An dielectric-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. The dielectric constant is K. A. Find the energy U_1 of the dielectric-filled capacitor. The capacitor remains connected to the battery I was able to figure out this one: (epsilon_0*K*A*V^2)/(2d) But I cannot figure out the rest... B.The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U_2 of the capacitor at the moment when the capacitor is half filled with the dielectric. Express your answer in terms of A,d,V,K,epsilon_0. C.The capactor is now disconnected from the battery, and the dielectric plate is then slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U_s.Express your answer in terms of A,d,V,K,epsilon_0 D. In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work W is done by the external agent acting on the dielectric.Express your answer in terms of A,d,V,K,epsilon_0 ANY help would be greatly appreciated.... Thanks in advance
When the capacitor is half filled with the dielectric, you can imagine two capacitors having area A/2 each and connected in parallel. One of them is with dielectric and other without dielectric. Calculate the equivalent capacitance and charge on it. By this hint you can solve C and D
C1 = epsilon_0*A/2*/d and C2 = epsilon_0*K*(A/2)/d. Since the battery is still connected the potential difference is same in C1 and C2.