1. Aug 9, 2013

### benjamin198

1. The problem statement, all variables and given/known data

Derive the equation of motion of a particle of mass m and charge q moving in three dimensional space under a Coulombic attraction toward a fixed center.

I have the answer for a plane but a i need it for the space

2. Relevant equations

Mathematical Methods for Scientists and Engineers (Donald A. mcquarrie), chapter 20,.

3. The attempt at a solution

I did the exercise in polar coordinates

1. The problem statement, all variables and given/known data

2. Aug 9, 2013

### SteamKing

Staff Emeritus
The OP shows thumbnails of images, but you can't open the images themselves. We can't see the problem statement or your work.

3. Aug 9, 2013

### benjamin198

4. Aug 10, 2013

### Telemachus

You should use spherical coordinates. You don't have any constraints, so, just write the Langrangian, and apply Lagrange equations.

$L=T-V\\ x=r\sin \phi \cos\theta \\ y=r\sin \phi \sin \theta \\ z=r\cos \phi$
$T=\frac{1}{2}m \left[ \dot {x}^2 + \dot {y}^2+\dot {z}^2 \right]=\frac{1}{2}m \left[ \dot r^2 +r^2\sin^2 \phi\dot \theta^2+r^2\dot \phi^2 \right]$, $V=\frac{1}{4 \pi \epsilon_0}\frac{q}{r}$

You'll have three equations, one for each of the generalized coordinates $r,\theta,\phi$.

For example, r:

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot r}\right)-\frac{\partial L}{\partial r}=0 \\ \therefore m\ddot r-mr(\dot \phi^2+\sin^2\phi\dot\theta^2)+\frac{1}{4 \pi \epsilon_0}\frac{q}{r^2}=0$

(I didn't check if what I did is right, you should do it).

Last edited: Aug 10, 2013
5. Aug 14, 2013

### Dick

A charged particle moving a coulombic potential in space does move in a plane. The plane is defined by the center of attraction and the vectors corresponding to the initial displacement of the charge and the initial velocity of the charge. You don't have to resolve the dynamical problem. You just have to figure out how to transform the planar solution into 3d coordinates.

6. Aug 15, 2013

### Telemachus

Yes, that's because of the conservation of angular momentum, that you can derive from the equations of motion from the Lagrangian I gave (if it was set right, of course). As the force is central, there are no external torques, so the angular momentum is conserved. If you set an initial speed and an initial position, you can get from the vector product the direction on which the angular momentum goes, and define the plane of movement from it. So, it can actually be properly done on cylindrical coordinates instead of polar coordinates, but you can also derive all this from the equations of motions in spherical coordinates.