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Distance of particle drop in magnetic field (probably geometry)

  1. Jul 9, 2015 #1
    1. The problem statement, all variables and given/known data
    IZi8obz.png

    Basically there are 3 plate capacitors, with a distance a between the first two and b between the last two. In between the first two plates is voltage V. A particle of charge q is released from the first plate, and when it leaves the second plate it is no longer accelerating. There is a magnetic field of B in the second part, and then a hole of radius r at the third plate, find the distance that the particle drops due to the magnetic field. I neglected to outline this specifically in the picture, but basically it is the distance of the bottom black line and could be called "d"

    2. Relevant equations
    U=K
    F=qv x B
    Sum F = ma
    a=v^2/r
    E=qV
    3. The attempt at a solution

    First I had to find the velocity of the particle at the point of the first plate, and did this using conservation of energy.

    U=K
    qV=.5mv^2
    v=sqrt(2qV/m)

    Second, is finding the radius of the circle in the second part, using F=ma and angular acceleration

    Then Sum F = ma, only force is magnetic
    qv x B = ma

    qv x B = m(v^2/R)

    R=m*v^2 / q*v*B

    R=m*(2qV/m) / q*B*sqrt(2qV/m)

    R=2V/Bsqrt(2qV/m)

    At this point I have the radius R of the circular path that the magnetic field induces, but I'm totally lost at this point. I'm assuming it's just geometry of some fashion. Really would appreciate help.
     
  2. jcsd
  3. Jul 9, 2015 #2
    Because my picture was pretty potato quality, here is a new version. Also I found a simplification of R, where it is mv^2/qvB, leading to mv/qB, which is m*sqrt(2qV/m) / qB

    https://photos-1.dropbox.com/t/2/AABdnf-4ewEkuSufX_wYb_F1zWxQY_4lbn-mX5OImOW3dg/12/51054650/jpeg/32x32/1/1436472000/0/2/2015-07-09%2013.36.16.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/LLxgaJo81ulvhjEAskZsAlZZK8XGg4aTgh9X-QeV0m4?size=1024x768&size_mode=2 [Broken]
     
    Last edited by a moderator: May 7, 2017
  4. Jul 9, 2015 #3

    TSny

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    Welcome to PF!

    Yes, it's just geometry. Your red circular path looks a little sloppily drawn. Try to make it more accurate. In particular, make sure you have the center of the circular path placed properly and make sure that the circular path passes through the second plate such that the tangent line to the path is in the correct direction.

    I presume that there is a little hole in the second plate for the charge to pass through and that the distance between the second and third plates is known.

    Your expression for R looks good, but you can simplify it a bit.
     
  5. Jul 9, 2015 #4
    Perhaps my second post answered these questions? I'm assuming that the center of the circle is in line with the second (middle) plate
     
  6. Jul 9, 2015 #5

    TSny

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    OK, I did not see your second picture. I see that b represents the distance between the second and third plates and that you have the direction of the path at the second plate in the right direction (horizontal). I'm still not too sure where you are placing the center of the circular path.
     
  7. Jul 9, 2015 #6

    TSny

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    Yes.
     
  8. Jul 9, 2015 #7

    TSny

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    In your second picture, you have drawn a radius R. Try making that the hypotenuse of a right triangle.
     
  9. Jul 9, 2015 #8
    That's still an assumption on my part. Between the second and third plates, the only force that should be acting on the particle is the magnetic force, which is F=q*v x B, and that it should move in a circle, and reasoning (albeit crappy explanation it is) thought it should be where it is, is there a different spot it should be?
     
  10. Jul 9, 2015 #9

    TSny

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    You are correct that the center of the circle should be in line with the second plate. But neither of your pictures seems to show this. Looks to me that you have the center a little bit to the right of the second plate.
     
  11. Jul 9, 2015 #10
    Sorry, that is uncertainty and sloppiness in action there.

    Is this what you meant by making a triangle? If you can find the hypotenuse created by the red line, using some kind of circle magic, you have the side b, and the last side should be d, correct? If this is the right course of action then what circle magic is done?

    https://photos-1.dropbox.com/t/2/AACYhnNE5kOBe9cXiXds1vuFNx8SE31Le9ybSvxPspgdoA/12/51054650/jpeg/32x32/1/1436472000/0/2/2015-07-09%2013.51.02.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/hgbzOISA7LFxF96Sk24ynqGuBtMLIhtL66D0AEny48I?size_mode=5 [Broken]
     
    Last edited by a moderator: May 7, 2017
  12. Jul 9, 2015 #11

    TSny

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    Draw a right triangle with the hypotenuse as the solid line that you drew and labeled R. Forget the red line that you drew.
     
  13. Jul 9, 2015 #12
    Surely it can't be that simple? d=√R^2-b^2 ?
     
  14. Jul 9, 2015 #13
    It looks kind of legit if you make that line and flip it around and upside down
     
  15. Jul 9, 2015 #14

    TSny

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    Not quite that simple.
     
  16. Jul 9, 2015 #15

    TSny

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    ???
     
  17. Jul 9, 2015 #16

    TSny

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    Here's the right triangle I was suggesting.
     

    Attached Files:

  18. Jul 9, 2015 #17
    So given your triangle, how would I use that to find d?

    I was just assuming something like this. Don't know how picture got flipped.

    https://photos-5.dropbox.com/t/2/AADxYk-aonmJAYOZcE2sYaAXx2xtx0obeTzHz0YEMyG5RA/12/51054650/jpeg/32x32/1/1436475600/0/2/2015-07-09%2014.07.38.jpg/CLqQrBggASACIAMgBCAFIAYgBygBKAIoAygH/DqJgHWIuRBypec8tKOV3Tk3svKkHv0AeV478BINibRc?size=1024x768&size_mode=2 [Broken]
     
    Last edited by a moderator: May 7, 2017
  19. Jul 9, 2015 #18
    https://photos-3.dropbox.com/t/2/AADWd9kWZiwMFH1Svwi-bbLvApEmtOjBYuDxYjJxCWD5JQ/12/2668640/jpeg/32x32/1/1436475600/0/2/2015-07-09%2021.04.34.jpg/CODwogEgASACIAMgBCAFIAYgBygBKAIoBw/mATI66H3HO_k1EIE84VJevokK97mR2xUrjAyfX-h5Yk?size_mode=5 [Broken]

    Someone else worked it out like this, is this allowed? (Ignore the smaller red circle), where his d is my b, and his a is my d, and the triangle created by the circle's radius and "b" is radius - d, and then work from there?
     
    Last edited by a moderator: May 7, 2017
  20. Jul 9, 2015 #19
    Where R =d+x and x = Sqrt(R^2 - b^2), so d = R - Sqrt(R^2 - b^2) ?
     
  21. Jul 9, 2015 #20

    TSny

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    Looks good to me.
     
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