How Do You Calculate the Charge on a Capacitor in Series?

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SUMMARY

The discussion focuses on calculating the charge on a 9 μF capacitor connected in series with two other capacitors (8 μF and 5 μF) to a 4 V battery. The correct method involves first determining the equivalent capacitance using the formula 1/C = 1/C1 + 1/C2 + 1/C3. The equivalent capacitance for the series connection is approximately 2.0 μF, leading to a charge of 8 μC on each capacitor, including the 9 μF capacitor. The initial calculation of 1.74 μC was incorrect due to a miscalculation of the equivalent capacitance.

PREREQUISITES
  • Understanding of capacitor series connections
  • Familiarity with the formula for equivalent capacitance (1/C = 1/C1 + 1/C2 + 1/C3)
  • Knowledge of the relationship between charge, capacitance, and voltage (Q = C*E)
  • Basic circuit diagram interpretation skills
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Homework Statement



Three capacitors of capacitance 8, 9, and 5 μF respectively, are connected in series to a 4 V battery. What is the charge on the positive plate of the 9 μF capacitor?


Homework Equations



Q = C*E
1/C = 1/C1 + ..

The Attempt at a Solution



I drew out the circuit it would make. I know that to calculate C it is 1/C = 1/C1+1/C2...
I took the general equation for Q= C*E. I did this and got 1.74 uC. But, this is not correct. Could you tell me where I am going wrong?
 
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Hi purduegirl,

You calculation of the equivalent capacitance is wrong. Your post does not have many numbers, but you seem to have calculated that the equivalent capacitance was about 0.435 [itex]\mu[/itex]F, but that is not what 1/C = 1/C1+1/C2... gives.

Do you see what went wrong, and what the equivalent capacitance should be?
 

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