A capacitor with a capacitance of 100 F is charged to a potential difference of 24 V . After disconnecting the charging battery, the capacitor is connected to a different battery of emf 12 V such that the positive plate of the capacitor is attached to the positive terminal of the battery. Calculate the decrease in the electrostatic field energy of the capacitor.

Step-by-step solution

Initial energy of the capacitor is given by: U_i = 1 2 CV₁^2 Substituting the given values, C = 100 F and V₁ = 24 V : U_i = 1 2 100 10⁻⁶ (24)^2 = 28.8 10⁻³ J = 28.8 mJ When the capacitor is connected to a 12 V battery, the final potential difference across it becomes V₂ = 12 V . Final energy of the capacitor is: U_f = 1 2 CV₂^2 U_f = 1 2 100 10⁻⁶ (12)^2 = 7.2 10⁻³ J = 7.2 mJ The decrease in the electrostatic field energy of the capacitor is: U = U_i - U_f U = 28.8 mJ - 7.2 mJ = 21.6 mJ