0)^2 J U = 1 2 30.0 11.0 (36.0)^2 J U = 15 11 1296 J U = 19440 11 J 1767.27 J U 1.77 mJ