Homework Answers
given
V = 550 V
C = 2900 pF
= 2900*10^-12 F
Potential energy stored, PE = (1/2)*C*V^2
= (1/2)*2900*10^-12*550^2
= 4.39*10^-4 J
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Constants Part A 550 V is applied to a 2900-pF capacitor. How much energy is stored?...
Physics 2 Help PartA 600 V is applied to a 2900-pF capacitor How much energy is...
Physics 2 Help
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Physics 2 Help
PartA 600 V is applied to a 2900-pF capacitor How much energy is stored? Express your answer using three significant figures and include the appropriate units. | PB- İ Value Units
Review | Constants A parallel-plate capacitor consists of two plates, each with an area of 29 cm2 separated by 3.0 mm. The charge on the capacitor is 43 nC. A proton is released from rest next to the positive plate Part A Express your answer to two significant figures and include the appropriate units. Voluee sec Submit Request Answer Provide Feedback Next >
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