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Electrion moving between two parallel plates

  1. Jan 25, 2015 #1
    1. The problem statement, all variables and given/known data
    At what angle will the electrons in the image leave the uniform electric field at the end of the parallel plates (point P) ? Assume the plates are .049 m long, E = 5x10^3 N/C and v0 = 1x10^7 m/s. Ignore fringing of the field.
    GIANCOLI.ch21.p59.jpg
    2. Relevant equations
    E = kQ/r^2
    F=ma
    F= kQ1Q2/r^2

    3. The attempt at a solution
    I am having difficulty figuring out how to get started. I know the equation for the motion of the electron is
    y= -(eEx^2)/(2mv0^2) but I don't know if this helps me. I think they want the angle the path of the electron makes with the x axis, but I am not sure. Can someone get me started or give me a hint on how to relate the angle to everything else please?
     
  2. jcsd
  3. Jan 25, 2015 #2

    TSny

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    The angle at which the electron is moving is given by the direction of the velocity vector of the electron. So, try to figure out the x and y components of the velocity at point P.

    Note: Two out of the three equations that you listed under "relevant equations" are not actually relevant to this problem. Can you spot them?
     
  4. Jan 25, 2015 #3
    I tried plugging in .049 m for x into the equation of motion as well as the mass of an electron for m and all the other given information and got
    y = - [(1.602x10^-19)(5x10^3)(.049)^2]/[2(9.1x10^-31)(1x10^7)^2] = -.011 m
    which is how far below the x axis it is so the angle would be tan^-1(-.011/.049) = -13 degrees?
     
  5. Jan 25, 2015 #4
    I guess F= ma is the relevant one?
     
  6. Jan 25, 2015 #5

    TSny

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    Right. But you'll need some other equations, too.
     
  7. Jan 25, 2015 #6
    ok so to find the velocity vector at point P ... there is no x component of acceleration, so vx= v0
    vy = vy0 - at
    vy0 = 0 so
    vy = - at
    a = F/m = qE/m = -eE/m
    so
    vy = eEt/m ... am i on the right track?
     
  8. Jan 25, 2015 #7

    TSny

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    Yes, you are!
     
  9. Jan 25, 2015 #8
    I think I am now stuck though because I dont have any information for t
     
  10. Jan 25, 2015 #9

    TSny

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    Any way to use the information about the length of the plates?
     
  11. Jan 25, 2015 #10
    ah thank you! so it constantly travels 1x10^7 m/s in the x direction for the length of .049 m.
    v = d/t so t= d/v = .049 m / 1x10^7 m/s = 4.9 x 10 ^-9 s
    so
    vy = eEt/m = (1.602x10^-19)(5x10^3)(4.9x10^-9)/(9.1x10^-31) = 4313077 m/s right?
     
  12. Jan 25, 2015 #11

    TSny

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    That looks right. I haven't actually grabbed a calculator and checked your numbers.
     
  13. Jan 25, 2015 #12
    ok so that means theta = tan^-1(4313077/(1x10^7)) = 23.3 degrees but -23.3 degrees because it is going clockwise from the x axis
     
  14. Jan 25, 2015 #13

    TSny

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    Yes. Or you can just say 23.3o below the horizontal. (That way, whoever sees your answer doesn't have to decipher the meaning of the negative sign.) But, if you are having to plug your answer into some sort of grading program, then you might need to include the sign.
     
  15. Jan 25, 2015 #14
    thank you so much!
     
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