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If a charged particle approached another particle of same charge held by some external force, then by coulomb's law the particle will start electrostatic repulsion. Thus coming to a closest distance and stopping and thereafter, tracing its path back (the free particle). I know that energy is conserved here. The kinetic energy turning to Electrostatic Potential energy and thus back to its former state.

But my question is whether the momentum be conserved or not?

Edit:- Ignore Gravity, Radiation, Electromagnetic Field etc.

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  • $\begingroup$ Of course momentum is conserved. A core principle of any scattering theory. $\endgroup$
    – Jon Custer
    Aug 16, 2019 at 15:19
  • $\begingroup$ @JonCuster: I wouldn't be immediately dismissive; the role of momentum in electrodynamics is pretty subtle in some cases, particularly when radiation is taken into account. $\endgroup$ Aug 16, 2019 at 15:42
  • $\begingroup$ @MichaelSeifert - well, I didn’t see any hints in the Q that details like that were in play here. So, I’ll stick in a Rutherford-like context and say yes, momentum is conserved. $\endgroup$
    – Jon Custer
    Aug 16, 2019 at 15:49
  • $\begingroup$ Umm I would not consider taking radiation into account, especially for this one. I ll make an edit in the question. $\endgroup$
    – Asad Ahmad
    Aug 16, 2019 at 15:53

2 Answers 2

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If the stationary particle is "held by an external force", then the momentum of the system comprising the two particles will not be conserved. The momentum of a system is only conserved if there are no external forces acting on it.

It is true that the magnitude of the momentum must be the same before and after in this situation, since the kinetic energy of a particle can be written as $$ K = \frac{1}{2} m |\vec{v}|^2 = \frac{|\vec{p}|^2}{2m}. $$ So if the initial and final values of the kinetic energy are equal, then the value of $|\vec{p}|$ will be the same initially and finally as well. But the direction of $\vec{p}$ will not remain constant. For example, if the stationary particle is at the origin, and the moving particle approaches moving in the negative $x$-direction (i.e., approaching from the right), then after it bounces off of the stationary particle, it will be moving in the positive $x$-direction. Since the direction of $\vec{p}$ is not constant, this does not qualify as a situation where "the momentum of the system is conserved".

If the particle originally at rest is not held in place, then the momentum of "the system" will be conserved. However, what you mean by "the system" sometimes has to include electromagnetic field momentum. Effectively, whenever a charge accelerates, it can potentially create electromagnetic waves, and these waves have momentum (and energy) of their own. This is usually negligible if the acceleration of the particles is small, so if you bounce high-mass, low-charge particles off of each other you can typically ignore this effect; in which case the total momentum of the two particles before & after will be the same.

(As an aside, such approximations are usually implicit in the problems you encounter in introductory physics classes; the effects of radiation aren't typically covered until upper-level undergraduate courses or graduate-level courses. So if the above paragraph was gibberish to you, come back and revisit it in a few years.)

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  • $\begingroup$ Okay instead of external force, what if I say that it is connected to a spring i.e the stationary particle is now connected to a spring. Will now the momentum be conserved? Since the particle is free to move up to a distance where the electrostatic force equals the spring force. (Ofcourse the spring is connected to a wall) $\endgroup$
    – Asad Ahmad
    Aug 16, 2019 at 15:45
  • $\begingroup$ Also can you explain in your answer, how is the direction not same after it traces its path back? $\endgroup$
    – Asad Ahmad
    Aug 16, 2019 at 15:47
  • $\begingroup$ @AsadAhmad: Then the wall must exert a force on the spring to keep it in place. Or the wall could start moving, I suppose. See the comment in dmckee's answer about how you have to account for the momentum of the object providing the external force. $\endgroup$ Aug 16, 2019 at 15:47
  • $\begingroup$ Definitely the wall isn't gonna move. So we can say that since the spring exerts an external force on the particle attached to it, an external force comes into play and thus the momentum isn't conserved anymore. Am I right? $\endgroup$
    – Asad Ahmad
    Aug 16, 2019 at 15:51
  • $\begingroup$ @AsadAhmad: Yup, that's correct. $\endgroup$ Aug 16, 2019 at 15:54
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Momentum is definitely conserved, but “the system” you must consider includes:

  1. the incident particle
  2. the target particle
  3. the object providing the “external force”
  4. the electromagnetic field

I trust that you can understand why (3) is there, and (4) is there because accelerating charged bodies radiate and radiation carries momentum.

The need to deal with radiation makes the problem rather more difficult than the corresponding pure mechanics problem.

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  • $\begingroup$ Definitely the system is consisting 3 points. But as for the 4th one, we can just ignore it. $\endgroup$
    – Asad Ahmad
    Aug 16, 2019 at 15:55
  • $\begingroup$ Ignoring the radiation contribution of bulk neutral objects is generally safe, but you have two explicitly charged particles in the problem statement; so ignoring the radiation contribution depends on keeping the accelerations small enough. But without more information about the situation you can't guarantee it. $\endgroup$ Aug 16, 2019 at 15:58

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