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=C epslion` appears on one plate and -Q on the other. When the polarity is reversed, a charge -Q appears on the first plate and +Q on the second. A charge 2Q, therefore, passes through the battery from the negative terminal to the positive terminal. The battery does a work equal to
`W=(2Q)epsilon=2C epsilon^2`
in the process. The potential energy stored in the capacitor is the same in the two cases, thus the work done by the battery appears as heat in the connecting wires. THe heat produced is, therefore `2C epsilon^2`.
One might wonder that where did his heat energy go. Its is not mentioned in the problem that the connecting wires had some resistance, which would imply that the wires get heated up as the charge passes through them The answer is that even if the wires were superconductor some amount of energy will definitely be dissipated because the charges would get accelerated in the process and according to Maxwell.s equations. accelerating charge would always radiate some energy. The surprising part is that no matter what the resistance of the circuit is the energy, dissipated would turn out to be the same, the only difference is that one process would be very quick and another rather slow.