Question

A 6.00-?F parallel plate capacitor has a charge of +40.0 ?C and -40.0 ?C on each plate,...

A 6.00-µF parallel plate capacitor has a charge of +40.0 µC and -40.0 µC on each plate, respectively. The potential energy stored in this capacitor is:

a) 123 µJ
b) 133 µJ
c) 113 µJ
d) 143 µJ
e) 103 µJ
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Answer #1

PE = q^2 / 2C

= 40? * 40? / (2*6?)

= 133.3333 ?J

answered by: Tarah
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Answer #2
v=c/q=40/6
U=0.5cv^2
U=0.5x6^2x40=133


b
answered by: spike
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Answer #3
E = q^2/2C = 133 uJ
option B
answered by: Tynia
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Answer #4

b) 133 µJ

answered by: donna phan
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Answer #5

PE = q^2 / 2C

= 40? * 40μ/ (2*6j)

= 133.3333 μJ

b is correct

answered by: JL
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