A parallel plate capacitor of capacitance C₀ is charged to a potential difference V₀ by a battery. The battery is then disconnected. A dielectric slab of thickness equal to half the plate separation and dielectric constant K=3 is inserted between the plates. The final electrostatic energy stored in the capacitor is :

Step-by-step solution

Since the battery is disconnected, the charge on the capacitor remains constant. Initial charge, Q = C₀ V₀ Let the plate separation be d . The initial capacitance is C₀ = ₀ A d . When a dielectric slab of thickness t = d 2 and K=3 is inserted, the new capacitance C_f is: C_f = ₀ A d - t + t K = ₀ A d - d 2 + d 2 3 C_f = ₀ A d 2 + d 6 = ₀ A 4d 6 = 3 2 ₀ A d = 3 2 C₀ The final electrostatic energy stored in the capacitor is: U_f = Q^2 2 C_f = (C₀ V₀)^2 2 ( 3 2 C₀ ) = C₀^2 V₀^2 3 C₀ = 1 3 C₀ V₀^2 Answer: 1 3 C₀ V₀^2