Finging the charge on each capacitor in series

[CogitoEAS]

Finding the charge on each capacitor in series

Homework Statement

In the circuit in the figure below, assume the resistance values are R1=1,400Ω and R2=2,500Ω, C1=55μF and C2=15μF, and the emf=5.5 V. The switch is labeled S.

Consider the circuit in the figure below and assume the switch has been closed for a very long time.

What are the charges on the two capacitors?

Homework Equations

C=Q/V, Loop Rule - 5.5-(Q1/55e-6)-(Q2/15e-6)=0, 1/Ceq=(1/C1)+(1/C2)

The Attempt at a Solution

I think I'm just blanking on something here because it's either a matter of finding the potential across each capacitor or finding another equation relating Q1 and Q2 so that I can solve for both. Maybe Qtotal=CV(1-e^(-t/tau))=Q1+Q2?

 

[ehild]

Homework Statement

In the circuit in the figure below, assume the resistance values are R1=1,400Ω and R2=2,500Ω, C1=55μF and C2=15μF, and the emf=5.5 V. The switch is labeled S.

Consider the circuit in the figure below and assume the switch has been closed for a very long time.

Homework Equations

C=Q/V, Loop Rule - 5.5-(Q1/55e-6)-(Q2/15e-6)=0, 1/Ceq=(1/C1)+(1/C2)

The Attempt at a Solution

I think I'm just blanking on something here because it's either a matter of finding the potential across each capacitor or finding another equation relating Q1 and Q2 so that I can solve for both. Maybe Qtotal=CV(1-e^(-t/tau))=Q1+Q2?

What is the question?

How are the charges on series capacitors related?

ehild

 

[CogitoEAS]

Yes, sorry, I've added the central question "What are the charges on the two capacitors?" to the original post. I'm also 100% sure I spelled FINDING right in the title but for some reason the d keeps getting changed to a g?!

 

[ehild]

Two capacitors in series are connected to a battery. How are their charges related? Check your lecture notes, book, or see http://www.electronics-tutorials.ws/capacitor/cap_7.html

ehild

 

[CogitoEAS]

Thank you!

 

[epenguin]

And we hen they say it has been switched on for a very long time they are hoping you will reason out 'the capacitors are as charged as they ever will be, so no current isnflowingr, now the voltage drop across resistors when there is no current is...'