A parallel plate capacitor of capacity 100 F is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance, the additional energy given by the battery to the capacitor in joules is

Step-by-step solution

C=100 F , V=50 volt Capacitance of parallel plate capacitor C= ₀ A d d= separation between the plates Initial energy aligned E₁ & = 1 2 C V^2 & = 1 2 (100 10⁻⁶ ) (50)^2 & =50 10⁻⁶ 2500 & =125 10⁻³ ~J aligned When distance between the plates becomes double, then capacitance C^ = C 2 = 100 2 =50 F Final energy E₂= 1 2 C^ V^2 aligned & = 1 2 50 10⁻⁶ (50)^2 & =25 10⁻⁶ 2500 & =625 10⁻⁴ & =625 10⁻³ ~J aligned aligned & Additional energy =E₁-E₂ & =125 10⁻³-625 10⁻³ & =62.5 10⁻³ & = 125 2 10⁻³ ~J & aligned