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hw7 - Homework 7 Solutions Spring 2010 Required Problems...

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Homework 7 Solutions Spring 2010 Required Problems Chapter 24 8. ( a )The total charge on the combination of capacitors is the sum of the charges on the two individual capacitors, since there is no battery connected to them to supply additional charge, and there is no neutralization of charge by combining positive and negative charges. The voltage across each capacitor must be the same after they are connected, since each capacitor plate is connected to a corresponding plate on the other capacitor by a constant-potential connecting wire. Use the total charge and the fact of equal potentials to find the charge on each capacitor and the common potential difference. 1 1 1 2 2 2 1 1 final 2 2 final initial initial initial initial final final Total 1 2 1 2 1 1 2 2 1 final 2 final initial initial final final initial initial 1 1 2 2 initial i final Q CV Q C V Q CV Q C V Q Q Q Q Q CV C V CV C V CV C V V = = = = = + = + = + = + + = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 6 nitial 6 1 2 1 2 6 3 1 1 final final 6 3 2 2 final final 2.70 10 F 475V 4.00 10 F 525V 6.70 10 F 504.85V 505V 2.70 10 F 504.85V 1.36 10 C 4.00 10 F 504.85V 2.02 10 C C C V V Q CV Q C V - - - - - - - × + × = + × = = = = = × = × = = × = × ( b )By connecting plates of opposite charge, the total charge will be the difference of the charges on the two individual capacitors. Once the charges have equalized, the two capacitors will again be at the same potential.
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1 1 1 2 2 2 1 1 final 2 2 final initial initial initial initial final final Total 1 2 1 2 1 1 2 2 1 final 2 final initial initial final final initial initial 1 1 initial final Q CV Q C V Q CV Q C V Q Q Q Q Q CV C V CV C V CV C V = = = = = - = + - = + - = ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 6 6 2 2 initial 6 1 2 1 2 6 4 1 1 final final 6 3 2 2 final final 2.70 10 F 475V 4.00 10 F 525V 6.70 10 F 122.01V 120V 2.70 10 F 122.01V 3.3 10 C 4.00 10 F 122.01V 4.9 10 C V C C V V Q CV Q C V - - - - - - - × - × = + × = = = = = × = × = = × = × 30. C 1 and C 2 are in series, so they both have the same charge. We then use that charge to find the voltage across each of C 1 and C 2 . Then their combined voltage is the voltage across C 3 . The voltage across C 3 is used to find the charge on C 3 . ( 29 ( 29 1 2 1 2 1 2 1 2 3 1 2 3 3 3 12.4 C 12.4 C 12.4 C ; 0.775V ; 0.775V 16.0 F 16.0 F 1.55V ; 16.0 F 1.55V 24.8 C Q Q Q Q V V C C V V V Q C V μ μ μ μ μ μ μ = = = = = = = = = + = = = = From the diagram, C 4 must have the same charge as the sum of the charges on C 1 and C 3 .
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