A parallel-plate air capacitor of capacitance 100 pF is charged to 200 V and then disconnected from the source. A dielectric slab of dielectric…

Initially $U_i=\tfrac12C_0V_0^2=\tfrac12(100\times10^{-12})(200)^2=2.0\ \mu\mathrm{J}$. With a slab of thickness $t=d/2$ and $\kappa=4$, $C=\dfrac{\varepsilon_0A}{d-t+t/\kappa}=\dfrac{\varepsilon_0A}{d(\tfrac12+\tfrac18)}=1.6\,C_0$. The capacitor is isolated, so the charge $Q$ is fixed and $U=Q^2/2C$ falls to $U_f=U_i/1.6=1.25\ \mu\mathrm{J}$. The slab is pulled into the field by the capacitor itself, so the external agent does negative work equal to the change in energy: $W=U_f-U_i=-0.75\ \mu\mathrm{J}$. The positive value $+1.20\ \mu\mathrm{J}$ would be the energy change at constant potential difference (battery connected, $U_f=3.2\ \mu\mathrm{J}$). Because the plates are isolated, the field in the air gap stays $E=Q/(\varepsilon_0A)$ while the field inside the slab falls to $E/4$, so the potential difference drops from $200\ \mathrm{V}$ to $200\times\tfrac58=125\ \mathrm{V}$, consistent with the lower stored energy.