A parallel-plate air capacitor of capacitance C is charged to a potential difference V0 and then disconnected from the battery. It is connected…

The second capacitor is two capacitors in series: an air layer of thickness $d/2$ with capacitance $2C$ and a dielectric layer with capacitance $2\kappa C=6C$. Its capacitance is $C_2=\frac{2C\cdot 6C}{8C}=1.5C$, which equals $\varepsilon_0A/(d/2+d/6)$. Charge is conserved on the joined plates, so the common potential is $V=\frac{CV_0}{C+1.5C}=0.4V_0$. Final energy $U_f=\frac12(2.5C)(0.4V_0)^2=0.2CV_0^2$ while initial energy $U_i=0.5CV_0^2$. Energy lost $=0.3CV_0^2$, which is $60\%$ of $U_i$. This is lost in the connecting wires as heat and radiation, even though the wires are ideal. Remember that the slab lies against one plate, but the order of the two layers does not change the series capacitance. The result depends only on the ratio of the capacitances, not on the plate area.