A parallel-plate capacitor with a plate area of 400 cm ^2 and a separation of 1.0 mm is connected to a 100 V power supply. While the supply remains connected, a dielectric slab with a thickness of 1.0 mm and a dielectric constant of 5.0 is introduced into the gap. Calculate the increase in the electrostatic energy of the capacitor.

Step-by-step solution

The initial capacitance of the parallel-plate capacitor is C₀ = ₀ A d . Substituting A = 400 cm ^2 = 4 10⁻² m ^2 and d = 1.0 mm = 10⁻³ m : C₀ = (8.85 10⁻¹²) (4 10⁻²) 10⁻³ = 3.54 10⁻¹⁰ F = 354 pF . The initial electrostatic energy is U_i = 1 2 C₀ V^2 = 1 2 (3.54 10⁻¹⁰) (100)^2 = 1.77 10⁻⁶ J = 1.77 J . When the dielectric slab of thickness 1.0 mm (which completely fills the gap) is inserted, the new capacitance is C_f = K C₀ = 5(354) = 1770 pF . Since the power supply remains connected, the voltage V is constant. The