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cyanfly142Lv1
11 Dec 2019
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 (which is connected across the capacitor terminals).
A.)Find 
, the the energy dissipated in the resistor.
Express your answer in terms ofand 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. The battery is then disconnected and the capacitor is discharged. For this situation,what is , the energy dissipated in the resistor?
Express your answer in terms ofand other given quantities.
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 (which is connected across the capacitor terminals).
A.)Find , the the energy dissipated in the resistor.
Express your answer in terms ofand 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. The battery is then disconnected and the capacitor is discharged. For this situation,what is , the energy dissipated in the resistor?
Express your answer in terms ofand other given quantities.
Supratim PalLv10
5 Oct 2020
