# Electric field generated by a point charge moving at the speed of light

As you see, this is the electric field generated by a point charge moving at constant speed v. I know that when $v$ -> 0, $E$ is just the Coloumb Law. But how do you interpret $E$ when $v$ -> $c$ ?

Can I just interpret it as the field of electromagnetic wave, because it moves at the speed of light?

• @BMS The question given to me is very vague. I don't think I need to do the Taylor expansion, right? – Lawerance May 20 '14 at 5:58
• The homework question asks about a limit, whereas the title of the question refers to a charge moving at c. These are two different things. It's not possible for a charge to move at exactly c. All charged particles have mass, and massive particles can't move at c. – Ben Crowell Jul 24 '14 at 18:30
• But it is easy to imagine e.g. a massless Dirac field with electric charge. – Robin Ekman Aug 28 '14 at 9:45

• When $v^2/c^2$ is small, the $\sin^2 \theta$ doesn't matter much, but when $v^2/c^2$ gets close to 1, the $\sin^2 \theta$ will become very important. You'll find that the field will be small ahead and behind of the particle and much larger along the sides, where $\sin^2 \theta$ is closer to 1. I'll leave the math to you, but the effect is that the fields get flattened in the plane perpendicular to the charge's travel. – krs013 May 20 '14 at 6:27
• I got your idea. Basically, because $\sin^2\theta$ is symmetric, we can just first consider 0-90 degrees. And the rest is just arguing the theta. But why this is somehow predicted as stated in the problem? Is it because of the contribution of retarded position of the charge? – Lawerance May 20 '14 at 7:33