Capacitor with a Dielectric: Energy and the Pull-In Force

A parallel-plate capacitor stores charge $Q = CV$ with capacitance $C = \varepsilon_0 A/d$, and slipping a dielectric slab into the gap raises that capacitance by the dielectric constant $\varepsilon_r$. When the slab covers a fraction $x$ of the plate area it acts as two capacitors in parallel, $C(x) = C_0\,[1 + (\varepsilon_r - 1)x]$, and more charge crowds onto the plates where the dielectric sits. The slab is not just passive: the fringing field at the plate edge pulls it in. Whether that lowers or raises the stored energy depends on what you hold fixed. Disconnect the battery and the charge is fixed, so $U = Q^2/2C$ falls as the slab enters and the field itself does the work of pulling it in. Keep the battery connected and the voltage is fixed, so $U = \tfrac12 CV^2$ rises, but now the battery supplies the energy while the slab is still pulled in. Either way the dielectric is sucked into the capacitor, a small force with a clear sign.