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2 identical capacitors, given potential in joules

  • Thread starter jendead
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  • #1
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Homework Statement


Capacitors A and B are identical. Capacitor A is charged so it stores 4J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now ____.


Homework Equations


U = q^2/2C

The Attempt at a Solution


Ok, I'm not sure if this is allowed.. I already know the answer is 2J. My professor gave us an answer key with the problem done out, but I don't understand what is happening. I'm not sure where 2J went. It says 2J of energy were required to move charge from the charged capacitor to the uncharged one.

I know initial potential of A is 4J, and initial potential of B is 0. For final potential, she has U_final = q^2/2C = 1/2*q^2/2C. Where did the second 1/2 come from?

I would really appreciate some clarification.. I'm really confused.
 

Answers and Replies

  • #2
Dick
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When the two capacitors are connected in parallel the capacitance doubles, while Q remains the same.
 
  • #3
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I understand that, but I don't see how that relates here. From the look of the answer key, I don't understand where total potential is halved when being spread out over 2 capacitors.
 
  • #4
Dick
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The potential is halved because the same charge is being spread out over two capacitors. What the relation between potential and charge for a capacitor?
 
  • #5
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It's U = q^2/2C.. so if I plug in (q/2)^2/2(2C), which is halving the charge and doubling the capacitance, I'd have q^2/16C.. which still makes no sense.

Am I missing something conceptually or algebraically?
 
  • #6
Dick
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The U in that equation is energy stored the capacitor, not potential between the plates (volts). Is that your problem?
 
  • #7
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Yes, I'm trying to find the potential energy (in joules), not the voltage. I may have been unclear when I said potential.
 
  • #8
Dick
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If you understand why capacitance doubles, then why can't you just change the C in Q^2/(2C) to 2C and conclude 4J changes to 2J?
 
  • #9
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Ok, that's starting to make sense.. but why isn't Q also changed to Q/2 since it's being halved at the same time?

I apologize for being so dense - I tend to have a really hard time with these concepts sometimes.
 
  • #10
Dick
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Q is total Q, between both capacitors. Total charge doesn't change. C changes to C/2. It does change.
 
  • #11
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Ok, that makes sense. Thank you for being patient. :)
 

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