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Charging capacitors

  • #1

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 [Broken]


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.

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
275
0
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
 
  • #3
gneill
Mentor
20,793
2,773
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.
 
  • #4
Are 6q and 3q capacitances then? I feel really stupid that I just can't work out how to solve this.
 
  • #5
gneill
Mentor
20,793
2,773
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.
 
  • #6
So then the first one won't change, the second one will increase, and the third one will decrease?
 
  • #7
gneill
Mentor
20,793
2,773
It is so.
 
  • #8
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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