Energy radiated from a charge in electric field

[987oscar]

Homework Statement

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 v_0. It comes in, decelerates to v=0 and returns out to infinity. What fraction of its initial energy (\frac{1}{2}mv_0^2) is radiated away? Assume v_0<<c and that you can safely ignore the effect of radiative losses on the motion of the particle. Answer: \frac{16}{45}\frac{q}{Q}\left(\frac{v_o}{c}\right)^3

Homework Equations

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

The Attempt at a Solution

In this problem acceleration depends on x: from Coulomb's law and Newton second Law: F=\frac{1}{4\pi \epsilon_0}\frac{qQ}{x^2}=m\ddot{x}. We don't know the time spent by particle q in its travel in order to compute total energy radiated.
From linear momentum p=m\dot{x}, we have \dot{p}=m\ddot{x} and \dot{p}=\frac{dE}{dx}.
E is the energy of particle, the sum of kinetic and electrostatic potential energy: E=\frac{1}{2}m\dot{x}^2+\frac{1}{4\pi \epsilon_0}\frac{qQ}{x}.
But \frac{dE}{dx}=\frac{dE}{dt}\frac{dt}{dx} and I can write Larmor formula as:
P=\frac{\mu_0 q^2}{6\pi cm^2}\left(\frac{dE}{dt}\frac{dt}{dx}\right)\left(\frac{dE}{dx}\right)
The fraction asked is:
f=\frac{P}{\frac{dE}{dt}}=\frac{\mu_0 q^3Q}{24\pi^2 \epsilon_0 cm^2x^2\dot{x}}
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!

 

[TSny]

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!]

 

[987oscar]

Hello TSny:
Yes! as \ddot{x}=\frac{1}{4\pi\epsilon_0}\frac{qQ}{mx^2}, I can write Larmor formula as:\frac{dE_{rad}}{dx}\dot{x}=\frac{\mu_0q^4Q^2}{96\pi^3\epsilon_0^2cm^2x^4}.
But from Energy Conservation: \frac{1}{2}mv_0^2=\frac{1}{2}m\dot{x}^2+\frac{1}{4\pi\epsilon_0}\frac{qQ}{x}, so I find \dot{x}=\sqrt{V_0^2-\frac{1}{2\pi\epsilon_0 m}\frac{qQ}{x}}.
Particle goes from infinity to x_0=\frac{qQ}{2\pi\epsilon_0mv_0^2} and returns out to infinity. I have integrated E_{rad} from x_0 to infinity and finally I could get the correct answer.
Thank you very much!

 

[TSny]

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.

 

[987oscar]

Ok. Iwill try that way too...
Thanks again