# Change in electric field of charged particle

1. Jan 21, 2010

### arul_k

It is known that the motion of a charged particle creats a magnetic field and also at speeds approaching c there will be an increase in its relativistic mass, but what effect does motion of a charged particle have on the magnitude of its electric field (Couloumb force). From what I know at velocities much less than c there appears to be no effect on the magnitude of the electric field but as its velocity approaches c the entire electric field converts to a magnetic field. Is this correct?

2. Jan 21, 2010

### tiny-tim

Hi arul_k!
Not quite …

the magnetic field becomes stronger, but E2 - B2 is a constant, and since it starts positive, it stays there, and |E| is always greater than |B|.

(also, E.B is constant, and it starts zero, so E and B are always perpendicular)

3. Jan 21, 2010

### Bob S

The transverse electric field Eperp is multiplied by a factor γ as the velocity approaches c. The longitudinal component of E is not changed. There is also a transverse magnetic field. See the last four Lorentz-transformation equations in

http://pdg.lbl.gov/2009/reviews/rpp2009-rev-electromag-relations.pdf

Bob S

4. Jan 21, 2010

### GRDixon

You might find it helpful to index "the fields of a point charge moving with constant velocity" (or something to that effect) in an EM text or two. Yes, assuming the charge's velocity is constant, E differs little from the electrostatic field when v<<c. At speeds approaching c, E increases "to the sides" of the moving charge. And of course there is a magnetic field when the charge moves. As v approaches c, E and B "to the sides" approach infinity! If you own or can borrow a copy of "Introduction to Electrodynamics", 2nd Edition, David J. Griffiths, then the exact formula for E is provided in Eq. 10.103.

5. Jan 22, 2010

### arul_k

Thanks for the replies. Would it be right to say that the magnitude of E has no relation to the formation of the magnetic field of a moving charged particle i.e the magnitude of E and B are independent of each other.

6. Jan 22, 2010

### GRDixon

No, that would not be right. For a charge moving with constant velocity, B is related to E by B = v X E / c^2, where v is the charge's velocity.

7. Jan 22, 2010

### arul_k

Right, I didn't quiet word my question correctly, what I wished to ask was does the creation of the magnetic field result in a change in the magnitude of the electric field of the charged particle.

8. Jan 22, 2010

### arul_k

9. Jan 22, 2010

### Bob S

If you look at the Lorentz transformation of fields in the LBL PDG website (see last fourl ines):
http://pdg.lbl.gov/2009/reviews/rpp2009-rev-electromag-relations.pdf
you will see (from the second of the four equations) that the transverse electric field is increased by a factor γ, while the longitudinal field is unchanged. What is not shown is that the relativistic contraction of longitudinal length causes angles to increase, and this in turn compresses the transverse electric field from isotropic (for a stationary charge) to a purely transverse electric field with an opening angle of 1/γ.
For a particle traveling in the direction x, an angle θ transforms as

tan θ' = tan y'/x'= tan γθ = tan γy/x

Thus at very relativistic velocities, the electric field is a flat disc normal to the velocity.

Bob S.

[added] Although it may be counter-intuitive, a relativistic charged particle does not "drag" the electric field like a shock wave. The peak electric field is at right angles to the velocity, as seen above. If the electric field were dragged, then the Poynting vector would imply that the charged particle were losing (or radiating) energy, which is not the case with a constant velocity particle in free space. In the case of a relativistic charged particle inside a vacuum tube, there are image currents in the vacuum conducting tube wall. If the vacuum tube wall is resistive, there is power loss in the vacuum tube wall, and the electric field is dragged, meaning that the Poynting vector is pointing outward (as it must)

If there is no magnetic field B in the unprimed system, then the electric field is exactly as I described above. The electric field in the unprimed system does produce a transverse magnetic field in the primed system, but that was not part of your question.

Last edited: Jan 22, 2010
10. Jan 22, 2010

### tiny-tim

Yes …

the strength of the E field is greater than that of the B field,

but the forces are qE and qvxB/c, where v is the velocity of the particle, and c is the speed of light …

so the force generated by the E field depends only on the charge of the particle, but the force generated by the B field depends also on the speed and direction of the particle, but since v/c must be < 1, |qE| is always greater than |qvxB/c|.