# Distance between two point charges as a function of time.

1. Feb 17, 2010

### Chubigans

Edit: Apparently Coulomb's law only holds for electrostatics... in that case I understand that this problem is physically impossible. I'd still like to solve it if we assumed that coulomb's law did hold for moving charges

If there are two point charges that are initially separated by (x10-x20), which exert a force kq1q2/(x1-x2)^2 on each other, how can I find delta x (x1 - x2) as a function of time?

When I use f = mx'', I get kq1q2/(x1-x2)^2 = mx''

or if i introduce a new variable U = kq1q2/m, it's U/(x1-x2)^2 + x1'' = 0 for the first and U/(x1-x2)^2 - x2'' = 0 for the second force.

I don't know where to go from here. I want to solve for (x1-x2) given the initial conditions. I'm not well versed in differential equations, because I'm only now starting my first semester of differential equations, but I want to say that this is a linear second order partial differential equation? I feel like it might be not difficult to solve because of the symmetry.

Can it be solved? Because the question has been haunting me for months, I'm just so darn curious. Obviously if they are attractive, delta x will approach 0 at an incredibly fast acceleration. At what time point does delta x equal 0? Or will it just approach 0 forever?

I'm new here and this is my first post, but I thought that rather than go bother my differential equations professor I would try posting it here first!

Last edited: Feb 17, 2010
2. Feb 17, 2010

### torquil

You already have these (I haven't double-checked anything):

U/(x1-x2)^2 + x1'' = 0
U/(x1-x2)^2 - x2'' = 0

Just add the two equations, and defind d(t) = x2(t)-x1(t).

This gives you an quation

2U/d^2 = d''

which is a separable differential equation.

Torquil