31_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

31_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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Electrical Energy and Capacitance 65 The final charge on each of the capacitors is ()( ) ( ) 2 11 4.0 F 44.4 V 1.8 10 C QC V µ =∆= = × and 22 2.0 F 44.4 V 89 C = 16.63 (a) 123 2 e ee kQ V V xd x =++= + +− () ( VV ) 2 e xx d x d xx d  −− − + +  =  which simplifies to 32 kQd d d == V (b) When and , then xdx >> 2 2 2 e d = V becomes 2 3 2 e x V 16.64 The energy required to melt the lead sample is 6 3 6.00 10 kg 128 J kg C 327.3 C 20.0 C 24.5 10 J kg 0.383 J Pb f
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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