Finding the charge on each capacitor

[Femme_physics]

Homework Statement

http://img832.imageshack.us/img832/6442/cabalb.jpg

E = 12V
C₁ = 8 [uF]
C₂ = 6.571 [uF]
C₃ = 6 [uF]

Find:
1) C_T
2) Q₁, Q₂, Q₃ on each of the capacitors.

The Attempt at a Solution

http://img221.imageshack.us/img221/478/ctcap.jpg

C_T was easy. I wasn't sure how to go about finding Q1, Q2 and Q3.


The idea I had was to simply find the voltage across each of them, but that means I have to ohm's law. But I can't use ohm's law with capacitors. If these were resistors I could. But, I don't have a formula that relates voltage, current and capacitance.

I only have Q = VC

Any clues?

 

[rl.bhat]

Total charge extract from the cell is Q = C_T*E.
Across C2 potential difference is V = E.
so Q2= V*C2.
In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.

 

[Femme_physics]

Total charge extract from the cell is Q = C_T*E.
Across C2 potential difference is V = E.
so Q2= V*C2.

Really? I wasn't sure if E=V in C2, but come to think about it it makes sense since the flow is uninterrupted.

So far so good. Simple enough. :)

In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.

A lot more simple than I would've first imagined. .

My thanks!

 

[Ouabache]

In series combination, charge are the same in each capacitor. Hence Q1 = Q3 = Q - Q2.

A lot more simple than I would've first imagined. .

If you would like to more fully appreciate this concept, there is nice visual information here (last diagram on that page): http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capac.html" With your embedded image, you've inspired an interesting task to see if I can translate. How about this?

(Net 10) a. The following circuit which includes three capacitors, is connected to a power source E equal 12 Volts.

(Find):
1. The equivalent capacity of the circuit.
2. The electric charge Q1, Q2, Q3, located over each (capacitor) C1, C2, C3 - respectively.