A dielectric-filled parallel-plate capacitor has plate area A = 20.0 ccm2 , plate separaton d = 10.0 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 . Find the energy U1 of the dielectric-filled capacitor. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. 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?

In (Figure 1) C1 = 6.00 μF, C2 = 6.00 μF, C3 = 12.0 μF, and C4 = 3.00 μF. The capacitor network is connected to an applied potential difference Vab. After the charges on the capacitors have reached their final values, the voltage across C3 is 40.0 V. What is the voltage across C4? What is the voltage Vab applied to the network? Please explain everything in steps.

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