# Force on charge particle with constant velocity

1. Oct 12, 2014

### elsafo

I have a problem. Anyone can help me?
Two equal charges q move with equal velocity v. What is the force acting between two charges?
The distance between charges is R and the angle between R and v is A.

Last edited: Oct 12, 2014
2. Oct 12, 2014

### Orodruin

Staff Emeritus
Why would it be zero? What is the electromagnetic field associated with a moving point charge?

3. Oct 12, 2014

### elsafo

The electric force cancel each other and the magnetic field exerted by each charges are opposite but opposite direction. How?

4. Oct 12, 2014

### Orodruin

Staff Emeritus
I believe that what is sought is not the net force on the two-patricle system but the force on one of the particles induced by the other.

5. Oct 12, 2014

### elsafo

So how to solve it?

6. Oct 12, 2014

### Orodruin

Staff Emeritus
Start by trying to answer the second question of my first post.

7. Oct 12, 2014

### zoki85

8. Oct 12, 2014

### Philip Wood

I'll deal just with the case of A = $\frac{\pi}{2}$, that is the velocity is at right angles to the line joining the charges. The easiest way to do this is to start in the frame of reference in which the charges are stationary. The force between them in this frame is simply the ordinary Coulomb's law force. Now transform this force to the lab frame, in which the charges are moving. The force transformation is very easy as

transverse force = change in transverse momentum / time taken to change.

The change in transverse momentum is the same in both frames, as it is a Lorentz invariant. The time is dilated by the usual gamma factor in the lab frame, compared with that in the frame in which the charges are at rest. So the force between the charges in the lab frame is, in SI units
$$\frac{\sqrt{1 - \frac{v^2}{c^2}} Q^2}{4 \pi \epsilon_0 d^2}.$$
This reduction in the force can be interpreted as a magnetic (Ampère) attractive force coming into play in the opposite direction to the electric (Coulomb) repulsive force. But it's not as simple as that… As Zoki's equations show, the electric field is also changed. You'll note that Zoki's equations boil down to give the result I've derived above, in the special case of velocity at right angles to the line joining the charges. When the angle is not a right angle things get more complicated, but, again, you can either go for a relativistic force transformation approach (treating components parallel to, and transverse to, the velocity separately), or you can use Zoki's equations (having first derived them?)

Last edited: Oct 13, 2014