A parallel-plate capacitor featuring a plate area of 400 cm ^2 and a separation of 1.0 mm is initially connected to a 100 V power supply, and a dielectric slab of thickness 1.0 mm and dielectric constant 5.0 is placed inside the gap. Subsequently, the power supply is disconnected, and the dielectric slab is extracted from the gap. Determine the further increase in the electrostatic energy of the system.

Step-by-step solution

With the dielectric slab inserted and the 100 V power supply connected, the capacitance is C_f = 1770 pF and the stored energy is U_f = 8.85 J . The charge on the capacitor is Q = C_f V = (1770 10⁻¹²)(100) = 1.77 10⁻⁷ 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, C₀ = 354 pF . The new electrostatic energy is U_ final = Q^2 2 C₀ . U_ final = (1.77 10⁻⁷)^2 2 (3.54 10⁻¹⁰) = 3.1329 10⁻¹⁴