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.

 

[rwisz]

I would suggest using the formula Q = CV to solve for the charge on each capacitor. This formula relates the charge (Q) on a capacitor to its capacitance (C) and the voltage (V) across it. In this case, we know the capacitance of each capacitor and the overall voltage across the circuit (12V). Using this formula, we can calculate the charge on each capacitor as follows:

1) CT = C1 + C2 + C3 = 8uF + 6.571uF + 6uF = 20.571uF

2) Q1 = C1V = (8uF)(12V) = 96uC
Q2 = C2V = (6.571uF)(12V) = 78.852uC
Q3 = C3V = (6uF)(12V) = 72uC

Therefore, the charges on each capacitor are Q1 = 96uC, Q2 = 78.852uC, and Q3 = 72uC. It is important to note that these are the charges on each capacitor at this specific moment in time, and they may change over time as the capacitors discharge or recharge.