Finding the charge on a capacitor

[diredragon]

Homework Statement

[/B]
$E_1=6V$
$E_3=5V$
$R_1=150Ω$
$R_2=R_3=100Ω$
$R_4=50Ω$
$R_5=300Ω$
$C_1=1.5 mF$
$C_2=0.5mF$
Text:
In the first condition when the switches 1 and 2 are open the capacitor 2 has charge on it. First the switch 1 closes and the charge flow of $q_1=1 mC$ is established to flow through the branch as shown in the picture. Then the switch 2 closes and the flow is now $q_2=-1.4mC$. Find the initial charge on capacitor 2 ($Q_{20}$).

Homework Equations

3. The Attempt at a Solution [/B]
My only problem here is with the charge division. Correct me if I'm wrong here:
When the switch 1 closes the voltage across 1 and two must be the same so we have:
$q_{11}+q_{12}=q_1$
$\frac{q_{11}}{C_1}=\frac{Q_{20}+q_{12}}{C_2}$
I can't find the voltage they're equal to because i don't have $E_2$ but i do have the change when the switch 2 closes. So:
$\frac{q_{11}+q_{21}}{C_1}=\frac{Q_{20}+q_{12}+q_{22}}{C_2}$
The change is then just:
$ΔU_{c1}=\frac{q_{21}}{C_1}$
$ΔU_{c2}=\frac{q_{22}}{C_2}$
And $q_{12}+q_{21}=q_2$ and from there i can find the $q_{12}$ and $q_{22}$ and have the change in the voltage while my Δ-circuit looks like this:

Where finding the current is trivial. Is this the right track?
Thanks

 

[diredragon]

I continued along this road and i got a wrong answer. I will now post my full work so you can see where I'm wrong.
Strarting from the delta circuit where I am trying to find the current $ΔI_g$ i have this:
1) $ΔU=\frac{q21}{C1}=\frac{q22}{C2}$
2) $q21+q22=q2=-1.4$
$q21=\frac{C1}{C2}q22$
and from this i get $q21=-0.35$,$q22=-1.05$
so $ΔU=-0.7 V$ and i can continue finding the current $ΔI_g$
Using the current divider i see that:
$ΔU=ΔI_gR4+R2ΔI_d=ΔI_gR4+R2\frac{R3}{R2+R3+R5}ΔI_g$
and from here $ΔI_g=-0.01A$
Now goind back to the original circuit, where i have the current through the branch when the switch closes i have enough information to find $E_2$ which later i can use to find the voltages across the capacitors. From mesh current analysis get that the loops that make $ΔIg$ are some $IB and IA$
So by arbitrary choice:
$ΔI_g=I_b-I_a$ and by means of mesh i get $I_b$ because i only need that one. I could go the other way but i need E2 then.
Here is the picture of the work:

I get $Q=2mC$ while the answer is $Q=-0.1mC$
Where and what is wrong?

 

[magoo]

I'm not sure why E2 is not given.

Nonetheless, with Switch 1 open, C2 is just dangling. There is no voltage across it so it cannot have a charge on it. On the other hand, C1 will have a charge on it.

This changes most of what you did initially and will affect everything beyond that point.

 

[doktorwho]

I'm not sure why E2 is not given.

Nonetheless, with Switch 1 open, C2 is just dangling. There is no voltage across it so it cannot have a charge on it. On the other hand, C1 will have a charge on it.

This changes most of what you did initially and will affect everything beyond that point.

It shouldn't change anything initially since the change is considered when the switch 2 closes so $ΔU$ remains as it is regardless of the initial charge of Q1 and it makes appearance only at the end, but i supposed that it is uncharged since its stated that C2 has charge on it.

 

[magoo]

OK, I see the point about having a trapped charge on C2. Sorry about that.