# How to approximate trajectories and movement of two oppositely charged particles?

Imagine a single, stationary charged atomic ion, say a Lithium anion or cation (Li+ or Li-). Now imagine another a single free, oppositely charged particle--perhaps an electron or Hydrogen ion (H+)--passing by the first stationary atomic ion at a "classical" non-relativistic speed. For simplicity, imagine they are both in a vacuum and both in zero gravity and free from outside electrical noise or other forces.

What is a good equation to use to calculate the trajectories and/or movement of both of these particles, knowing the velocities and masses of each?

UPDATE:

While Ana V's answer below is very good, I'm really looking for an actual approximate answer of how close the two particles need to be to have measurable movement based on the forces. I'm not looking for a high level of accuracy. Just trying to get a sense of the scale of impact on the trajectory of the moving particle and the movement (if any) of the stationary particle as you vary the closest distance between them. Would they need to be very close to each other (say, less than a micron)? Or would the stationary ion still exert enough force at macro-scale distances (say, a meter?) to measurably change the moving particle's trajectory and/or push away the stationary ion? How close together do they need to be for the electrostatic forces to move one or both of them measurably? Thank you!

• Define "little" – Jimmy360 May 30 '15 at 4:25
• Come on, really? Small can mean different things. Is small a deviation of Planck length or a nanometer? – Jimmy360 May 30 '15 at 4:48
• Depends on the energy of the electron. – CuriousOne May 30 '15 at 4:48
• @CuriousOne in what way? if the energy is higher, is the repulsive force higher or lower? – Brad Cooper - Purpose Nation May 30 '15 at 4:54
• The force is the same, the momentum transfer will be smaller for a higher energy electron. You are basically looking at a Kepler problem if you are happy with the classical description of the motion. – CuriousOne May 30 '15 at 4:57

This may help your, it comes from Rutherford scattering by which they determined that the atom has a hard core. It is positive alphas against positive nucleus, but the math is the same. Determining the closest approach to the nucleus amounts to calculating the minimum distance for the hyperbolic orbit which is produced by the coulomb repulsive force. The expression for the closest approach as a function of the impact parameter b is given by: Which has a limit for back-scattering which is just determined by setting the initial kinetic energy equal to the final potential energy. This corresponds to stopping the projectile and sending it backward, so at its closest approach, all the energy is in electric potential energy. If you are truly interested enter the values in the calculating entries of the link and get hard numbers.

edit after edit

One has to realize that the Coulomb potential goes to infinity and a unique number cannot be given for a quantity that has a functional dependence. In this case the momentum and the angle of scattering have to be defined to determine r_min, which makes for a complicated algebraically analysis dependent on the specific masses and momenta involved .

• Anna V: Thank you, perfect. This formula applies to the projectile, so I wonder if there is equal and opposite formula for the target? With Rutherford experiment, the target is fixed. In mine, the target is free (like a Hydride gas, I suppose). With both the projectile particle and the target being free, assume the formula would also include the target and it's mass/momentum (it is stationary in my example). – Brad Cooper - Purpose Nation May 30 '15 at 13:01
• Interestingly enough, Rutherford derived the scattering expected from a generic inverse square law. It applies equally well to repulsive or attractive forces, although the trajectories are of course different for the two. For example, the cross section and scattering angles for comets and the sun are described by the generic Rutherford formula. And, no, Rutherford's math does not 'fix' the nuclei - that is why the center of mass term is in there. – Jon Custer May 30 '15 at 15:22
• Jon Custer: Thank you. What I mean by "fixing" the nuclei is that Rutherford was not interested in any movement of the target atoms -- in your astronomical example, how much the Sun is moved by the passing comet. Guessing a comet cant move the sun much more than a He+ ion hitting gold atom? But that is the 2nd part of my question--how do determine how much the stationary ion is moved by the passing particle. – Brad Cooper - Purpose Nation May 30 '15 at 17:53
• you have to do a kinematic transformation. Define the ion initially at rest, laboratory frame, and see the changes in time. farside.ph.utexas.edu/teaching/336k/Newtonhtml/node52.html – anna v May 30 '15 at 18:17
• @PurposeNation You can view this from the center of mass of both particles, such that the other particle will would trace the mirrored trajectory of the first particle but scaled by the mass ratio (such that the center of mass would not move). – fibonatic Jun 6 '15 at 4:34