Need help with part B,C,D PCC -PHY 152-85 58820576&offset; next Spring 2016 dy 1

Question

Need help with part B,C,D PCC -PHY 152-85 58820576&offset; next Spring 2016 dy 152 Lab 02 t Energy or a Capacitor in the Presence of a Dielectric soned inai Lewis Energy of a Capacitor in the Presence of a Dielectric Part B pacitor, which remains c A delectric filled parallel-plate capacitor has plate area A moment when the capacitor is harflled wth the dielectric 30 0 cm plate separation d 500 mm and dielectric Express your answer numerically in joules. constant k 5.00. The capacitor is connected to a battery s a constant voltage V 10.0 V problem, use eo 8.85x10 12 C /N.me Hints Mh Answers Give up Review Part Incorrect Try Again; 4 attempts remaining Part C slowly removed the rest of the way out of the capactor Find the new energy of the capacitor Express your answer numerically in joules.

Explanation / Answer

A = (30 cm²) (100 cm m)² = 0.003 m²

d = 0.0095m

C = (ø)•()•A d
C = [8.85 ×10^(-12)]•(5)•(0.005) (0.005)
C = 4.425 ×10^(-11) F


U = (½)•C•V²
U = (½)•[4.425 ×10^(-11)]•(10²)
U = 221 ×10^(-12) Joules

______________________________________...

At the halfway point there are essentially two capacitors in parallel

C = (ø)•A d...(air region)
C = [8.85 ×10^(-12)]•(0.0015) (0.005)
C = 2.655×10^(-12) F
C = 2.655 pF

C = (ø)•()•A d...(dielectric region)
C = [8.85 ×10^(-12)]•(5)•(0.0015) (0.005)
C = 1.327 ×10^(-11) F


C = C + C
C = 1.475 + 4.425 = 1.58*10^11F

U_2 = (½)•C•V²
U_2 = (½)•[1.58 ×10^(-11)]•(10²)
U_2 = 7.9 ×10^(-10) Joules

______________________________________...

Q is conserved in this case:

Q = C•V
Q = [1.58 ×10^(-11)]•(10)
Q = 1.58×10^(-10) C

C = (ø)•A d...(completely air)
C = [8.85 ×10^(-12)]•(0.005) (0.005)
C =8.85 ×10^(-12) F


U_3 = (½)•C•V²
U_3 = (½)•C•(Q C)²
U_3 = (½)•Q² C
U_3 = (½)•[1.58 ×10^(-11)]² 8.85 ×10^(-12)
U_3 = 1.63*10^24J
_________________________________

Work = U = (163.1*10^12J) (7.9 ×10^(-10)J)

Work = 16.31*10^12 pJ

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