An air-filled parallel-plate capacitor has plate area A and plate separation d .

Question

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.

A)Find the energy U0 stored in the capacitor.

Express your answer in terms of A, d, V, and 0. Remember to enter 0 as epsilon_0.

B) 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 U1 of the capacitor after this process.

Express your answer in terms of A, d, V, and 0.

C)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 U2 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 0.

Explanation / Answer

A) Capacitance is C = eo*A/d

Uo = 0.5*C*V^2 = (0.5*eo*A/d)*V^2

B) now new capacitance is Cnew = eo*A/(3d)

U1 = 0.5*Cnew*V^2 = 0.5*eo*A*V^2/(3d)

C) U2 = 0.5*C*V^2 = 0.5*K*eo*A*V^2/d

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