problem26_81

problem26_81 - Q Q dt t dQ f f at current the So RC Q T Q...

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26.81: a) If the given capacitor was fully charged for the given emf, = = CV Q max . C 10 12 . 6 ) V 180 )( F 10 4 . 3 ( 4 6 - - × = × Since it has more charge than this after it was connected, this tells us the capacitor is discharging and so the current must be flowing toward the negative plate. The capacitor started with more charge than was “allowed” for the given emf. Let f RC t f f Q e Q Q t Q t Q t Q Q t Q + - = = = = = - ) ( ) ( , all For . ) ( and ) 0 ( 0 0 from and C 10 15 . 8 ) ( ; time some at given are We 4 - × = = = T t Q T t Q = = = = × = - - - ) ( , At . ) ( current The . C 10 12 . 6 above ) ( ) ( 4 0 T Q T t e t I Q RC t RC
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Unformatted text preview: Q Q dt t dQ f f . ) ( ) ( at current the So . ) ( ) ) ( ( ) ( RC Q T Q RC T RC Q Q f RC T f f f e T I is T t Q e Q Q-----=-= = +-plate). negative the (toward A 10 24 . 8 ) ( Thus 3 ) 10 40 . 3 )( 10 25 . 7 ( C 10 12 . 6 C 10 15 . 8 6 3 4 4-× Ω × × + ×-×-= =---F T I b) As time goes on, the capacitor will discharge to C 10 12 . 6 4-× as calculated above....
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This homework help was uploaded on 04/19/2008 for the course PHYS 4 taught by Professor Stuart during the Spring '08 term at UCSB.

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