- #1

meteorologist1

- 100

- 0

[tex] \vec{E} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}(\frac{c+v}{c-v})\hat{x} [/tex] and [tex] \vec{B} = 0 [/tex]

What are the fields on the axis to the LEFT of the charge?

I'm thinking that I need to use the two formulas for E and B which are derived from the Lienard-Wiechert potentials.

See here: http://scienceworld.wolfram.com/physics/PointCharge.html

Thanks.