A parallel-plate air capacitor of capacitance C0 is charged to potential V by a battery, and the battery stays connected. A dielectric slab of…

Initially $U_0=\tfrac12 C_0V^2$. With the battery connected the slab raises the capacitance to $3C_0$ at the same potential $V$, so the charge becomes $Q=3C_0V$. The battery is then disconnected, so $Q$ stays fixed at $3C_0V$ while the slab is withdrawn, and the capacitance falls back to $C_0$. The final energy is $U_f=\dfrac{Q^2}{2C_0}=\dfrac{(3C_0V)^2}{2C_0}=\tfrac92 C_0V^2$. Dividing, $U_f/U_0=\dfrac{\tfrac92 C_0V^2}{\tfrac12 C_0V^2}=9$. The extra energy is supplied by the external agent that pulls the slab out against the attraction of the induced charges; this is the energy-conservation check. The common error is to assume constant $V$ during the withdrawal and obtain $1$.