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Parallel-plate capacitor problem

  1. Sep 7, 2006 #1
    Question I need to solve:

    I started off by getting the equation U = (1/2)QV = (1/2)CV^2 = (Q^2)/2C. Then I realized that I was given A and d, so I reworded the equation into U = (dQ^2)/(2EoA). At this point I was stuck because I do not have a Q to plug in.
     
  2. jcsd
  3. Sep 7, 2006 #2

    Astronuc

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  4. Sep 7, 2006 #3
    Accidentally made new post instead of editting this post...whoops!
     
    Last edited: Sep 8, 2006
  5. Sep 8, 2006 #4
    I used U = (CV^2)/2 = (kEoAV^2)/2d, using k = 1.00059, Eo = 8.85 x 10^-12 C^2/Nm^2, A = 3.15 x 10^-4 n^2m d = 0.00275 m, and V = 575 V, obtaining an answer of 1.6768 x 10^-5 J as my answer for the first part. I tried the same method for part (b), but my answer of 4.192 x 10^-6 J is incorrect. Any ideas of where I went wrong?
     
    Last edited: Sep 8, 2006
  6. Sep 8, 2006 #5
    Bump! Is anyone able to help me figure this problem out?
     
  7. Sep 9, 2006 #6

    Astronuc

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    You did use U = (1/2)C V2, right?

    Then C = [tex]\frac{k\,\epsilon_o\,A}{d}[/tex].


    Also, look at "A = 3.15 x 10^-4 n^2m" the units are strange. Area should be square of the basic length dimension, as in m2. So, 315 cm2 = 315 x 10-4 m2 = 0.0315 m2.

    Also, since C = a/d where a is just [itex]k\,\epsilon_o\,A[/itex], then

    C1d1 = C2d2. If one solves part a, then one can use this relationship for the second part.
     
    Last edited: Sep 9, 2006
  8. Sep 9, 2006 #7
    Sorry, I actually meant to put 315 x 10^-4 m^2 = A. I don't know why I put those units that I put lol. But using your method, I got the same answer I got before, 4.192 x 10^-6 J, which is incorrect. Using your equation, I solved for C2 = C1d1/d2 = (1.0143 x 10^-10 F)(0.00275 m)/(0.011 m) = 2.5358 x 10^-11, which I plugged into U = (1/2)CV^2, getting 4.192 x 10^-6 J, which again is incorrect. What am I doing wrong?
     
  9. Sep 9, 2006 #8

    Doc Al

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    Check your arithmetic.
    The capacitance changes as Astronuc explained. But what remains the same when the plates are separated?
     
  10. Sep 9, 2006 #9
    The voltage remains the same when the plates are separated, right?
     
  11. Sep 10, 2006 #10

    Doc Al

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    No. Hint: After charging the capacitor, the battery was disconnected.
     
  12. Sep 10, 2006 #11

    Astronuc

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    I mislead you with the U = (1/2)C V2.

    Looking at U = (1/2)QV = (1/2)CV^2 = (Q^2)/2C, and thinking about what Doc Al mentioned, what changes and what remains the same as the distance is changed and the capacitor is isolated?
     
  13. Sep 10, 2006 #12
    As the distance is changed and the capacitor is isolated, U stays the same, and V, C, and Q are changed, right? What do I do next?
     
    Last edited: Sep 10, 2006
  14. Sep 11, 2006 #13

    Doc Al

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    V, C, and Q are not all changed. Once the plates are disconnected from the battery, charge cannot enter or leave.
     
  15. Sep 11, 2006 #14
    Hmm, okay. That makes sense. So V remains the same. Does this also mean that after the separation increased to 11 mm, the same amount of energy is stored as before, which is 1.6768 x 10^-5 J?
     
  16. Sep 11, 2006 #15

    Doc Al

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    V does not remain the same!

    Reread my last post for a hint about what does remain the same as the plates are separated.
     
  17. Sep 11, 2006 #16
    Sorry, I quickly assumed charge referred to V, so it means that Q remains the same?
     
  18. Sep 11, 2006 #17

    Doc Al

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    That's right.
     
  19. Sep 11, 2006 #18
    Okay, I figured it out. Thank you for your help!
     
    Last edited: Sep 11, 2006
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