Does a Moving Charged Particle in a Dielectric Create a Magnetic Field?

[Ene Dene]

Homework Statement

Suppose that the entire region below the plane z=0 is filled with uniform linear dielectric material of permittivity epsilon. In t=0 a particle of mass m and charge e travels with velocity v in x direction and is parallel with the z=0 plane, and is distance d above the plane. I need to calculate where will the particle fall on to the z=0 plane.

The Attempt at a Solution

Now, I have calculated the force on a particle in t=0 due to bound charge in statical case. I also calculated the bound charge in statical case. Statical case would be if v=0.
What I'm interested in is, if the particle travels with velocity v in x direction, that means that bound charge configuration will follow the particle. In some sense it looks like surface current traveling at velocity v in x direction, right? So my guess is that it will produce a magnetic field in -y direction.
Now, if this is true, then I'm in trouble, I have no idea how to calculate that magnetic field.
Surface current would be:
$$\vec{K}=\sigma*\vec{v}$$
Magnetic field would be in such direction that it would affect the falling of a particle. This bound charge is the result of polarization, It's not really traveling but as particle travels it constantly changes it's configuration, so to me it looks like it's traveling, but does that produce a magnetic field? If so, how do I calculate it, I can't do it by Bio-Savart law since I can't get the B out of the integral.

 

[Jhero]

Thank you for your interesting question. I can offer some insights into your problem and provide a potential solution.

First, let's consider the situation where the particle is at rest (v=0). In this case, the bound charge distribution will be static and the electric field it produces will be directed downwards, towards the z=0 plane. This will create a force on the particle in the opposite direction, causing it to fall towards the plane.

Now, let's consider the situation where the particle is moving with velocity v in the x direction. As you correctly pointed out, this will cause the bound charge distribution to also move with the particle. This will result in a changing electric field, which in turn will produce a magnetic field according to Maxwell's equations. However, this magnetic field will not affect the falling of the particle, as it will be perpendicular to the velocity of the particle and thus will not exert any force on it.

To calculate the final position of the particle on the z=0 plane, you can use the equation for motion under constant acceleration: z(t) = z(0) + v(0)t + (1/2)at^2. In this case, the acceleration will be due to the electric force on the particle, which can be calculated using the electric field produced by the bound charge distribution.

I hope this helps you in your calculations. Keep up the good work in your studies of electromagnetism!