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Unformatted text preview: 2 But also V1 0 V2 0 V3 # V # 36 V, so U # 1 QV # 1 (76 $ C)(36 V) # 1.4 " 10!3 J. 2 2 EVALUATE:! We could also use U # Q 2 / 2C and calculate U for each capacitor. (c) IDENTIFY:! The charges on the plates redistribute to make the potentials across each capacitor the same. SET UP:! The capacitors before and after they are connected are sketched in Figure 24.61b. Figure 24.61b EXECUTE:! The total positive charge that is available to be distributed on the upper plates of the three capacitors is Q0 # Q01 0 Q02 0 Q03 # 3 . 76 $ C / # 228 $ C. Thus Q1 0 Q2 0 Q3 # 228 $ C. After the circuit is completed the charge distributes to make V1 # V2 # V3 . V # Q / C and V1 # V2 so Q1 / C1 # Q2 / C2 and then C1 # C2 says Q1 # Q2 . V1 # V3 says Q1 / C1 # Q3 / C3 and Q1 # Q3 . C1 / C3 / # Q3 . 8.4 $ F/4.2 $ F / # 2Q3 Using Q2 # Q1 and Q1 # 2Q3 in the above equation gives 2Q3 0 2Q3 0 Q3 # 228 $ C. 5Q3 # 228 $ C and Q3 # 45.6 $ C, Q1 # Q2 # 91.2 $ C Q1 91.2 $ C Q 91.2 $ C Q 45.6 $ C # # 11 V, V2 # 2 # # 11 V, and V3 # 3 # # 11 V. C1 8.4 $ F C2 8.4 $ F C3 4.2 $ F The voltage across each capacitor in the parallel combination is 11 V. (d) U # 1 QV1 0 1 Q2V2 0 1 Q3V3 . 21 2 2 Then V1 # But V1 # V2 # V3 so U # 1 V1 . Q1 0 Q2 0 Q3 / # 2 24.62. 1 2 .11 V / . 228 $ C / # 1.3 " 10!3 J. EVALUATE:! This is less than the original energy of 1.4 " 10!3 J. The stored energy has decreased, as in Example 24.7. !A Q IDENTIFY:! C # 0 . C # . V # Ed . U # 1 QV . 2 d V 3 2 SET UP:! d # 3.0 " 10 m . A # , r , with r # 1.0 " 103 m . ! A (8.854 " 10!12 C 2 /N + m 2 ), (1.0 " 103 m) 2 EXECUTE:! (a) C # 0 # # 9.3 " 10!9 F . d 3.0 " 103 m Q 20 C (b) V # # # 2.2 " 109 V C 9.3 " 10!9 F V 2.2 " 109 V (c) E # # # 7.3 " 105 V/m d 3.0 " 103 m (d) U # 1 QV # 1 (20 C)(2.2 " 109 V) # 2.2 " 1010 J 2 2 EVALUATE:! Thunderclouds involve very large potential differences and large amounts of stored energy. Capacitance and Dielectrics 24.63. 24-19 I...
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.

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