36 uF capacitor is being charged by a voltage source and the energy stored is equal to 247.41 uj. While the capacitor is still connected to the voltage source, a dielectric material is ncreases by 324.67 uj. The dielectric constant of this material is equal to

A 15 µF capacitor is connected to a 50V battery and becomes fully charged. The battery is removed and a slab of dielectric that completely fills the space between the plates is inserted. If the dielectric has a dielectric constant of 5.0, (a) what is the capacitance of the capacitor after the slab is inserted? (b) What is the voltage after the dielectric is inserted? (c) How much energy is stored after the dielectric is inserted?

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2cm2 , plate separation d = 5.00 mmmm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 VV . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2C2/N⋅m2 . 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 is done by the external agent acting on the dielectric?

The capacitor is charged to a voltage of 4.00 kv using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kv despite the reduction in the potential difference between the plates. Calculate the initial capacitance value of the system.