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A particle inside eletric field

  1. Mar 24, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    Sorry about my english , i'm still learning.

    I did it by energy, but i want to solve by integration

    QUESTION : A charged and massive particle runs with 66 m/s, 3^10-6 coulomb and 6^10-3 Kg towards a fixed particle with 4,5^10-6 coulomb, separated by 4,3 meters . What's the distance between them which the initial velocity is zero ?

    ANSWER : 0,00761 meters .

    I started that way :

    1) a = F/m ; F= kQq/r² ; 2) v²=vo² - 2.a.d; and i substituted the acceleration gives by 1 in equation 2 . So i isolated d and integrated 4,2 and (4,2-d) interval. It results in a cubic equation which doesn't make sense .
     
  2. jcsd
  3. Mar 24, 2016 #2

    BvU

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    Hello Alois, :welcome:

    This is not a matter of uniform acceleration ! You can't use the SUVAT equations here.
    And you don't need to integrate either. That is being taken care of perfectly by using the electric potential.
    Or did you do that already and did you find the 8 mm that way ?

    Could you show your calculations in a bit more detail ? I can't follow what you mean with (4,2) and (4,2-d) ?
     
  4. Mar 24, 2016 #3
    I already dit it by using eletric potential and cinetic energy (as Wolfgang Bauer's book did) . By integration i didn't find the correct answer .
    I thought even the acceleration is not uniform , is given us how it changes with the distance , and we can evaluate by integrating all points of it in the path . SUVAT equation was the only way i saw to associate velocity , acceleration and distance .

    (4,2) and (4,2-d) are the higher and lower integral limits


    http://[ATTACH=full]200062[/ATTACH] [ATTACH=full]200063[/ATTACH]
     
    Last edited by a moderator: May 7, 2017
  5. Mar 25, 2016 #4

    BvU

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    The equation of motion for uniform acceleration integrates a constant acceleration ##\vec a = {\vec F \over m}##. Here you have to deal with a force that depends on ##| \vec r - \vec r' | ## . In the simplest form ( ## F = - {1\over r^2} ## ) this gives a ## 1\over r ## as you find in the expression for the potential.

    I did not understand the underlined part ?
     
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