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Homework Help: Charge to mass ratio of an electron shot through a cathode-ray oscilloscope

  1. Jan 27, 2005 #1
    I've been working on this problem for awhile but I just can't seem to hack it. (This is problem 76 from Serway 3rd for the first electricity chapter if you happen to have that book and want a visualization)
    Given: an electron with charge, -e, and mass, m, is projected with speed v, at a right angle to a uniform electric field that is flowing in the negative y direction. The oscilloscope has length d, and after the particle exits the field it travels a length, L, and smacks a screen. It was displaced in the positive y direction to a height of h when it has hit the screen. Assuming d is much less than L in magnitude, and ignoring gravity, show the charge-to-mass ratio is given by:
    e/m = hv^2/(ELd)

    First, I set Ee = ma and divide through to get e/m = a/E

    Solving for a in terms of L, d, and h, I know the electron is travelling with constant horizontal velocity, and it travels a distance of (L + d) in time t, or

    v = (L+d)/t so t = (L+d)/v

    I also know that it is accelerating in the y direction, so

    h = vyt + 1/2at^2 (there is no initial y component to the velocity so...)

    h = .5at^2

    plugging in for t and solving for a I get...

    2h/(L+d)^2/v^2 = a which simplifies to 2hv^2/(L^2+2dL+d^2) = a

    I then plug it into the original ratio equation

    e/m = 2hv^2/(E(L^2+2dL+d^2) which assuming d is much less than L,

    e/m = 2hv^2/(E(L^2 +2dL) which is close but not the answer

    If anyone can help me I would be greatly appreciative. I know I have probably made a stupid mistake somewhere, but for the life of me, I can't find it. Thank you.
  2. jcsd
  3. Jan 27, 2005 #2

    Doc Al

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    Staff: Mentor

    Realize that the electron is only accelerated during its travel across d, not d + L.

    Try this: Find the impulse given to the electron, then the change in its vertical speed. Then use that to find the angle the electron is traveling when it leaves the field. (Ignore the change in height during the acceleration; assume the electron starts its straight line journey at y = 0.)
  4. Jan 27, 2005 #3

    Thank you so much, I was killing myself over this problem
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