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Separating ions. Acceleration problem

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    You are trying to separate 2 types of ions (fast and slow). The fast ones travel at 1000m/s and the slow ones travel at 100m/s. At the beginning you have 2 plates. One has + and the other one has negative charge. The size of the plates is 5cm and they are 3 cm apart. What will the acceleration have to be between the 2 plates that will put the fast and slow ions 2 cm apart at 50cm once out of the plates.

    Remember once the ions are out of the plates, they will not accelerate any more.
    3. The attempt at a solution

    I drew it out.
    The reds are the ions. The blue box at the bottom is the emitter. The 2 gray ones are the charged plates that will accelerate the ions.

    Ions.gif

    From what i know there will be 2 different times since the ions travel at different speeds
    For the fast one i calculated
    .5m / 1000m/s = .0005s
    for the slow one i got
    .5m / 100m/s = .005s

    From here on out, i'm stuck and have no idea what to do next.
    Thanks in advance.
     
  2. jcsd
  3. Feb 7, 2009 #2

    LowlyPion

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    Homework Helper

    The key is how long the fast or the slow will be traveling through the plates..

    By establishing a charge between the plates, they will be traveling across the 5 cm distance and will be accelerated (deflected) according to normal kinematics. The slower ones being between the plates will be subjected to an accelerating force for a longer interval and hence will be deflected more.

    Start with how long each will be traveling across the plates and see what occurs to you remembering that as usual F = m*a.
     
  4. Feb 7, 2009 #3
    Thank you for the feedback sir. So i got that. 5cm = .05 m

    Since their initial velocity is in the Y direction during the acceleration towards X will not change the Y, I do this to find their times.
    Fast = .05m / 1000m/s = .00005s
    Slow = .05m / 100m/s = .0005s

    Then i dont know where to plug it in or how. Where does the 2cm difference end up going?
    I assume i'm going to have 2 equations. One for the acceleration here and another once its out of the plates but i have no idea what formulas to use.

    Since both are going to curve, i dont know what the angle in between will be which is where i'm confused.

    I've never done a problem like this before so i'm confused what to do next.
     
  5. Feb 7, 2009 #4

    LowlyPion

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    Homework Helper

    The 2 cm distance is at 50 cm.

    What component of velocity is required to get the slow ion to move farther to the right by 2 cm by the time it reaches 50 cm than the fast one?

    You know V = a*t where t is the time the particle is subjected to acceleration between the plates.

    That velocity times the time T to get to 50 cm at the original speed then would be the deflection I'd think for each particle. The additional constraint will be that 1 is 2 cm further along when it gets there. One distance to the right will be 2 cm more than the other.
     
  6. Feb 9, 2009 #5
    alright so here is what i got.

    position after the .05m (between plates) is
    Xf = 0ms + 0t + .5At^2

    slow beam..
    Xfs = .5a(.0005s^2)
    fast beam
    Xff = .5a(.00005s^2)

    Next velocity at the end
    Slow beam
    Vfs = a*.0005s
    Fast beam
    Vff = a*.00005s


    Next, X2-X1 = .02m
    so
    .02m = (.5a(.0005s^2) + (a*.0005s)*.005) - (.5a(.00005s^2) + (a*.00005s)*.0005)
    A = 7696.00 m/s^2
     
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