# Time it takes two oppositely charged particles to collide

I think this is supposed to be a simple problem but I am having a hang up converting it to a one-body problem. It's one-dimensional. +q and -q a distance d apart, held stationary then let go at t=0. The potential is V(x)=kq^2/x. If I turn it into a one body problem, then m-->m/2, but how do i interpret the new x? Both particles are moving toward each other, so they travel a distance d/2 before colliding. I am guessing the relevant equation will be $t = {\sqrt{\frac{m}{2}}} \int \frac{dr}{\sqrt{E - V(x)}}$

What concepts am I lacking? I think this is supposed to be really easy, but it's not for me.

edit, so x is now the relative distance between the two particles so it should be like one particle traveling the whole distance d ? I get a negative value, but is that acceptable? Something like

$t=\frac{\sqrt{m}}{2} \int_d^0 \sqrt{\frac{d}{kq^2}} \frac{dx}{\sqrt{1-d/x}}$ And that isn't giving me a very good answer when I calculate it.

• I do hope your calculations would take into account the radiation reaction self force.
– guru
Commented Sep 28, 2013 at 22:26
• They won't collide. They will, at large times, exponentially repel each other. Commented Sep 28, 2013 at 22:28
• I found this physics.stackexchange.com/questions/54655/… , does this explain the situation better? Commented Sep 28, 2013 at 22:34

In a 1D problem, you can write $$t_f-t_i=\int_{x_i}^{x_f}\frac{\text dx}v,\ \text{where}\ v^2=\frac2m(E-V(x)).$$ In the present case, though, with $x$ decreasing as $t$ increases, you need to take the negative sign for $v=\cfrac{\text dx}{\text dt}$. Thus if the particles are a distance $d$ at time zero and collide at time $T$, you should write $$T=-\int_d^0\frac{\text dx}{\sqrt{\tfrac2m(E-V(x))}}>0.$$ Easy!
• I looked the integral up wolfram and got a logarithm for an answer, $t= \frac{\pi d} {q} \sqrt{2\pi\epsilon_0md}$ should be the right answer, not sure how to get there though. Commented Sep 28, 2013 at 22:48
• Keep your eye on the signs: $V(x)<E<0$. Commented Sep 28, 2013 at 23:01