# capacitor transfer

Hi,

I have 2 capacitors c1, c2

c1=.05 uF
c2=.1 uF

Q=CV

c1 is fully charged and connected to a .4V battery in a parallel circuit
c2 is uncharged

c2 was unconnected initially to c1 but once joined c2 can receive a charge from c1.

charge on C1=2 X 10^-8

what is the % charge transferred from c1 to c2 once joined in the circuit

I thought the answer is 100% but it is 70%.

I  dont get it?
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Commented:
What is the total capacitance of the new circuit Ctotal? (with c1 and c2 connected: parallel or series from the question's assertion .15uF or .098uF ?)
What is the charge on the Combined Ctotal?
Then disperse the charge to the individual members.
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Commented:
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Commented:
>>  c1 is fully charged and connected to a .4V battery in a parallel circuit

If the battery is still part of the circuit when c2 is connected, then c2 charges from the battery.
No charge is transferred from c1.
But usually you charge the cap and remove the battery in this sort of problem.

>>  What is the % charge transferred from c1 to c2 once joined in the circuit.
I thought the answer is 100% but it is 70%.

Before the capacitors are connected, one of them has all the voltage and all the charge.
After they are connected, the charge is conserved and flows until the voltages on the capacitors are equal.
If the capacitors were equal, each would have 50% of the original charge.
There is no way to transfer 100% of the charge.

The equation you have to solve is     c1*V  +  c2*V  =  c1*[0.4]
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Commented:
You had all the info to get to the equation d-glitch provided.

Redrawing he circuit often makes it clearer.
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