A charge of 3.40μC is held fixed at the origin. A second charge of 3.40μC is released from rest at the position (1.25m, 0.570m). If the mass of the second charge is 2.49 m, and its speed at infinity is 7.79 m/s, at what distance from the origin does the second charge attain half the speed it will have at infinity?
I really am unsure which equations to use. However, I think that, when the second charge attains half the speed it will have at infinity, half its energy will be kinetic and half its energy will be potential. (Right? No? Maybe?) So some relevant equations would be:
Ki + Ui = Kf + Uf
F = kQ/r^2
ΔK = 1/2mvf^2 - 1/2mvi^2
ΔV = qΔV
You are solving for r.
The Attempt at a Solution
My issue in physics always is that I'm unsure how to set up the problem, and I never know which equations to pick. Lately this has been compounded by the fact that I've lost class and study time due to severe migraines, so I've been teaching myself almost everything (and this part of physics is something I've been struggling to pick up on my own).
Anyway, my attempt at a solution was this:
1/2mvi^2 + qVi = 1/2mvf^2 + qVf
1/2mvi^2 + q(kq/ri^2) = 1/2mf^2 + q(kq/rf^2)
I plugged in the numbers given in the problem (using 1.37 for ri, calculated from the initial position of the second charge). I got an rf value of 5.37. Any help would be appreciated!