A parallel-plate capacitor featuring a plate area of 400 cm^2 and a...

With the dielectric slab inserted and the $100 \text{ V}$ power supply connected, the capacitance is $C_f = 1770 \text{ pF}$ and the stored energy is $U_f = 8.85 \mu\text{J}$.

The charge on the capacitor is $Q = C_f V = (1770 \times 10^{-12})(100) = 1.77 \times 10^{-7} \text{ C}$.

When the power supply is disconnected, this charge $Q$ remains constant.

Upon removing the dielectric slab, the capacitance returns to its initial value without the dielectric, $C_0 = 354 \text{ pF}$.

The new electrostatic energy is $U_{final} = \dfrac{Q^2}{2 C_0}$.

$U_{final} = \dfrac{(1.77 \times 10^{-7})^2}{2 (3.54 \times 10^{-10})} = \dfrac{3.1329 \times 10^{-14}}{7.08 \times 10^{-10}} = 4.425 \times 10^{-5} \text{ J} = 44.25 \mu\text{J}$.

The further increase in energy is $\Delta U' = U_{final} - U_f = 44.25 - 8.85 = 35.4 \mu\text{J}$.