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Energy radiated from a charge in electric field

  1. Sep 17, 2014 #1
    1. The problem statement, all variables and given/known data
    This is problem 11.13 from Griffiths Introduction to Electrodynamics 4th Ed.
    A positive charge q is fired head-on at a distant positive charge Q (which is held stationary) with a initial velocity [itex]v_0[/itex]. It comes in, decelerates to v=0 and returns out to infinity. What fraction of its initial energy [itex](\frac{1}{2}mv_0^2)[/itex] is radiated away? Assume [itex]v_0<<c[/itex] and that you can safely ignore the effect of radiative losses on the motion of the particle. Answer: [itex]\frac{16}{45}\frac{q}{Q}\left(\frac{v_o}{c}\right)^3[/itex]

    2. Relevant equations

    Larmor formula
    [tex]P=\frac{\mu_0 q^2a^2}{6\pi c}[/tex]

    3. The attempt at a solution
    In this problem acceleration depends on x: from Coulomb's law and Newton second Law: [itex]F=\frac{1}{4\pi \epsilon_0}\frac{qQ}{x^2}=m\ddot{x}[/itex]. We don't know the time spent by particle q in its travel in order to compute total energy radiated.
    From linear momentum [itex]p=m\dot{x}[/itex], we have [itex]\dot{p}=m\ddot{x}[/itex] and [itex]\dot{p}=\frac{dE}{dx}[/itex].
    E is the energy of particle, the sum of kinetic and electrostatic potential energy: [tex]E=\frac{1}{2}m\dot{x}^2+\frac{1}{4\pi \epsilon_0}\frac{qQ}{x}[/tex].
    But [itex]\frac{dE}{dx}=\frac{dE}{dt}\frac{dt}{dx}[/itex] and I can write Larmor formula as:
    [tex]P=\frac{\mu_0 q^2}{6\pi cm^2}\left(\frac{dE}{dt}\frac{dt}{dx}\right)\left(\frac{dE}{dx}\right)[/tex]
    The fraction asked is:
    [tex]f=\frac{P}{\frac{dE}{dt}}=\frac{\mu_0 q^3Q}{24\pi^2 \epsilon_0 cm^2x^2\dot{x}}[/tex]
    I have differentiate E with respect to x.
    I don't know how to evaluate "f" and I don't know if my approach is correct.
    Any help? Thanks in advance!
  2. jcsd
  3. Sep 17, 2014 #2


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    Hello 987oscar. Welcome to PF!

    ##\small P## is the rate of creation of radiation energy: ##\small P## ## = \frac{dE_{rad}}{dt}##. The energy of the particle, ##\small E##, is assumed to remain essentially constant according to the phrase "you can safely ignore the effect of radiative losses on the motion of the particle".

    Note ##\small P## ## = \frac{dE_{rad}}{dt} = \frac{dE_{rad}}{dx} \small \dot{x}##. Use this for the left side of the Larmor formula and try to see where to go from there.

    [EDIT: Alternately, you can try the substitution ##\small P## ## = \frac{dE_{rad}}{dt} = \frac{dE_{rad}}{dv} \small \dot{v}##, where ##\small v## is the speed of the particle. I believe this substitution will make life easier!]
    Last edited: Sep 18, 2014
  4. Sep 18, 2014 #3
    Hello TSny:
    Yes!! as [itex]\ddot{x}=\frac{1}{4\pi\epsilon_0}\frac{qQ}{mx^2}[/itex], I can write Larmor formula as:[tex]\frac{dE_{rad}}{dx}\dot{x}=\frac{\mu_0q^4Q^2}{96\pi^3\epsilon_0^2cm^2x^4}[/tex].
    But from Energy Conservation: [tex]\frac{1}{2}mv_0^2=\frac{1}{2}m\dot{x}^2+\frac{1}{4\pi\epsilon_0}\frac{qQ}{x}[/tex], so I find [itex]\dot{x}=\sqrt{V_0^2-\frac{1}{2\pi\epsilon_0 m}\frac{qQ}{x}}[/itex].
    Particle goes from infinity to [itex]x_0=\frac{qQ}{2\pi\epsilon_0mv_0^2}[/itex] and returns out to infinity. I have integrated [itex]E_{rad}[/itex] from [itex]x_0[/itex] to infinity and finally I could get the correct answer.
    Thank you very much!!!! :smile:
  5. Sep 18, 2014 #4


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    Great! That's the way I did it at first, also. If you want, you can try the other substitution ##\small P ## ##= \frac{dE_{rad}}{dt} = \frac{dE_{rad}}{dv} \dot{v} =\frac{dE_{rad}}{dv} a##. I found the integration easier in this case.
  6. Sep 18, 2014 #5
    Ok. Iwill try that way too...
    Thanks again
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