- 天同Lv 78 年前最佳解答
I suppose the capacitor is initially charged.
(a) The insertion of a piece of dielectric increases the capacitance. Because the capacitor is isolated, no charge would flow. The energy stored in the capacitor thus decreases, as in according to the formula: energy stored = (1/2).(Q^2/C), where Q is the charge stored and C is the capacitance.
(b) Because the voltage across the capacitor is now fixed, the increase of capacitance leads to a flow of charge from the battery so that the charge stored is increased.
Hence, the energy stored in the capacitor is now increased.