Asked by lo
Capacitors C1 = 5.0 µF and C2 = 2.0 µF are charged as a parallel combination across a 150 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.
Answers
Answered by
drwls
C1 acquires C1*V = 7.5*10^-4 C and
C2 acquires C2*V = 3.0*10^-4 C, initially. When rewired, there is a total of 4.5*10^-4 C on the positive-plate sides and -4.5*10^-4 C on the negative-plate sides. The charges redistribute to maintain the same potential on each capacitors.
(2/7)*4.5*10^-4 = 1.29*10^-4C will be on the smaller capacitor and
(5/7)*4.5*10^-4 = 3.21*10^-4 C will be on the larger capacitor.
C2 acquires C2*V = 3.0*10^-4 C, initially. When rewired, there is a total of 4.5*10^-4 C on the positive-plate sides and -4.5*10^-4 C on the negative-plate sides. The charges redistribute to maintain the same potential on each capacitors.
(2/7)*4.5*10^-4 = 1.29*10^-4C will be on the smaller capacitor and
(5/7)*4.5*10^-4 = 3.21*10^-4 C will be on the larger capacitor.
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