What is the final charge and potential difference on each capacitor?

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SUMMARY

The final charge on capacitor C1 is calculated to be 1.4 x 10-5 C when a voltage of 30 V is applied across the circuit containing capacitors C1 (1.0 µF), C2 (1.9 µF), and C3 (2.8 µF). The charge on capacitors C2 and C3 is determined to be equal, denoted as q, with a value of 15.9 µC. The relationship between the charges and capacitances is established using the equation Q=CV, and the final potential differences across each capacitor are derived from the total voltage and individual charge distributions.

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ooohffff
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Homework Statement


In the circuit shown in the figure, C1 = 1.0 µF, C2 = 1.9 µF, C3 = 2.8 µF, and a voltage Vab = 30 V is applied across points a and b. After C1 is fully charged the switch is thrown to the right. What is the final charge and potential difference on each capacitor?

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Homework Equations



Q=CV

The Attempt at a Solution



I found everything (V1, Q2, V2, Q3, V3) except for Q1. I tried doing Q1 = CeffV1, but the answer was wrong.
 
Last edited:
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Ok so since V1=V2+V3, I did Q1/C1 = Q2/C2 + Q3/C3. Solving for Q1, I get Q1 = C1[ Q2/C2 + Q3/C3].

Q2=Q3=q

Q1=C1*q[1/C2 + 1/C3]

Plugging in the numbers, when q = 15.9μC, I get Q1 = 1.4*10-5 C. Does that look right?
 
Last edited:
ooohffff said:
Plugging in the numbers, when q = 15.9μC, I get Q1 = 1.4*10-5 C. Does that look right?
That looks right for the final charge on C1, yes.
 
gneill said:
That looks right for the final charge on C1, yes.
Thank you!
 

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