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Electron Acceleration & movement between a uniform magentic field

  1. Dec 19, 2011 #1
    1. The problem statement, all variables and given/known data
    An electron is fired at 4.0x106 m/s horizontally between the parallel plates as shown, (see diagram) starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x102 N/C. The seperation of the plates is 2.0 cm

    A) find the acceleration of the electron
    b)find the horizontal distance traveled by the electron when it hits the plate
    c) find the velocity of the electron as it strikes the plate

    2. Relevant equations

    b) Δdy=V1yΔT + 1/2 aΔT2
    ΔDx= V1xΔT
    c)V2x=√V1x + 2aΔdy Really unsure about this, using x and y components in the same equation?
    qelectron= -1.6x10-19
    Melectron= 9.11x10^ -31 kg
    3. The attempt at a solution

    A) ma=εq
    a= (4.0x10^2)(1.6x10^-19)/(9.11x10^-31)
    a= 7.03 x10^13 m/s^2 [down]

    B)Δdy=V1yΔT + 1/2 aΔT2
    0.02=0ΔT +1/2 (7.03x1013ΔT2 (the 0Δ T cancels?)
    ΔT= √(0.02)(2)/(7.03x1013
    ΔT= 2.3x10-8s

    ΔDx= V1xΔT
    ΔDx= (4.0x10^6)*(2.3x10^-8)
    =0.092 m
    =9.2 cm

    C) V2x=√V1x + 2aΔdy
    V2= √(4.06x10^6)+2(7.03x10^13)(0.02) <--(should I use horizontal or vertical distance here)
    V2= 4.34x10^6 m/s ( it gets faster which is a good sign lol)

    My biggest confusion is when to use vertical distance 2 cm or horizontal distance 9.2 cm .

    I took a lot of time to write this up so any help is appreciated immensely. I feel like I did everything right I just need re assurance. Im a perfectionist and I want to get full marks.

    Sorry forgot to attach diagram, its crude but its exactly what is happening diagram.png
    Last edited: Dec 19, 2011
  2. jcsd
  3. Dec 19, 2011 #2
    You know the electric field strength = 400N/C so you can find the force on an electron.
    If you know the Force and the mass of an electron then you can calculate the acceleration.
    You know the horizontal velocity of the electron so the question is a sort of projectile question..... does this help??
  4. Dec 19, 2011 #3
    Uhh to be honest, It has only made me second guess myself lol :P
  5. Dec 19, 2011 #4
    OK... I will go through your numbers
    I used F = E x q to get Foce on electron = 6.4 x 10 ^-17N
    Then used F= ma to get acceleration a = 7.05 x 10^13 .... same as you
    I got time to move between the plates (use 2cm as S in equation S = 0.5at^2) to be
    2.39 x 10^-6 s (you got x 10^-8).... I will check mine again
    (I think you have made a mistake where you have written 'the ΔT cancels'....it does not !!!)

    I will be back soon with my complete calculation....

    I made a mistake... also got the time to be 2.38 x 10^-8 s....
    back soon

    I got the same as you for the velocity..... the length of the plates does not come into the calculation.... they tell you the electron hits the plates and that confirms it....
    Last edited: Dec 19, 2011
  6. Dec 19, 2011 #5
    You got it right..... do you want a challenge?????
    At what angle does the electron strike the plate ????
    I am surprised they did not ask you that :smile:
  7. Dec 19, 2011 #6
    Thank you very much for the help, as for the challenge I have no real idea how to even begin to find the angle..

    I appreciate the input however, you seem to be helping me a lot lately !

    Good vibes :)
  8. Jan 12, 2012 #7
    The question asked for Velocity. Not speed. Velocity has direction.
    in this case, the direction would be the angle at which it strikes the plate.
    evidently they did ask you for that :P

    Tan(theta) = V_f_y / V_f_x
    (theta) = 23degrees

    The correct full answer being:

    "Therefore the final velocity of the electron was 4.3 x 10^6 m/s [ E 26(degrees) S ]"

    P.S. Also remember to show in your diagram which direction is North. (You can use up,down,left,right, or [Towards negative plate] if you wish
    Last edited: Jan 12, 2012
  9. Jan 12, 2012 #8
    How are you getting Vfx I used your answer of 23 degrees and solved backwards to get a vfx that should equal about 1.08x10-5 but I can't seem to get that answer as my vfx for the life of me.

  10. Jan 16, 2012 #9

    I hope im allowed to bump threads i need an answer !
  11. Jan 16, 2012 #10
    I said I would post my calculations and forgot !!!!! Here are the vital steps.
    Force on electron = 6.4 x 10^-17 N
    Gives vertical acceleration = 7 x 10^13 m/s^2
    using separationof plates as distance travelled this gives vertical velocity = 1.67 x 10^6 m/s
    Horizontal velocity is constant therefore velocity of electron is given by:
    v^2 = (4 x 10^6)^2 + (1.67 x 10^6)^2
    gives velocity = 4.34 x 10^6 m/s
    Just as Vanished has pointed out
    The angle of the velocity is given by Tanθ = vertical velocity/horizontal velocity
    Tanθ = 1.67 x 10^6/4 x 10^6 = 0.4175 gives θ = 23 degrees to horizontal
    Hope this is what you wanted.... ask again if you are not sure of anything
  12. Jan 16, 2012 #11

    Uh, how is vertical velocity found, its driving me insane. I know you sort of explained it but Ive tried to find it, and tried to get the same number as you but I can't.

    If you can explain it a little further that would be great !

  13. Jan 17, 2012 #12


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    Homework Helper

    Vertically you have constant force so constant accelerated motion, so use Vf = Vi + a*t.
    Vi is zero vertically. Never mind signs; it accelerates down and you know the velocity is downward.

    I think you will be disappointed with those answers because they are not very accurate after part (a). You have 2.3 x 10^-8 for the time. I have 2.386. If you want 2 digit accuracy in final answers after this point you must start with 3 digit accuracy for the time. If you want 3 in final answers, keep 4 along the way.
  14. Jan 17, 2012 #13
    Do you see the vertical acceleration is 7.0 x 10^13 m/s^2 ?
    I then used the distance between th 2 plates (2 x 10^-2 m) to find the vertical component of velocity using
    V^2 = u^2 + 2aS and got v =1.7 x 10^6 m/s (1.68 to 3 figures)

    As far as significant figures are concerned you are not given any numbers to more than 2 significant figures so it is not legitimate to give an answer to more than 2 figures
    I used 3 figures in my calculations
    by definition any more than 2 figures is insignificant

    Hope you see the physics behind the calculation
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