A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _____ μJ .

Step-by-step solution

The formula to calculate the initial energy stored in the first capacitor is given by U i = 1 2 C V 2       . . . 1 The formula to calculate the final energy stored when two capacitors are connected is given by U f = 1 2 C V 2 2 × 2 = 1 4 C V 2       . . . 2 Subtract equation (2) from equation (1) to obtain the energy loss ∆ U . ∆ U = 1 2 C V 2 - 1 4 C V 2 =   C V 2 4       . . . 3 Substitute the values of the known parameters into equation (3) to c