A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm^2, plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm^2, plate separation d = 5.00 mm and dielectric constant K = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 10.0 V. Throughout the problem, use epsilon_0 = 8.85×10-12 C^2/N cdot m^2.

1. 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, U_3.

2.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?

Explanation / Answer

a) where K = dieelctric constant = 5            A = area = 20 cm 2 = 20 * 10 -4 m2             d = separation between the plates = 5 mm = 5 * 10 -3m substitue values weget C value where V = 10 V capacitance of the capacitor C = K o A / d = 17.7 * 10^-12 F energy stored in in the capacitor U = ( 1/ 2) C V2 = 8.85 * 10^-10 J The new energy stored in the capacitor, U3 = U/K = 1.77 * 10^-10 J b) We know that the difference in the energy is equal to the workdone. So the difference in potential energy = U - U3 = 7.08 * 10^-10 J This difference in the potential energy is equal to the work done by the external agent acting on the dielectric capacitance of the capacitor C = K o A / d = 17.7 * 10^-12 F capacitance of the capacitor C = K o A / d = 17.7 * 10^-12 F energy stored in in the capacitor U = ( 1/ 2) C V2 = 8.85 * 10^-10 J The new energy stored in the capacitor, U3 = U/K = 1.77 * 10^-10 J b) We know that the difference in the energy is equal to the workdone. So the difference in potential energy = U - U3 = 7.08 * 10^-10 J This difference in the potential energy is equal to the work done by the external agent acting on the dielectric                                         

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