E&M fields of a moving point charge

[meteorologist1]

Hi, I need help on the following question: Suppose a point charge q is constrained to move along the x-axis. Show that the fields at points on the axis to the RIGHT of the charge are given by

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

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.

 

[AvroArrow]

The fields at points on the axis to the LEFT of the charge are given by \vec{E} = \frac{q}{4\pi\epsilon_0}\frac{1}{r^2}(\frac{c-v}{c+v})\hat{x} and \vec{B} = 0

 

[thepatientmental]

Hello! Thank you for reaching out for help with this question. The equations you have provided for the electric and magnetic fields of a moving point charge are correct. They are derived from the Lienard-Wiechert potentials, which describe the electromagnetic fields of a moving point charge in terms of its position, velocity, and acceleration.

To show that these equations are valid, we can use the method of images. This method involves creating a "mirror charge" that is equal in magnitude but opposite in sign to the original charge, placed at a distance r from the original charge on the opposite side of the x-axis. This creates an electric field that is equal and opposite to the original charge's field, canceling it out on the x-axis.

Using this method, we can see that the electric field on the axis to the LEFT of the charge is also given by the equation you provided, but with a negative sign. This makes sense, as the electric field should be pointing in the opposite direction on the left side of the charge compared to the right side.

As for the magnetic field, it is zero on both sides of the charge. This is because a moving point charge only produces a magnetic field when it is accelerating, and since it is constrained to move along the x-axis, there is no acceleration.

I hope this explanation helps you understand the fields of a moving point charge along the x-axis. If you have any further questions, please don't hesitate to ask. Good luck with your studies!