Equivalent Capacitance of a network

[Canopuswr]

Homework Statement

To find the equivalent capacitance of the following network (image attached at the end of the post). The blue capacitors have capacitance of 2C and the black ones have a capacitance of C.

Homework Equations

$$Q = CV$$ where Q is the charge on a capacitor, C it's capacitance and V the potential difference through it's terminals.
$$Q_t=C'*V_n$$, where Qt is the total charge stored by the network (I need help with this, I think Qt is the charge on capacitor 1 + the charge on capacitor 3, but I am not sure), Vn the potential difference across the network and C' the equivalent capacitance, which is what I'm seeking
$$\frac{1}{C'} = \frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$$ for the equivalent capacitance C' of n capacitors in series
$$C' = C_1+C_2+...+C_n$$ for the equivalent capacitance of n capacitors in parallel
$$\sum_{i=0}^n\ V_i = 0$$
if Vi is the voltage across the capacitors and if you follow them in a loop (also being a little careful with signs) (this is the same as closed loop integral of E*dl = 0)

The Attempt at a Solution

Not much of an attempt, since it needs some more before i might get something valid from it, but, not knowing how to do it just with series/parallel, I'm doing it using what I know about charges on the plates and potentials. going through the capacitors in loops i get
$$V_I+V_I_I=V_I_I_I+V_I_V$$
$$V_I+V_V=V_I_I_I$$
$$V_V+V_I_V=V_I_I$$
I was thinking of using the fact that the charges on the "inner plates" need to add up to zero in order for charge to be conserved (no charge moves through the capacitors), but I'm not sure on which side will Capacitor V have postive charge and on which side it will have negative charge (this changes the result, I think) and I'm assuming all the other capacitors will have postive charge on their "left" plate and negative on the "right" plate (I'm solving this from right to left, even though it's symmetric that way).
Assuming Capacitor V has positive charge on the upper plate, i think the equations should be
$$2CV_I=2CV_V+CV_I_I$$
$$CV_I_I_I=-2CV_V+2CV_I_V$$

And that would completely determine the problem (the system is to have infinitely many solutions), since I can find the charge on capacitors 1 and 3, their sum (I think) being Qt and I can add potential differences V1 and V2 to get the potential difference across the network and that's that. I am however shaky on the last two equations and I don't know if the Qt is the sum of charges on capacitors 1 and 3.
I appreciate the help.

 

[Canopuswr]

Oh god, I'm sorry, i forgot to attach the file. there it is.

 

[kerimek]

Dear student,

Thank you for sharing your attempt at solving this problem. It seems like you have a good understanding of the equations and concepts involved in finding the equivalent capacitance of a network. However, I would suggest that instead of trying to solve it using the charges and potentials, you can use the equations for equivalent capacitance in series and parallel.

In this case, the capacitors are arranged in a combination of series and parallel connections. The blue capacitors (2C) are in parallel with each other, and the black capacitors (C) are in series with each other. Therefore, you can first find the equivalent capacitance of the two blue capacitors, which would be C' = 2C/2 = C. Then, you can find the equivalent capacitance of the two black capacitors in series, which would be C' = C + C = 2C. Finally, you can find the equivalent capacitance of the entire network by combining the two equivalent capacitances in parallel, which would be C' = C + 2C = 3C.

I hope this helps. Good luck with your homework!