A parallel plate capacitor with plate separation d is charged to a certain potential using a battery, and then the battery is disconnected. An uncharged metal slab of thickness d 3 and area equal to that of the plates is then fully inserted between the plates. What is the ratio of the initial electrostatic energy stored in the capacitor to the final electrostatic energy?

Step-by-step solution

Let the initial capacitance be C_i = ₀ A d . When a metal slab of thickness t = d 3 is inserted, the new capacitance becomes: C_f = ₀ A d - t = ₀ A d - d 3 = ₀ A 2d 3 = 3 2 C_i Since the battery is disconnected, the charge Q on the capacitor remains constant. The electrostatic energy stored in a capacitor is given by U = Q^2 2C . The ratio of initial to final energy is: U_i U_f = Q^2 2C_i Q^2 2C_f = C_f C_i = 3 2 C_i C_i = 3 2 Thus, the ratio is 3:2 . Answer: 3:2