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

[purduegirl]

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?

 

[alphysicist]

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 $\mu$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?

 

[Moetasim]

Hello,

Thank you for your question. It seems that you have correctly set up the equation for calculating the equivalent capacitance in a series circuit. However, there may be a mistake in your calculation for the charge on the positive plate of the 9 μF capacitor.

To find the charge on a specific capacitor in a series circuit, we can use the formula Q = C*V, where Q is the charge, C is the capacitance, and V is the voltage. In this case, the voltage across all three capacitors is 4 V, so we can use this formula to find the charge on the 9 μF capacitor.

Q = (9 μF)*(4 V) = 36 μC

Therefore, the charge on the positive plate of the 9 μF capacitor is 36 μC.

I hope this helps clarify your understanding of effective capacitance and how to calculate the charge on a specific capacitor in a series circuit. If you have any further questions, please don't hesitate to ask.

Best regards,