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Electron and proton between two plates

  1. Dec 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Between two plates (distant 4cm) there's an electric field (E = 100N/C).
    A proton starts moving from the positive plate and an electron starts moving from the negative plate.
    At what distance from the positive plate do they meet? (is "meet the right verb here? :biggrin:)

    2. Relevant equations
    ## a = qE/m ##
    ## v = at ##
    ## U = - \frac{e^2}{4\pi\epsilon_0 r}##
    ## U_i + K_i = U_f + K_f ##

    3. The attempt at a solution
    I calculated ##U_i##.
    It is ##-5.76x10^{-27} J ##.

    Now,
    ## U_i = Kf ##
    and
    ## Kf = \frac{1}{2} m_p v_p^2 - \frac{1}{2}m_e v_e^2##

    I replace ## v = at = \frac{eEt}{m} ##

    ## U_i = [\frac{1}{2} e^2E^2 ( \frac{1}{m_p} - \frac{1}{m_e})] t^2 ##
    ## t = 6.39x10^-12 s ##

    ## x(t) = \frac{1}{2} at^2 ##

    I obtain a very small value of x. 10^-13 or so.
    There's a big difference (4 orders of magnitude) between the acceleration of the electron, that's true.
    But they meet at very very very small distance from the positive plate.

    Could you tell me if this is right?

    Thank you so much
     
  2. jcsd
  3. Dec 16, 2014 #2

    ehild

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    10^-13 in what units???
    No, it is not right.
    Think: Forces of the same magnitude act on both particles, F=qE. And you correctly wrote that the acceleration is a=qE/m.
    me=9,11x10-31kg. mp=1.67x10-27kg.
    The ratio of the masses mp/me=1833. What is the ratio of the accelerations?
    Again you wrote correctly, that the distance travelled by one particle is x=a/2 t2. They travel for the same time. So what is the ratio of the distances they travelled?
     
  4. Dec 16, 2014 #3
    Sorry, I forgot to say the units. 10^-13 m!

    The ratio of accelerations is ##\frac{a_p}{a_e} = \frac{1}{1833} ##
    The ratio of distance is the same, because t is the same.

    I don't understand how to use it, though
     
  5. Dec 16, 2014 #4

    ehild

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    The proton travels x distance and the electron travels 1833 x distance. The sum of their distances is 4 cm. What is the distance travelled by the proton?
     
  6. Dec 16, 2014 #5
    Oh, ok

    ## 0.04 = x(1833+1) ##

    x is ##2.18x10^{-5}##m.

    What was my mistake then?
     
  7. Dec 16, 2014 #6

    ehild

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    I could not understand what you did.
     
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