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How will the particle move?

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    http://prntscr.com/6d250v Hello, ignore the greek language. It says that this particle is in a homogenous gravitational and electrical field (from left to right the letters are a,b,c,d,e)

    2. Relevant equations
    the question is: Which is the most likely route it'll follow?

    3. The attempt at a solution
    i cannot understand why it cannot be the b.), I suppose its due to the gravity and the Force cause they stay constant thus it wont move in a parabolic orbit but in a straight line..However i learnt that in a electrical field between 2 bars where we ignore the gravitational field the particle moves in a parabolic way...i think im missing something...
     
  2. jcsd
  3. Mar 5, 2015 #2
    If the particle is subatomic, I would have thought you would be permitted to ignore the gravitational force, since this is absolutely tiny. However, all the options have a downward component, so I guess not. The left/right direction depends on the charge of the particle, and remember that gravity and the E field will both cause an acceleration through Newton's second law. But one is (much) stronger than the other, so think about how that would affect the shape.
     
  4. Mar 5, 2015 #3
    It's my mistake i didnt mention. The particle has negative charge. I suppose that it's an electron..The answer says its C ( γ) in the picture
     
  5. Mar 5, 2015 #4
    anyone? :)
     
  6. Mar 5, 2015 #5

    PeroK

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    What can you say about each of the gravitational and electrical accelerations? Given that the fields are homogeneous.
     
  7. Mar 5, 2015 #6
    they are constant?
     
  8. Mar 5, 2015 #7

    PeroK

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    Yes. Can you write the equations for the x and y displacements? Assuming accleration g in the y direction and a in the x-direction?
     
  9. Mar 5, 2015 #8
    Assuming the particle has no initial velocity. ## y=\dfrac{1}{2}gt^2 ## and ## x=\dfrac{1}{2}at^2 ## where from the ## F=E|q| ## ##<=> #### a=\dfrac{E|q|}{m}##
     
  10. Mar 5, 2015 #9

    PeroK

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    Why are the equations different? ##t^2## in one and ##t## in another?
     
  11. Mar 5, 2015 #10
    sorry i didnt notice i was thinking about velocity
     
  12. Mar 5, 2015 #11

    PeroK

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    Okay. Now plot x against y on a graph.
     
  13. Mar 5, 2015 #12
    if i did this right in a straight line where it starts from the beginning of the axes
     
  14. Mar 5, 2015 #13

    PeroK

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    It would be a straight line, yes.
     
  15. Mar 5, 2015 #14
    i got it now :) thank you very much for your time
     
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