# Homework Help: Finding charge in a circuit with capacitors

1. Oct 3, 2012

### nautola

1. The problem statement, all variables and given/known data
http://screencast.com/t/9KET4sNSAQWj
If the picture isn't showing up, the order of the capacitors is 1 -> 2 & 4 (parallel) -> 3, with a battery in the circuit.

There's a dielectric in C2 with k = 2.71.
There's a battery with v = 5.33v in the picture.
Capacitors:
C1 = 11.2 μF
C2 = 4.04 μF
C3 = 13.1 μF
C4 = 3.32 μF

The question wants to know the charge on the capacitor with the dielectric (C2).

It also asks how much work is needed to remove the dielectric from the capacitor after the battery is removed.

2. Relevant equations
C = Q/V
capacitor circuit relationships
U = 1/2 Q2 / C
W = -U

3. The attempt at a solution
I got a Ceq for 2 and 4, and that gave me a charge, Q, for the parallel part of the circuit, but I don't know how to separate it from there.

As for the work, I'm pretty sure I can't do that part until I finish this first part. But even then I'm not entirely sure what to do.

Last edited: Oct 3, 2012
2. Oct 3, 2012

### tiny-tim

hi nautola!

(http://screencast.com/t/9KET4sNSAQWj)
you know that 2 and 4 have the same voltage, and you know the ratio of their capacitances, so use your capacitor equation C = Q/V to find the ratios of their charges

3. Oct 3, 2012

### nautola

I got the total charge on the center, and the ratio of the capacitances and set up a system of equations where the net charge (et charge is the charge of capacitor 1 or capacitor 3, or the middle equivalent capacitor) equals the sum of the charges. So I solved it and got that the charge on C2 should be the net charge times the ratio of C2 to C2 + C4.
But that's wrong and I don't know why.

Last edited: Oct 3, 2012
4. Oct 4, 2012

### tiny-tim

hi nautola!

(just got up :zzz:)
that should work
i don't understand this … what sum?

(if you're still not getting it, show us your equations )