One dimensional inverse square law - motion of charged particles

Question

I'm working on a simple case of two oppositely charged, equal-mass particles in a 1-dimensional system. It might as well be an electron and positron, but without the case of annihilation on contact. The actual system is two magnetic domain walls of opposite magnetic charge (think south and north pole) in a ferromagnetic nanowire.

I situate one particle at x = 0 and it remains fixed there. The second particle starts at x = +x_0 with some velocity v_0 directed away from the first particle in the positive x-direction.

The second particle feels an attractive Coulombic force from the first particle, proportional to some constant 'k' * 1/x^2. I would like an equation of motion for the position 's' of the second particle at a given time 't'.

My current attempt is: s = v_0*t + 1/2 ('k'/m * 1/s^2)*t^2

Due to the 1/s^2 term on the RHS I'm having trouble finding a usable solution, any help much appreciated! Thanks :)

Answer 1

The $\frac12 a t^2$ formula is for constant acceleration, which you don't have.

By integrating the Coulombic force for distances $s$ from any particular position $x$ out to infinity, you get $$ \int_{x}^\infty -\frac{k}{s^2} \,ds = -\frac{k}{x}, $$ which is your potential energy when the second particle is at $x.$ If the total energy of the particle is $E,$ the kinetic energy at position $x$ is $ E + \frac kx. $ We know that the kinetic energy at $x = x_0$ is $\frac 12 mv_0^2,$ hence $$ E + \frac k{x_0} = \frac 12 mv_0^2. $$ So the kinetic energy at any point is $$ \frac12 m\left(\frac{dx}{dt}\right)^2 = E + \frac kx \tag1$$ where $$ E = \frac 12 mv_0^2 - \frac k{x_0}. $$ From Equation $1$ we get $$ \frac{dt}{dx} = \frac{1}{\sqrt{\frac 2m\left( E + \frac kx \right) }}$$ when the second particle's $x$ coordinate is increasing, and therefore \begin{align} t &= \int \frac{dx}{\sqrt{\frac 2m\left( E + \frac kx \right) }} \\ &= \frac{\sqrt{mx (E x + k)}}{\sqrt2 E} - \frac{k\sqrt m \log(2 \sqrt{E x (E x + k)} + 2 E x + k)}{2\sqrt2 E^{3/2}} + \mathrm{constant} \end{align} where the constant is determined by the fact that $t=0$ when $x=x_0.$

I don't think there's a closed form solution for $x$ in terms of $t,$ but you could solve for it numerically given a value of $t,$ keeping in mind that if the total energy $E$ is negative there will also be times when the second particle's $x$ coordinate is decreasing.

And that's assuming the magnetic field does not cause the particle's path to curve, which is part of the problem setup that was not clear to me.