Question: Why does energy stored in a capacitor decrease when a dielectric is inserted after disconnecting the battery?

Reasoning Chain:

Battery disconnected → Charge Q on plates is fixed (no current can flow in or out) → Dielectric slab is inserted between plates → Polar molecules in dielectric align with the external field (polarization) → Polarization creates an internal field opposing E (E_dielectric = $E_{0}$/K) → Net field between plates decreases by factor K → Since V = E × d, the voltage decreases (V' = V/K) → New energy U' = $Q^{2}$/(2C') = $Q^{2}$/(2KC) = U/K → energy decreases by factor K → Where did the energy go? → The electric force pulled the dielectric slab inward (mechanical work done ON the slab) → This work came from the stored electrical energy → Energy decreased = work done on dielectric slab

Contrast — Battery Connected: Battery connected → V = constant → Dielectric inserted → C increases to KC → Battery pushes more charge (Q' = KQ) to maintain V → New energy U' = ½(KC)$V^{2}$ = KU → energy increases by factor K → Extra energy = charge pumped in by battery