Motion of a point charge near an electric dipole

[hd90001]

I was hoping people might be able to provide some insight into the following fairly basic scenario (not homework/school related). Imagine a positive point charge on the x-axis. Also on the x-axis, to the right, is an electric dipole, with finite spacing, pointed along the x axis, with the positive charge on the right. This is a diagram of the situation:

-----(+)-------------------------(-)--------(+)----------------------------------

Imagine the dipole charges to be fixed. What happens to the velocity and potential energy of the test particle on the left when it is released from rest, at the following positions A-G?

-----(+)A-----------------------B(-)C--D--E(+)F----------------------G----------

Note: Distance A-D is the same as distance D-G.

Does the point charge have enough energy to make it beyond point D (or somewhere between D and E)? Is it reflected and does it oscillate around the negative charge before point D?

Now imagine the same scenario, but the particle has an initial positive velocity to the right. I believe that it needs some initial velocity to get beyond D, but I may be wrong. Imagine that it does have just enough velocity to get up and over point E. What is the velocity and energy at G compared to point A in this case? Will it continue accelerating in the positive x direction forever in this case?

Thanks a lot for your insight! If anybody has any links to helpful websites/animations/applets, that would be really great too!

 

[kuruman]

Since you are concerned about what happens when the test charge gets close to the other two, the dipole approximation is not valid and you might as well consider the two fixed charges as point charges. Now look at your drawing. The test charge will experience a net attractive force that gets larger as it moves because it is closer to the negative charge and farther away from the positive charge. Strictly speaking, when the point test charge is on top of the negative charge at zero distance from it, the attractive $1/r^2$ force will be infinite while the repulsive force from the positive charge will be finite. This means that the test charge will be stuck on the negative charge having fallen in an infinite potential well. Let's not worry about the singularity and assume that the test charge can zip through the negative charge, say because the negative charge is not a mathematical point, but has some finite extent.

In that case it will be possible for the test charge to reach the region between the charges. If the test charge starts at infinity with speed $v_0$, its total mechanical energy at any distance will be $ME=\frac{1}{2}mv_0^2$. At finite distance $x$, $(x<d)$ to the right of the negative charge, its potential energy will be $$U=-\frac{kq}{x}+\frac{kq}{d-x}=kq\frac{2x-d}{x(d-x)}$$
To calculate the speed $v$ at that point, we use mechanical energy conservation
$$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2+kq\frac{2x-d}{x(d-x)}$$The minimum initial speed required to reach point $x$ between the charges is obtained when you set $v=0$ in the above expression.
$$v_{0,min}=\left[\frac{2kq}{m}\times \frac{2x-d}{x(d-x)}\right]^{1/2}.$$
The expression says that if you want the test charge to reach the midpoint ($x=d/2)$ with zero speed, all you have to do is release it from rest at infinity. That's because the electric potential for this distribution is zero at both infinity and $x=d/2.$