A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is____ nJ .

Step-by-step solution

Initial charge on first capacitor will be, Q = C V As the second capacitor is identical to the first one, hence potential drop across both capacitor will be equal to V 2 . Now, loss of energy, Δ H = U i - U f ⇒ Δ H = 1 2 C V 2 - 1 2 2 C × V 2 2 ⇒ Δ H = 1 2 C V 2 - 1 4 C V 2 ⇒ Δ H = 1 4 C V 2 = 1 4 × 50 × 10 - 12 × 100 2 = 125 × 10 - 9   J = 125   nJ