Why does the energy decrease after insertion of dielectric into capacitor?
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The energy stored in a capacitor can be written as
U=12Q2CU=12Q2C
When the dielectric is removed, the charge on the plates, Q, does not change. The capacitance C decreases, so the energy must increase. (The capacitance decreases because the electric field increases without the polarized dielectric to cancel some of it out, or equivalently because the permittivity decreases to the vacuum permittivity.)
Where does the extra energy come from? The electric field in the capacitor separates the negative and positive charges in the dielectric (that's the definition of a dielectric material), so there is an attractive force between each plate and the dielectric. It takes work to overcome that force and remove the dielectric from the capacitor.
Note that this assumes the capacitor is not connected to anything and the distance between the plates remains constant when the dielectric is replaced with vacuum or air. If the plates of the capacitor are connected to an ideal battery, then the voltage, not the charge, would remain constant, and the energy stored would decrease. If the plates are allowed to touch once the dielectric is removed, then the capacitor isn't a capacitor any more.
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