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Particle in an electric field- formulae suggestions.

  1. Mar 26, 2009 #1
    1. A particle of charge q and rest mass [tex]m_0[/tex] is accelerated from rest at t=0 in a uniform electric field of magnitude E. All other forces are negligible in comparison to the electric field force. Show that after a time t the relativistic expression for the speed of the particle is:

    [tex] v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}[/tex]


    2. Relevant equations

    I think the equations relevant here have got to be [tex]E_k=(\gamma-1)m_0c^2[/tex] and [tex] E=\frac{F}{q} [/tex].

    I've tried rearranging the above equations and substituting them into each other but I can't get anything good. Does anyone know any other useful equations that I might be able to use?
     
  2. jcsd
  3. Mar 26, 2009 #2

    tiny-tim

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    Hi Alpha&Omega! :smile:

    (how come you have two different names? :confused:)
    Hint: with that v, what is (v/c)/√(1 - v2/c2) ? :wink:
     
  4. Mar 27, 2009 #3
    I felt insecure. XD

    Do you mean as a physical quantity?

    I would just do this:

    [tex]\frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    [tex]\sqrt{\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}[/tex]

    [tex]\sqrt{\frac{1}{1-\frac{v^2}{c^2}}-1[/tex]

    Is this what you meant? =S
     
  5. Mar 27, 2009 #4

    tiny-tim

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    No :biggrin:, I meant what is (v/c)/√(1 - v2/c2) if [tex] v=\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}[/tex] ? :wink:
     
  6. Mar 27, 2009 #5
    Ah, darn. Let me see:

    [tex]\frac{{\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}{c}}{\sqrt{1-\frac{(\frac{c}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}})^2}{c^2}}}[/tex]

    [tex]\frac{{\frac{c^2}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}}{\sqrt{1-\frac{(\frac{c^2}{{1+\frac{m_0^2c^2}{q^2E^2t^2}}})}{c^2}}}[/tex]

    [tex]\frac{{\frac{c^2}{\sqrt{1+\frac{m_0^2c^2}{q^2E^2t^2}}}}}{\sqrt{-\frac{m_0^2c62}{q^2E^2t^2}}}[/tex]

    Ummm, am I doing it correctly? I've got a complex number as the denominator! =S
     
  7. Mar 27, 2009 #6

    tiny-tim

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    uhhh? whadya do? :confused:

    should be qEt/m0c :wink:
     
  8. Mar 27, 2009 #7
    img040.jpg

    magic!

    This may sound weird, but how did you know to use [tex]\frac{\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]?
     
  9. Mar 27, 2009 #8

    tiny-tim

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    Because I knew that m0(v/c)/√(1 - v2/c2) is the … ? :smile:

    (btw, it would have save a lot of paper and time if you'd written X, say, instead of that fraction! :wink:)
     
  10. Mar 27, 2009 #9
    The momentum divided by c?

    So if momentum is MeV/c then in this equation we have MeV/c2 which is the mass!

    (I know it would have saved trees, but for some reason I decided to do it this long and complicated way =D).

    So I have to work from this equation to the equation posted in the thread.

    I just tried it and I noticed that the question doesn't mention whether the applied force is parallel to the x axis or perpendicular to it. Which expression for relativistic force should I use?
     
  11. Mar 27, 2009 #10

    tiny-tim

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    Yup! :biggrin:

    So the question is asking you to prove that after time t the momentum is qEt …

    what principle of physics would you use to do that? :wink:

    uhh? :confused: the x-axis is wherever you want it to be
     
  12. Mar 27, 2009 #11
    Would it be an integral of an expression for the force with respect to time?

    In my book it has two expressions for relativistic force. One is [tex]F=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}ma[/tex] if it's perpendicular to the x axis and the other is [tex]F=\frac{1}{(1-\frac{v^2}{c^2})^\frac{3}{2}}ma[/tex] if the force is in the same direction as the field.
     
  13. Mar 29, 2009 #12

    tiny-tim

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    I was thinking more along the lines of "Force = …"

    (work backwards from the answer! :wink:)
    ah …
    so the velocity will always be in the same direction as E :smile:

    btw, I don't normally approve of "relativistic mass", but in this case I find it easier to remember the equations as

    Fperp = mrela

    Fpara = mrela/(1 - v2/c2)

    so I remember the perpendicular force equation as being the same as Newton's :wink:
     
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