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Electron/proton released from charged parallel plates

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data
    Two parallel plates 1.40 cm apart are equally and oppositely charged. An electron is released from rest at the surface of the negative plate and simultaneously a proton is released from rest at the surface of the positive plate. How far from the negative plate is the point at which the electron and proton pass each other?

    Known:
    Qe = -1.6 x 10-19 C
    Qp = 1.6 x 10-19 C
    me = 9.11 x 10-31 kg
    mp = 1.6 x 10-27 kg


    2. Relevant equations

    a = qE/m
    d = 1/2*a*t2



    3. The attempt at a solution

    Acceleration of electron
    ae = qeE / me

    Acceleration of proton
    ap = qpE / mp

    Distance of electron
    de = aet2 / 2

    Distance of proton
    dp = apt2 / 2

    Since I don't know time, I tried to eliminate it by
    t2 = 2de/ae
    and plugging it into distance formula but from there, I'm stuck on what to do.
     
  2. jcsd
  3. Jan 21, 2017 #2

    TSny

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    Hello, and welcome to PF!

    You are given the distance D between the plates. Can you make use of this information?
     
  4. Jan 21, 2017 #3
    I'm not really sure how distance D relates to the problem, but my guess would be if electron is travelling from distance 0 to 1.4 cm and the electron is travelling from 1.4 cm to 0, if looking at it on an x-axis... then the distance the electron travels is S, while the distance the proton travel is 1.4 cm - S.

    Rewriting the equations would be...

    S = 1/2 * ae * t2

    and

    0.014 - S = 1/2 ap * t2

    How would I get the acceleration without knowing the electric field E?
     
  5. Jan 21, 2017 #4

    TSny

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    Good.

    Maybe you won't need actual values for the accelerations. Just keep working with symbols. What can you learn about the time from your equations above? For now, use D for the distance between the plates rather than use the number .014 m.
     
  6. Jan 21, 2017 #5
    Ah I see.
    I know that t would be the same for both because that's when they meet with each other.

    Would I solve for t and make them equal to each and solve for S?

    t2 = 2S / ae and t2 = 2(D-S) / ap

    And solving for S would be...

    S = D * ae / ap + ae

    Unless my my algebra is wrong.
     
  7. Jan 21, 2017 #6

    TSny

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    Did you mean S = D * ae / (ap + ae) so that both ap and ae are in the denominator?

    Can you write your result in terms of the ratio of ae to ap?
     
  8. Jan 21, 2017 #7
    Yes they are both in the denominator.

    Using the formula a = qE/m , I know that E should be the same for both the proton and electron, so I can set them equal.
    I'm not sure if this is what you were asking but I did
    (ae * me ) / qe = (ap * mp ) / qp

    Then solving for ae I got
    ae = ap * 1756.31
     
  9. Jan 21, 2017 #8

    TSny

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    OK. What do you get when you use this information in your result for S?
     
  10. Jan 21, 2017 #9
    Ah! I see.

    I plugging in everything I know, the accelerations got cancelled resulting in S = 0.01399 m. This makes sense because electrons move faster than protons.
     
  11. Jan 21, 2017 #10

    TSny

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    Looks good.
     
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