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Voltage adding

  1. Jun 23, 2011 #1
    1. The problem statement, all variables and given/known data


    A 3.0 µF capacitor charged to 40 V and a 5.0 µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other.


    b. Find the potential difference across the capacitors.



    3. The attempt at a solution

    Do you just add up the potentail differences? 18 + 40 = 58V?
     
  2. jcsd
  3. Jun 23, 2011 #2

    gneill

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    Nope. There will be charge movement (and cancellation) that happens when the capacitors are connected. Sketch a diagram showing the plates with their charges before connection, then work out what charges will move where when they are connected.
     
  4. Jun 23, 2011 #3
    Sum up the charges?

    3.0 µF * 40V = 1.2 *10^-4C

    5.0 µF * 18V = 9 * 10^-5 C
     
  5. Jun 24, 2011 #4

    gneill

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    Sum the charges taking into account their polarities on the capacitors. Did you draw the sketch that I suggested?
     
  6. Jun 24, 2011 #5
    I don't know how to draw it...

    They are connected in a way that they can be added up right? It's from - to + to - to +
     
  7. Jun 24, 2011 #6

    gneill

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    Can you draw one charged parallel plate capacitor?
     
  8. Jun 24, 2011 #7
    [PLAIN]http://img641.imageshack.us/img641/2150/unledgkx.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  9. Jun 24, 2011 #8

    gneill

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    Do the charges really sit off to the side of the plates?
     
  10. Jun 24, 2011 #9
    No the plus and minus just indicate which plate is positive and negative.
     
  11. Jun 24, 2011 #10

    gneill

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    Okay. So you've calculated the individual charges that exist on each capacitor. Draw two capacitors side by side (unconnected as yet) and label each plate with the charge that exists on it.

    When the capacitor plates are connected as specified, what will happen to the charges?
     
  12. Jun 24, 2011 #11
    Do they merge together to become one big charge? If so why is it leaking?
     
  13. Jun 24, 2011 #12

    gneill

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    Sorry, I don't understand your reference to "leaking".

    Charges combine -- They add or cancel depending upon their signs. You do remember that a + unit charge will cancel with a - unit charge, right?
     
  14. Jun 24, 2011 #13

    NascentOxygen

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    Last edited by a moderator: May 5, 2017
  15. Jun 25, 2011 #14
    OKay so once the capacitors are hooked up the charges combine to become one current in one branch

    so Q = 2.1 * 10^-4 C
     
  16. Jun 25, 2011 #15

    gneill

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    Can you draw a picture of that? How did you calculate Q?

    Any current that flows will be very brief as the charges sort out their new distribution. Note that in the end both capacitors will still have a charge, it just won't be the same charge that they started out with.
     
  17. Jun 25, 2011 #16
    [PLAIN]http://img37.imageshack.us/img37/2847/unledzz.jpg [Broken]

    I just added the charges from each plate. Should i divide it by 2 to average it out then?
     
    Last edited by a moderator: May 5, 2017
  18. Jun 25, 2011 #17

    gneill

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    Both terminals of each capacitor will be connected -- the capacitors end up in parallel (See attached figure). Which charges have access to each other via conduction paths once the connections are made? What will be the result?

    attachment.php?attachmentid=36716&stc=1&d=1309008311.gif
     

    Attached Files:

  19. Jun 25, 2011 #18
    Why is it connected in parallel and not series?
     
  20. Jun 25, 2011 #19

    gneill

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    With only two components, connected as shown, you could consider them connected both in series and in parallel if you wish. The net capacitance of the pair will be calculated using the parallel connection formula because the plates are in fact connected that way.
     
  21. Jun 30, 2011 #20
    [PLAIN]http://img811.imageshack.us/img811/5116/unledws.jpg [Broken]

    They don't look parallel now.
     
    Last edited by a moderator: May 5, 2017
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