Capacitator Circuit and determining unknown charges voltages

[liliacam]

Homework Statement

Suppose in the figure (Figure 1), that

C1 = C2 = C3 = 18.0μF and C4 = 28.2μF.

If the charge on C2 is Q2 = 37.4μC, determine the charge on each of the other capacitors
Determine the voltage across each capacitor.
Determine the Voltage Vab across the entire combination.

https://dl.dropbox.com/u/47465778/MP24.24.PNG

Homework Equations

Q=CV

Rules for Series:
Q is the same across a series
\frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + . . .
Vtotal = V1 + V2 . . .

Rules for Parallel:
Qtotal = Q1 + Q2 . . .
C_(eq) = C1 + C2 . . .
V is same across parallel

The Attempt at a Solution

Let 1 and 2 be in Series. Then Capacitance can be added as reciprocals
\frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18}
Ceq = 9μF for 1 and 2

Q1=Q2...because they are in series, Q is same. Q1 = Q2 = 37.4 μC

Q = CV. So,
37.4 = 9*V
V = 4.155 V

I figured that because Voltage is the same in parallel, then this voltage is the same across the board for 3 and 4. So I found C for 3 and 4

3 and 4 are in series with each other, so
\frac{1}{Ct} = \frac{1}{18} +\frac{1}{28.2}
Ct = 10.987...
Then Q = CtV
= 10.987*4.155 = 45.65 μC

Q3 = Q4 because Q's are equal in series.

Final Answer: Q1=Q2 = 37.4 and Q3 = Q4 = 45.65 all in microColoumbs (Wrong)

Although I'm fairly certain Q1 does equal Q2, but since I have to enter all of the answers at once, if one answer is wrong then the rest is wrong.

Where did I go wrong? Please help, I did my best but I'm pretty shaky on this stuff. Also this is my first time posting so sorry if I did anything weird with the formatting.

 

[tiny-tim]

welcome to pf!

hi liliacam! welcome to pf!

Q3 = Q4 because Q's are equal in series.

no, they're not in series, there's a junction between them!

also, what happens to the electrons that leave the top plate of Q₄ ?

they have to spread over the left plates of both Q₁ and Q₃ …

so how could they all be on Q₃ ? ​

 

[liliacam]

Hi! Thanks for responding so quickly

So I tried it again...I guess the diagram confused me. I redrew it here:

https://dl.dropbox.com/u/47465778/MP24Redrawn.PNG

With 3 still in parallel and 1, 2, 4 in series with each other. My current teacher still hasn't taught us how to redraw capacitors but I hope this is on the right track at least. So the electrons from 4 spread over 1 and 2, then 3?

If this drawing is correct, then:

Q1 = Q2 = Q4 = 37.4μC
Q3 = ?

\frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C4}
\frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18} + \frac{1}{28.2}
Ceq for 1,2,4= 6.822

New Capacitance:
Ceq total = Ceq for 1,2,4 + C3
Ceq total = 6.822 + 18 = 24.822 μF

New Voltage:
Q = CV
Q1 = C1V1
V1 = 2.077 V
If 1, 2, 4 is in parallel with 3, then V is the same across

V3 = 2.077 V

Q total = Ctotal*Vtotal
Qt = 24.822*2.077
Qt = 51.575 μC

Qt = Q3 + Q(1,2,4)
51.574μC = Q3 + 37.4
Q3 = 14.175 μC


Final Answers: Q1 = Q2 = Q4 = 37.4, Q3 = 14.75 all μC

Is this correct? I haven't had an official lecture on this yet, so it's just hard for me to visualize what exactly is going on.

 

[liliacam]

Nevermind, figured it out.