A capacitor of capacitance C is charged by connecting it to a battery of emf epsilon. The capacitor is now disconnected and reconnected to the battery with the polarity reversed. Calculate the heat developed in the connecting wires.

Explanation

When the capacitor is connected to the battery , a charge Q = Ce appears on one plate and - Q on the other . When the polarity is reveresed , a charge - Q appers on the first plate and +Q on the second . A charge 2Q, therefore passed through the battery form the negative to the positive terminal . The battery does a work.
`W = (2Q)e = 2Ce^(2)` in the process. The energy stored in the capacitor is the same in the tow case . Thus, the work done by the battery appears as heat in the connecting wires. The heat prodced is, therefore `2Ce^(2)`.