# Finding the charge on a capacitor

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1. Feb 10, 2017

### diredragon

1. The problem statement, all variables and given/known data

$E_1=6V$
$E_3=5V$
$R_1=150Ω$
$R_2=R_3=100Ω$
$R_4=50Ω$
$R_5=300Ω$
$C_1=1.5 mF$
$C_2=0.5mF$
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 ($Q_{20}$).
2. Relevant equations
3. The attempt at a solution

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

2. Feb 11, 2017

### 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 im 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?

3. Feb 11, 2017

### 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.

4. Feb 11, 2017

### doktorwho

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.

5. Feb 11, 2017

### magoo

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