A parallel-plate air capacitor of capacitance 2 muF is charged to 30 V and then disconnected from the battery. A dielectric slab of dielectric…

The charge $Q=C_0V=2\times30=60\ \mu\mathrm{C}$ stays on the isolated plates while the slab is inserted. With the slab covering half the area, the capacitor is two capacitors in parallel: $C=\dfrac{C_0}{2}+\dfrac{KC_0}{2}=\dfrac{2}{2}(1+3)=4\ \mu\mathrm{F}$. Connecting it to the uncharged $C_2=4\ \mu\mathrm{F}$ conserves charge; the common potential is $V'=Q/(C+C_2)=60/8=7.5\ \mathrm{V}$. Energy before: $\dfrac{Q^2}{2C}=\dfrac{3600}{8}=450\ \mu\mathrm{J}$. Energy after: $\dfrac{Q^2}{2(C+C_2)}=\dfrac{3600}{16}=225\ \mu\mathrm{J}$. The difference, $225\ \mu\mathrm{J}$, is dissipated as heat in the resistor. Equivalently $\Delta U=\dfrac{Q^2}{2}\dfrac{C_2}{C(C+C_2)}$. Note the work done while inserting the slab does not count: it changed the energy stored before the connection, which is already accounted for in the $450\ \mu\mathrm{J}$.