P26_062 - C and 60 µ ±). We do expect charge to be...

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62. We cannot expect simple energy conservation to hold since energy is presumably dissipated either as heat in the hookup wires or as radio waves while the charge oscillates in the course of the system “settling down” to its Fnal state (of having 40 V across the parallel pair of capacitors
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Unformatted text preview: C and 60 µ ±). We do expect charge to be conserved. Thus, if Q is the charge originally stored on C and q 1 , q 2 are the charges on the parallel pair after “setting down,” then Q = q 1 + q 2 C (100 V) = C (40 V) + (60 µ ±) (40 V) which leads to the solution C = 40 µ ±....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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