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 = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 12.5 V . Throughout the problem, use 0 = 8.85×1012 C2/Nm2 .

B.

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.

Express your answer numerically in joules.

C.

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.

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 numerically in joules.

Explanation / Answer

B)
One half of the parallel plate is filled with the dielectric and the other half with air. So we have two parallel plate capacitors in parallel.
Net capacitance = 0 x 0.5 A/d + k0 x 0.5A/d = 0.5 x (k+1) x 0A/d = 3 x 0A/d
0A/d = [(8.85 × 10-12) x (20 × 10-4)]/0.009 = 1.966 × 10-12 F
Net capacitance = 5.9 × 10-12 F
Since it is connected to battery, voltage is still 12.5 V
Energy, U2 = 0.5 x C x V2 = 4.61 × 10-10 J
Total charge, Q = CV = 73.75 × 10-12 C

C)
Charge is same as in B since there is no way charge could escape.
Capacitance = 0A/d = 1.966 × 10-12 F
Energy, U3 = Q2/2C = 13.83 × 10-10 J

D)
Work done is the change in energy
W = U3 - U2 = 9.22 × 10-10 J

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