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Thompson's cathode experiment

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known data
    plates.jpg

    What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
    Express your answer in terms of e, m, d, v0, L, and E0.

    2. Relevant equations
    kinematics equations
    F=qE=ma

    3. The attempt at a solution
    Split the problem into 2 sections, the first being with acceleration due to the electric field,
    t1 = when particle lease electric field
    t2 = when particle hits screen
    so at t1 a distance d

    d = v0t
    Δy1 = ½at2

    t values are unknown so I eliminate t
    t=d/v0
    and
    qE0=ma
    q=e
    (eE0)/m = a


    Δy1 = ½((eE0)/m) (d/v0)2

    now for the x and y components of velocity at this point
    using the equation
    v=v0+at

    vy = 0 + ((eE0)/m)(d/v0
    vx has not changed so = v[SUB]0[/SUB]

    Now is have the height at d and both components of velocity, I can use kinematics to find the particles motion from d to L using these values (at d) as the initial points.

    Δy2 = Δy1 + vy(t)

    this t value will be

    L= v0 t
    t= L/v0

    plugging it all in

    = ½((eE0)/m) (d/v0)2 +
    (eE0/m)d/v0*L/v0


    This is not the correct answer for Δy2
    once I know Δy2 I can find Δy = Δy2 + Δy1
    But I can not find my mistake in finding Δy2
     
    Last edited: Nov 27, 2016
  2. jcsd
  3. Nov 27, 2016 #2

    TSny

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    Did you forget to square the time?

    Why did you include Δy1 as part of Δy2?
     
  4. Nov 27, 2016 #3
    While typing up my question here I did, but I did not in my work and the answer I tried entering.

    I'm not sure what you mean, I wrote Δy2 as a function of Δy2 because of the kinematics equation x=x0 + v(t) when a=0 and Δy1 and Δy2 represent heights.
     
  5. Nov 27, 2016 #4

    TSny

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    In your diagram, the distance Δy2 does not include the distance Δy1.

    But your equation Δy2 = Δy1 + vy(t) contains Δy1 on the right hand side.
     
  6. Nov 28, 2016 #5
    I still do not see your point, Is that kinematic equation not applicable? Why?
    Lets consider this situation. I am in a car, on the x axis at x=5. I go east at 1 mile/hour. To find the position after 3 hours I would use that equation.
    x(t) = 1(t) + 5
    x(3) = 1(3) + 5 = 8

    I see no difference between that situation and this particle, at d its initial position is Δy1 and it is moving up at a rate of the velocity v. After a certain amount of time t it will be at Δy2.
     
  7. Nov 28, 2016 #6

    gneill

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    @TSny is making the observation that your use of variables has strayed from the definitions in the figure. In the figure (and the problem statement) the total displacement is given by ##Δy = Δy_1 + Δy_2##. You have essentially redefined ##Δy_2## to be the total displacement.
     
  8. Nov 28, 2016 #7
    Ah now I see what you mean, Δy2 is not measured with re
    respect to the zero line (whereΔy_1 was), my equation than should be
    ##Δy = ##Δy_1 +v(t)

    So how would I find Δy_2 before finding ΔY?
     
  9. Nov 28, 2016 #8

    gneill

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    Consider the "launch" conditions from the Δy1 location and use geometry (and similar triangles in particular).

    You should have both the x and y velocities of the electron at the Δy1 location.
     
  10. Nov 28, 2016 #9
    since
    Δy = Δy1 + Δy2
    and
    Δy= Δy1 + v(t)
    Δy1+Δy2 = Δy1 + v(t)

    Is that the correct approach?
     
  11. Nov 28, 2016 #10

    gneill

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    The velocity is 2D, that is, it has both x and y components. So unless the v in your equations is a vector, the equations are not correct.

    Start by finding the x an y components of the velocity at the "launch" point.

    EDIT: Perhaps I spoke too soon. If your v is the y-velocity component and t is the transit time from the launch point to the impact point, then your equations would work. You'll have to show how you get both v and t.
     
  12. Nov 28, 2016 #11
    I have found those, time is written in terms of x velocity.
     
  13. Nov 28, 2016 #12

    TSny

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    This would be correct if you replace the v(t) expression by the expression you wrote in your first post vy(t). I assume that by vy(t) you meant vy⋅t.
     
  14. Nov 28, 2016 #13
    Yes, because the only effect the x velocity is making it hit the plate before it can reach any higher.
     
  15. Nov 28, 2016 #14

    TSny

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    Yes. In your first post you have the correct expressions for vy and t to use for Δy2 = vy⋅t.
     
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