Mastering Physics: The Capacitor as an Energy-Storing Device

Mastering Physics: The Capacitor as an Energy-Storing Device

An air-filled parallel-plate capacitor has plate area A and plate separation d. The capacitor is connected to a battery that creates a constant voltage V.

Part A

Question: Find the energy U_0 stored in the capacitor. Express your answer in terms of A, d, V, and ϵ_0. Remember to enter ϵ_0 as epsilon_0.

Answer: U_0 =  ϵ_0 × A × (V)^2 / 2d

Part B

Question: The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. Find the new energy U_1 of the capacitor after this process.Express your answer in terms of A, d, V, and ϵ_0.

Answer: U_1 =  3 × ϵ_0 ×A ×(V)^2 / 2d

Part C

Question: The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved into the capacitor until the entire space between the plates is filled. Find the energy U_2 of the dielectric-filled capacitor. The capacitor remains connected to the battery. The dielectric constant is K.Express your answer in terms of A, d, V, K, and ϵ_0.

Answer: U_2 =  K × ϵ_0 ×A × (V)^2 / 2d