An alpha particle (the nucleus of a helium atom) is at rest at the origin of a Cartesian coordinate system. A proton is moving with a velocity of v towards the alpha particle in the xˆ direction. If the proton is initially far enough away to have no potential energy, how close does the proton get to the alpha particle? Your answer should be in terms of v, mp (mass of a proton) and e (charge of a proton).
Wext = [tex]- \Delta K[/tex]
[tex]\Delta U[/tex] = q(Vb-Va)
The Attempt at a Solution
I set the change in K equal to q(Vb - Va), with q being just e, and Va going to zero since the starting point has no potential. To find Vb I used the V for a point charge, (V = kq/r). From there I moved the variables around a bit, crossed out 2e on both sides, and got r = (mv^2)/k.
I wasn't sure if this was right because the distance is always changing. So I used the integral equation E dl, setting that equal to Vb, which was equal to the kinetic energy (I used gauss' law to find E as a function of r) and got the same answer. Something just seems wrong though. Also I definitely mixed up negative signs depending on the direction of the work. I am on the right track?