- #1
jordanl122
- 14
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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 traveling 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.
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 traveling 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.
