Calculate the acceleration of the particle toward the tube wall, 2eE/m
e is the electron charge and m is the mass of an alpha particle, 4 Mp
The time it takes to move and hit the side wall will be given by
R = (a/2) t^2
Solve for t.
The distance travelled will be V*t
The distance 0.849 m seems suspiciously short to me. That is a very weak E-field.
a = 2*1.6*10^-19*4.5*10^-4/[4*1.67*10^-27) = 2.16*10^4 m/s^2
t = sqrt(2R/a) = 6.81*10^-3 s
X = distance traveled = V*t = 8.5 m
Unless I made a mistake somewhere, I disagree with the answer you "know" by a factor of ten
"An alpha particle (a helium nucleus) is traveling along the positive x-axis at 1 250 m/s when it enters a cylindrical tube of radius 0.500 m centered on the x axis. Inside the tube is a uniform electric field of 4.50 X 10^-4 N/C pointing in the negative y direction. How far does the particle travel before hitting the tube wall? "
I know the answer is 0.849m I just don't understand how my teacher got that answer. Can someone please help "dumbify" it for me?
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