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## Homework Statement

Two capacitors,

*C*1 = 2910 pF and

*C*2 = 2230 pF, are connected in series to a 12.0 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? (Answer needs to be in units of Coulombs)

## Homework Equations

Q = C*V

V = Q/C

(Total capacitance): CTotal = 1/C1 + 1/C2

(Charge of capacitors in series is equal) Q of C1 = Q of C2

**3. The atempt at a solution**

I found the total Capacitance by doing the following:

1/C1 + 1/C2 = 1/( 1/2910 + 1/2230) = 11262.509 PF

I found the Charge of the Capacitors by doing the following:

Q = C*V = (1262.509pF)*(12V) = 1.5150108*10^-8 Coulombs (after conversion)

From here on is where I'm a bit confused. So what I did is:

V = Q/C =(1.5150108*10^-8 Coulombs)/(1.2625*10^-9F) = 12.00008554 Volts (this doesn't make sense to me because it is basically the voltage that was given in the problem)

From there:

q1 (Charge after disconnecting) = C*V = 1.2625 (12) = 1.5150108*10^-8 Coulombs (which is what the charge was before disconnecting)

I am assuming that the charge on the Capacitors changes after disconnecting? It doesn't make sense that my calculations say that the charge is still the same. Can someone please help me out and point me in the right direction?

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