A dielectric-filled parallel-plate capacitor has plate area A = 30.0 c m 2 , pla

Question

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 c m 2 , plate separation d = 8.00 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 C 2 /N m 2 .

Part A

Find the energy U1 of the dielectric-filled capacitor.

Express your answer numerically in joules.

Part 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

Part 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.

Part 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

A = (30 cm²) (100 cm m)² = 0.003 m²

d = 0.008 m
part A :
C = (ø)•()•A d
C = [8.85 ×10^(-12)]•(5)•(0.003) (0.008)
C = 16.59 ×10^(-12) F
C = 16.59 pF

U = (½)•C•V²
U = (½)•[16.59 ×10^(-12)]•(12.5²)
U = 1296 ×10^(-12) Joules
U = 1296 pJ
______________________________________...
partB:
At the halfway point there are essentially two capacitors in parallel

C = (ø)•A d...(air region)
C = [8.85 ×10^(-12)]•(0.0015) (0.008)
C = 1.659 ×10^(-12) F
C = 1.659 pF

C = (ø)•()•A d...(dielectric region)
C = [8.85 ×10^(-12)]•(5)•(0.0015) (0.009)
C = 8.295 ×10^(-12) F
C = 8.295 pF

C = C + C
C = 1.659+ 8.295 = 9.954 pF

U_2 = (½)•C•V²
U_2 = (½)•[9.954 ×10^(-12)]•(12.5²)
U_2 = 1555.31 ×10^(-12) Joules
U_2 =1555 pJ
______________________________________...

part C:

                  Q is conserved in this case:

Q = C•V
Q = [9.954 ×10^(-12)]•(12.5)
Q = 124.425 ×10^(-12) C
Q = 124.425 pC


C = (ø)•A d...(completely air)
C = [8.85 ×10^(-12)]•(0.003) (0.008)
C = 3.32 ×10^(-12) F
C = 3.32 pF

U_3 = (½)•C•V²
U_3 = (½)•C•(Q C)²
U_3 = (½)•Q² C
U_3 = (½)•[124.425 ×10^(-12)]² 3.32×10^(-12)
U_3 = 4463 pJ
_________________________________
part D:
Work = U = ( 4463 pJ) (1555 pJ)

Work = 2908 pJ

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