Understanding Charge Distribution in a Circuit with Multiple Capacitors

[RadiationX]

In the figure given the battery has a potential difference of V=10V and the five capacitors each have a capacitance of 10 micro Farads. What is the charge on (a) capacitor 1 and (b) capacitor 2 ?I totally know how to redraw the circuit but I don't have a clue as to how the capacitance equivalent relates to the rest of the ciruit. I need some help really bad. I have the soulution in my book but I don't understand it. How do I find these charges?

what I've done so far is find the capacitance equivalent which is
1.6 X 10^-5 Farad. Once I had that I found the total charge 1.6 X 10^-4 Couomb. I'm stuck as to how to continue now.

 

[RadiationX]

I hope the people who have looked at my post have looked at the attachment that goes with it.

 

[MTyler8]

I tried to look at it but it says "Attachments Pending Approval" and I can't open it.

 

[RadiationX]

ughh I dont' know what to do about that.

 

[kahless2005]

I see it, but unforutnately I am not good with Circuits.

I do know that the definition of capacitance is $Q/V=C_1 + C_2$

 

[Cyrus]

Capacitors in series are like resistors in parallel. Conversely, capacitors in parallel are like resistors in series.

What you have written is not true in general Kahless2005.

 

[kahless2005]

I have found a similar problem in my physics book, the difference is there are only 4 capacitors.

what you should do is find the $1/C_eq$ which would be
$1/((1/(C_1+C_3))+C_4)+1/C_5+1/C_2$

Then find the $Q_eq$ which is $C_eq*V$. That should be the charge across all the capacitors

 

[andrewchang]

When you do these kinds of problems, you need to understand how to simplify capacitors into equivalent ones. have you learned how to do that yet? parallel capacitors can be made into an equivalent capacitor by adding their capacitance, and series capacitors follow the same rule as resistors in parallel. just go through the problem capacitor by capacitor to get the total capacitance. once you know this, then C= Q/V.

 

[Cyrus]

That is wrong kahless2005.

 

[andrewchang]

In addition, it would be better for RadiationX to learn how to do this on his/her own.