A parallel-plate capacitor is charged by a battery and then the battery is removed and a dielectric of constant K is used to fill the gap between the plates. Inserting the dielectric changes the energy stored by a factor of

Question

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When a dielectric is inserted into a charged capacitor (with the battery removed), the capacitance increases by a factor of K, but the charge remains constant. The energy stored in the capacitor is given by U = Q²/(2C). Since the capacitance increases by K, the energy decreases by a factor of 1/K.

A Parallel-Plate Capacitor Is Charged by a Battery and Then