A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm2 , plate

Question

A dielectric-filled parallel-plate capacitor has plate area A = 20.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 4.00. The capacitor is connected to a battery that creates a constant voltage V = 7.50 V . Throughout the problem, use 0 = 8.85×1012 C2/Nm2 .

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.

Express your answer numerically in joules.

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 numerically in joules.

Explanation / Answer

Given; A = 20.0 cm^2, d = 9.0 mm, k = 4.00, V = 7.50 v, E0 = 8.85*10^-12 C^2/N.m^2

The capacitance when the dielectric is half filled

C = (1/2)*k*E0*A/d + (1/2)*E0*A/d

= (1/2)*4.00*(8.85*10^-12)*(20.0*10^-4)/(9.0*10^-3) + (1/2)*(8.85*10^-12)*(20.0*10^-4)/(9.0*10^-3)

= 3.933*10^-12 + 0.983*10^-12

C = 4.916*10^-12 F

the energy stored in the capacitor

U2 = (1/2)*C*V^2

= (1/2)*4.916*10^-12*7.50^2

U2 = 1.383*10^-10 J

The capacitor is now disconnected, and the charge stored will be constant

Q = C * V

= 4.916*10^-12 * 7.50

= 3.687*10^-11 C

new energy of the capacitor

U3 = (1/2)*Q^2/C

C = E0*A/d

= 8.85*10^-12 * 20.0*10^-4/9.0*10^-3

C = 1.96*10^-12 F

U3 = (1/2)*(3.687*10^-11)^2/1.96*10^-12

= 3.467*10^-10 J

the work done by removing the remaining half of the capacitor is the increase in energy stored in the capacitor

W = U3 - U2

= (3.467 -1.383)*10^-10

= 2.084*10^-10 J

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