Finding charge in a circuit with capacitors

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SUMMARY

The discussion focuses on calculating the charge on capacitor C2 in a circuit with capacitors C1, C2, C3, and C4, where C2 has a dielectric with a dielectric constant (k) of 2.71 and the circuit is powered by a 5.33V battery. The capacitors are arranged with C2 and C4 in parallel, followed by C3. The key equations used include C = Q/V for charge calculation and U = 1/2 Q² / C for energy. Participants emphasize the importance of understanding the voltage across capacitors in parallel and using capacitance ratios to determine charge distribution.

PREREQUISITES
  • Understanding of capacitor circuits and their configurations
  • Familiarity with the equations C = Q/V and U = 1/2 Q² / C
  • Knowledge of dielectric materials and their effects on capacitance
  • Ability to solve systems of equations related to electrical circuits
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  • Research the impact of dielectrics on capacitor performance and charge calculations
  • Learn about series and parallel capacitor combinations and their equivalent capacitance
  • Study the principles of energy stored in capacitors and work done in removing dielectrics
  • Explore advanced circuit analysis techniques, including Kirchhoff's laws
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Students studying electrical engineering, physics enthusiasts, and anyone involved in circuit analysis and capacitor behavior.

nautola
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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 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:
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hi nautola! :smile:

(http://screencast.com/t/9KET4sNSAQWj)
nautola said:
C = Q/V

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.

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 :wink:
 
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:
hi nautola! :smile:

(just got up :zzz:)
nautola said:
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? :confused:

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

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