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Parallel Plates

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63 x 10^4 m/s to the right. Its speed on reaching the other plate, 2.10 cm away is 4.15x10^4 m/s

    I got parts a and b - part C is what I'm having problems with.

    C) If the plates are square with an edge length of 25.4 cm, determine the charge on each.


    2. Relevant equations
    E=(4πkQ)/A


    3. The attempt at a solution

    I substituted for all the unknowns except for E and Q - but that leaves me with two unknowns; I'm completely flabbergasted with what to do next.
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    Can you post what parts a and b said, and what your answers were?
     
  4. Sep 13, 2007 #3
    A) What type of charge is on each plate?
    Left Plate Negative, Right Plate positive

    B) What is the direction of the electric field between the plates?
    From right to left
     
  5. Sep 13, 2007 #4

    learningphysics

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    For a capacitor Q = CV.

    Try to calculate C and V... then you can get Q.
     
  6. Sep 13, 2007 #5
    Thanks alot - i'll go do that. I didn't have that equation unfortunately xD.
     
  7. Sep 13, 2007 #6
    Ok I plugged in

    C=((8.85x10^-12)(9x10^9)(.064516))/(2.10x10^-2)=.2447

    Then Q=(.2447)(1.63X10^4)= 3988.61

    Q=(.2447)(4.15x10^4)= 10155

    I plugged each Q into E=(4pikQ)/A and I got two answers that were a bit off =\ No idea what I did wrong.

    I know the final answer is 1.13 x 10^-13 but i'm not getting anywhere near that.
     
  8. Sep 14, 2007 #7

    learningphysics

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    The V in Q = CV is voltage not speed. You need to find the voltage between the two plates of the capacitor... use energy to find voltage...
     
  9. Sep 14, 2007 #8
    Aaah sorry; just assumed what the variables were since I hadn't encountered this equation before - let me fix this now xD.
     
  10. Sep 14, 2007 #9
    Ok so V=Ue/Qe

    Ue= qeEd but E is unknown =\.

    I'm reading up on this equation (surprisingly in my book it isn't in the chapter that the problem was assigned from - it's in the next chapter) so I haven't had exposure to it.

    Am I doing something wrong? =\
     
  11. Sep 14, 2007 #10

    learningphysics

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    But you know that V = Ed...

    Ue = qe*V
     
  12. Sep 14, 2007 #11
    Unless i'm missing something (which I problem am due to my own ignorance ;_;) I still have multiple unknowns.

    UGH, it's irritating having an assignment which seems easy until you get to the last problem.

    Any other clues you can prompt me to? =\
     
  13. Sep 14, 2007 #12

    learningphysics

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    What is the change in kinetic energy of the electron? electric potential energy is being converted to kinetic energy...
     
  14. Sep 14, 2007 #13
    If I calculated it correctly I got 1.147x10^-26 - plugged that in for delta Ue; plugged Ue into the V equation - plugged V into the Q equation - which I plugged in for E and I got the wrong answer =s.
     
  15. Sep 14, 2007 #14

    learningphysics

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    That's not what I'm getting for change in kinetic energy... can you show your calculations?
     
  16. Sep 14, 2007 #15
    KEo= (1/2)(9.10938188 × 10-31)(1.63x10^4)
    KEf= (1/2)(9.10938188 × 10-31)(4.15x10^4)

    KEf-KEo=Delta KE which is where I got 1.148E-26

    I assumed the change in kinetic energy would be equal to the change in potential energy. So I took the magnitude of it and plugged in all the way through - lead me to a V of 7.17334E-8
    which then lead me to a Q of 1.7E-8 which I proceeded to plug into the initial equation; I got a number that I knew was off immediately so I didn't bother writing it down =\.
     
  17. Sep 14, 2007 #16

    learningphysics

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    you didn't square the velocties... (1/2) mv^2...
     
  18. Sep 14, 2007 #17
    Ouch >.<. K - correcting my mistake.
     
  19. Sep 14, 2007 #18

    learningphysics

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    also here... Capacitance = epsilon*A/d

    I think the k is a mistake.
     
  20. Sep 14, 2007 #19
    Ok corrected; that got my Ue to 6.55452E18 - which when plugged into my V got me 4.09658E37. That lead to Q being 1.00243E37 and E being some huge number =\.
     
  21. Sep 14, 2007 #20
    aaah ok i'll check on that - thanks; I knew something had to be wrong
     
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