# Energy radiated from a charge in electric field

1. Sep 17, 2014

### 987oscar

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 $v_0$. 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: $\frac{16}{45}\frac{q}{Q}\left(\frac{v_o}{c}\right)^3$

2. Relevant equations

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

3. 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!

2. Sep 17, 2014

### 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!]

Last edited: Sep 18, 2014
3. Sep 18, 2014

### 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!!!!

4. Sep 18, 2014

### 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.

5. Sep 18, 2014

### 987oscar

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

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