Charging Capacitors: Find Change in Charge

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Homework Help Overview

The problem involves analyzing three circuits with capacitors that are initially charged. Participants are tasked with determining how the charge on the left-hand capacitor will change after the switches are closed, focusing on whether it will increase, decrease, or remain the same.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to find voltages before and after closing the switches, with some expressing uncertainty about how to approach the problem. Others suggest that charge movement will continue until voltages across the capacitors equalize, and some question the interpretation of given values as charges or capacitances.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the relationship between charge, capacitance, and voltage, but there is no explicit consensus on the outcomes for each circuit.

Contextual Notes

There is confusion regarding the interpretation of capacitance values as charges, which may affect participants' understanding of the problem setup. The original poster expresses uncertainty about how to proceed with the calculations.

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Homework Statement


The figure shows three circuits, each consisting of a switch and two capacitors, initially charged as indicated (top plate positive). After the switches have been closed, in which circuit (if any) will the charge on the left-hand capacitor (a) increase, (b) decrease, and (c) remain the same?

[PLAIN]http://img714.imageshack.us/img714/3369/capacitance.png


Homework Equations


q = CV


The Attempt at a Solution


I really am not sure how to approach the problem. I know I should find the voltages before and after the switch is closed in each situation, but I'm not really sure how to do that with the "q"s in there.
 
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charges will move until the voltage over each of the 2 capacitors will be equal,
i.e. V will be constant over all the circuit and that means that there is no electric field so no more charges can be moved.

V=q/C
grad(V)=E
 
You only need to know what direction the current will flow when the switch is closed. You're given charge and capacitance for each capacitor (albeit symbolically). You should be able work out the relative sizes of the voltages on each cap.
 
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.
 
downwithsocks said:
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.

6q and 3q are charges on the capacitors of value 3C and C. You don't need to know what the actual values of q (coulombs) and C (farads) are.
 
So then the first one won't change, the second one will increase, and the third one will decrease?
 
It is so.
 
Thanks so much, I was confused because I thought the C values were charges in coulombs, so I didn't know how to use my knowledge of voltage given only charges.
 

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