(Figure 1) A parallel-plate vacuum capacitor is connected to a battery and charg

Question

(Figure 1) A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is U. The battery is removed, and then a dielectric material with dielectric constant K 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).

Part A Find Ur, the the energy dissipated in the resistor. Express your answer in terms of U and other given quantities.

Part B Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted.(Figure 2) The battery is then disconnected and the capacitor is discharged. For this situation, what is Ur, the energy dissipated in the resistor? Express your answer in terms of U and other given quantities.

Explanation / Answer

A)

The energy dissipated in the resistor is

Ur =(1/2)q^2/KC

Ur = (1/2)(1/K)q^2/C

Ur = (1/K) U

Ur = U/K

B)

The energy dissipated in the resistor is

Ur = (1/2)KCV^2

Ur = K (1/2)CV^2

Ur = K U

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