Consider the circuit shown in the figure. The switch S is open for a long time and subsequently closed. Determine the change in the energy stored in the capacitors.

Step-by-step solution

When the switch S is open, the two capacitors are connected in series across the battery. The equivalent capacitance of the circuit is C_ eq = C C C + C = C 2 . The initial energy stored in the capacitors is: U_i = 1 2 C_ eq E ^2 = 1 2 ( C 2 ) E ^2 = 1 4 C E ^2 When the switch S is closed, the capacitor on the right is short-circuited because both of its plates are connected to the same potential (the negative terminal of the battery). Therefore, it stores no energy. The capacitor on the left is now connected direc