Finding charge in a circuit with capacitors

[nautola]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

I got a C_eq 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.

 

[tiny-tim]

hi nautola!

(http://screencast.com/t/9KET4sNSAQWj)

C = Q/V
…
I got a C_eq 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.

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

 

[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.

 

[tiny-tim]

hi nautola!

(just got up :zzz:)

So I solved it and got that the charge on C2 should be the net charge times the ratio of C2 to C2 + C4.

that should work

… 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.

i don't understand this … what sum?

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

 

[Nickie]

To find the charge on capacitor C2, we can use the equation C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage. We know the values of C2 (4.04 μF) and the voltage (5.33V), so we can rearrange the equation to solve for Q: Q = C * V = (4.04 * 10^-6 F) * (5.33V) = 0.0216 C. Therefore, the charge on capacitor C2 is 0.0216 Coulombs.

To find the work needed to remove the dielectric from capacitor C2, we can use the equation U = 1/2 * Q^2 / C, where U is the potential energy, Q is the charge, and C is the capacitance. We already know the value of Q (0.0216 C), but we need to find the equivalent capacitance for the circuit without the dielectric. To do this, we can use the formula for capacitors in parallel: Ceq = C1 + (C2 * k), where k is the dielectric constant. Plugging in the values, we get Ceq = 11.2 μF + (4.04 μF * 2.71) = 22.73 μF. Now we can plug in the values to find the potential energy: U = 1/2 * (0.0216 C)^2 / (22.73 * 10^-6 F) = 2.48 * 10^-4 J. Therefore, the work needed to remove the dielectric from capacitor C2 is 2.48 * 10^-4 Joules.