Final electric potential difference in a circuit with two capacitors

  • Thread starter greg_rack
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  • #1
greg_rack
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Homework Statement:
FIGURE ATTACHED BELOW
##C_{1}=2.00\mu F, q_{1}=6.00\mu C, C_{2}=8.00\mu F, q_{2}=12.0\mu C##
The circuit gets closed and charge flows until the two capacitors have the same electric potential difference ##V_{F}## across its terminals.
-calculate ##V_{F}##.
Relevant Equations:
##q=CV##
IMG_4628.JPG
So, each capacitor must have a different potential difference, given by its capacity and charge... this would cause charge and current accordingly to flow in the circuit.
But how do I determine the final potential difference, which would of course be the same for both of them? I have tried writing down something, which I've found out to be unuseful to solve this problem.
 

Answers and Replies

  • #2
gneill
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The capacitors start out with different potentials across them. What are they?

If some amount of charge, say ##\Delta q## moves from the higher potential capacitor to the lower potential one, what expressions can you write for those new potentials?
 
  • #3
greg_rack
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The capacitors start out with different potentials across them. What are they?

If some amount of charge, say ##\Delta q## moves from the higher potential capacitor to the lower potential one, what expressions can you write for those new potentials?
Ok, so, since ##V_{1}>V_{2}##, I'll have a ##\Delta q## transferring from 1 to 2, so the final potential ##V_{f}## is going to be: ##V_{1f}=V_{2f}=V_{f}=\frac{q_{1}-\Delta q}{C_{1}}##... that makes sense, right?
 
  • #4
gneill
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Ok, so, since ##V_{1}>V_{2}##, I'll have a ##\Delta q## transferring from 1 to 2, so the final potential ##V_{f}## is going to be: ##V_{1f}=V_{2f}=V_{f}=\frac{q_{1}-\Delta q}{C_{1}}##... that makes sense, right?
Yes. But you're left with two unknowns: ##V_f## and ##\Delta q##. Write the expressions for both of the capacitors and you'll have two equations in those two unknowns.
 
  • #5
greg_rack
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Yes. But you're left with two unknowns: ##V_f## and ##\Delta q##. Write the expressions for both of the capacitors and you'll have two equations in those two unknowns.
Yep, sure!
Thank you very much :)
 

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