Question

Two capacitors, C1=5900pF and C2=2600pF, are connected in series to a 15.0 V battery. The capacitors...

Two capacitors, C1=5900pF and C2=2600pF, are connected in series to a 15.0 V battery. The capacitors are later disconnected from the battery and connected directly to each other, positive plate to positive plate, and negative plate to negative plate. What then will be the charge on each capacitor?

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Answer #1

in series, their equivalent C is 5900*2600/(5900 + 2600 ) = 1804.706 pF

With a 15 V battery, the charge is Q = CV = 1804.705*15 = 27070.588 pC

When they are disconnected, they each have that charge, but at different voltages. At that point the total charge is 2*27070.588 or 54141.176 pC

The charge is divided between the two Cs. Parallel equivalent of the two is 5900+2600 = 8500 pF, and voltage = Q/C = 54141.176 / 8500 = 6.369 volts.

Charge on 5900 pF = Q = CV = 5900*6.369 = 37580.345 pC
Charge on 1800 pF = Q = CV = 2600*6.369 = 16560.83  pC

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