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Acceleration of a charged particle

  1. Feb 15, 2008 #1
    1. The problem statement, all variables and given/known data
    What is the acceleration of a charged particle in a uniform electric field? Assume the particle moves along a straight line parallel to the electric field. Show that a particle starting from rest at x=0 and t=0 the speed and position are given by the following formulas.


    2. Relevant equations

    Vx=(eE/m)(t/Sqrt(1+(eEt/mc)^2)

    and

    x=(mc^2/eE)Sqrt(1+(eEt/mc)^2-1)



    3. The attempt at a solution
    I don't know where to start...
     
  2. jcsd
  3. Feb 15, 2008 #2

    malawi_glenn

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    what is the force?
     
  4. Feb 15, 2008 #3
    I don't know. It is just the "uniform electric field" of unspecified potential. This question is written exactly as it was given to me.
     
  5. Feb 15, 2008 #4

    malawi_glenn

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    But what course is this?

    [tex]\vec{F} = q\vec{E} [/tex], where q is the charge and E the electric field

    That formula you must know if someone wake you up in the middle of the night:P

    EDIT:
    Or w8 a minute, what is the relation between acceleration and veolcity? This must be a course in Newtonian dynamics ;)

    (this is not advanced physics, I cant belive someone has this as upper level undergraduate physics..)
     
    Last edited: Feb 15, 2008
  6. Feb 16, 2008 #5
    Yep, it's advanced physics. And this type of question is not even in the textbook! Another thing that bugs me is the little e. The ONLY mention of the little e in the text is for the rest energy of an electron as being e=.511MeV but looking at the relevant equations I'm given we have mass, time, Energy, velocity, and position (x) and the little e is all that remains to express the electric field which I would expect to be volts per meter. I would say E were the electric field but it is a scalar quantity and the conventions used in my textbook are that the scalar E is always energy.
     
  7. Feb 16, 2008 #6

    malawi_glenn

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    No the electric field is a vector quantity.

    And is this correct?

    x=(mc^2/eE)Sqrt(1+(eEt/mc)^2-1) ??

    It would reduce to

    x=(mc^2/eE)Sqrt((eEt/mc)^2) =(mc^2/eE)(eEt/mc)
     
  8. Feb 16, 2008 #7

    malawi_glenn

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  9. Feb 16, 2008 #8
    q is the electric charge of the particle (in coulombs)
    and
    E is the electric field (in volts per meter)and it equals...capital E as a scalar!

    Problem is there is still the question of what the little e is...
     
  10. Feb 16, 2008 #9

    malawi_glenn

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    charge of the electron, the elementary charge...

    sometimes they use e instead of q..

    And you should bewere that since the alphabet is very short in comparison with the number of physical quantities, you must look at the context. An E in an electro magnetism problem is in 99% of the cases the electric field.
     
  11. Feb 16, 2008 #10
    This is a special relativity question. From
    [tex] \vec{F}=\frac{d\,\vec{p}}{d\,t}\quad (1)[/tex]
    with
    [tex]\vec{F}=e\,\vec{E}[/tex]
    and
    [tex]\vec{p}=\gamma\,m\,\vec{u},\gamma=\frac{1}{\sqrt{1-(\frac{u}{c})^2}}[/tex]

    Since this is a 1-dimension problem, integrate (1) to get your result.
     
  12. Feb 16, 2008 #11

    malawi_glenn

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    why not wait til OP show some work? I already pointed out that he should look at the relation between velocity and acceleration. He must also show how to get to V_x and x
     
  13. Feb 16, 2008 #12
    @ malawi_glenn

    Is there a general rule on how someone must give instructions? Is it too muh trouble to remind someone the relations that he must apply?
     
  14. Feb 16, 2008 #13

    malawi_glenn

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    Depends on how you see it, the OP MUST show attempt to solution before any help can be recived. And since I already have kicked him in the correct direction by saying that he should look at the relation between veolcity and acceleration and that he should look at relativistic formula. So in fact, I myself broke the forum rules too.
     
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