# Two moving charges

Gold Member

## Main Question or Discussion Point

Here's something that i've been thinking about a while and haven't found a good answer to:
Lets say that there're two charges on the x-axis seperated by a certain distance and they're moving upwards along the y-axis. Lets look at charge A (on the right side), when it's at height y2 it's actually feeling the force of charge B from the retarded position y1 which is under y2 because it takes time for the E field of B to get to A(the distance between y1 and y2 is proportional to the distance that light goes in the time that it takes the charge to go from y1 to y2).
The force on A at y2 from the charge B at y1 is up and to the right because the field goes out radially from the position of the charge. By the same argument, the force on B from A is up and to the left. so it looks like the charges, one given a little initial speed, will just keep on going faster and faster untill the magnetic force pulls them together and the distance between them is 0. so in the end the speed along the y-axis will be bigger than in the beginning. how is momentum conserved?
Thanks.

Related Classical Physics News on Phys.org
I don't think your argument has used the correct field of a moving charge.

Gold Member
Isn't the field of the moving charge the same as a charge at rest just with a stronger field on the sides than in the direction of motion?

And the direction of the field lines?

Gold Member
they are still radial just they're more "bunched up" on the sides than in the direction of motion - but the important thing for the argument is that they spread out - they don't have to be strictly radial. at least i don't think so?
what do you think that the answer is?
And i thought of another strange thing, it can't be that moving charges apply forces to eachother in the direction of motion because it's obvious that in a frame where the charges are at rest there won't be a force in that direction!
so now i'm even more confused.

Isn't the field of the moving charge the same as a charge at rest just with a stronger field on the sides than in the direction of motion?
Isn't the electromagnetic force $E + v \times B$? So the moving charge creates additional force in a direction perpendicular to the direction of motion rather than all directions perpendicular to the direction of motion. That particular direction is "selected" by the magnetic field at the given point.

...when it's at height y2 it's actually feeling the force of charge B from the retarded position y1 which is under y2 ...
... it's obvious that in a frame where the charges are at rest there won't be a force in that direction!...
What if I said the field of a moving charge actually points in the direction of where the charge is "now" expected to be?

Daniel,

Absolutely speaking in a Maxwellian Basis:
F> = q E>+ q v> x B> + Fself> , where Fself> is a drag force on the moving charge that depends upon the acceleration vector.

If we are talking about description and understanding, then I can say:
This can be solved in the frame of the two charges, if they were initially moving with the same velocity and in the same line perp to the velocity vector.

Another way: As long as speeds are small rationed to c, then they at last be apart (assumed to be equal in magnitude and sign), due to the work done by the initial potential force. Magnetic field is very small with respect to electric field at that case.

You are right:a charge is being affect through force of the electromagnetic field at that place at that time, which must have transmitted through speed of light.

Yours,
Amr Morsi.

how is momentum conserved?
I think the description of the problem is correct. Most of the anwers were taking de Lorentz force for granted... $$\vec F = q (\vec E + \vec v \times \vec B )$$.
Your problem is that you are considering momemtum to be just $$m \vec v$$.
Remember that the EM field carries momentum as well... ;)

Last edited:
Meir Achuz