± Energy of a Capacitor in the Presence of a Dielectric A dielectric-filled para
ID: 1614862 • Letter: #
Question
± Energy of a Capacitor in the Presence of a Dielectric
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant k = 2.00. The capacitor is connected to a battery that creates a constant voltage V = 5.00 V . Throughout the problem, use 0 = 8.85×1012 C2/Nm2
A) Find the energy U1 of the dielectric-filled capacitor. ANSWER: U1= 7.38x10 ^11J
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. ANSWER: U2 = 5.53×10^-11 J
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.)
Previous Incorrect answer = 8.90*10^-11 J
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.)
Previous Incorrect answer = 3.05*10^-11 J
Explanation / Answer
B)There would be two cap in parallel now, one with air in core and other with dielectric
C1 = epsilon0 A/d
C1 = 8.85 x 10^-12 x 0.0015/0.009 = 1.48 x 10^-12
C2 = k epsilon0 A/d
C2 = 3 x C1 = 4.44 x 10^-12
Ceq = C1 + C2 = (1.48 + 4.44) x 10^-12
Ceq = 5.92 x 10^-12 F
U' = 1/2 Ceq V^2
U' = 0.5 x 5.92 x 10^-12 x 25 = 74 x 10^-12 J
Hence, U' = 74 pJ
C)Q is conserved
Q = CV
Q = 5.92 x 10^-12 x 5 = 29.6 x 10^-12 C
Q = 29.6 pC
C = epsilon0 a/d
C = 8.85 x 10^-12 x 0.003/0.009 = 2.95 x 10^-12 F = 2.95 pF
U'' = 1/2 Q^2/C
U'' = 0.5 x (29.6 x 10^-12)^2/2.95 x 10^-12 = 148 x 10^-12 J
Hence, U'' = 148 pJ
D)Work done will be:
W = 148 pJ - 74 = 74 pJ
Hence, W = 74 pJ
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