Question: A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is removed, and then a dielectric material with dielectric constant is inserted into the capacitor, filling the space between the plates. Finally, the capacitor is fully discharged through a resistor (a resistor is connected across the capacitor terminals).
A) Find , the energy dissipated in the resistor. Express your answer in terms of and other given quantities.
B)Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted. (Part B figure) The battery is then disconnected and the capacitor discharged. For this situation, what is , the energy dissipated in the resistor? Express your answer in terms of and other given quantities.