What is the distance to the nucleus when the particle's velocity is zero?

[photon_mass]

potential = 0
particle of mass m charge q is projected with kinetic energy K at a nucleus mass M charge Q that is at rest. it is shot with 'perfect aim' (along the x axis).
find the distance x to the nucleus when dx/dt of particle is zero.

i know that F = dK/dt = cqQ/x^2 where c = 1/4pi(epsillon naught)

.5m(d/dt)v^2 = cqQ/x^2 (1)

the coulomb force is position dependent force, so:

.5m(dv^2/dx)(dx/dt) = cqQ/x^2 (2)

.5m(dv^3) = cqQ(dx/x^2) (3)

this doesn't make any sense to me. it makes more sense to take the time derivative in (1) just giving Newtons 2nd, and then doing the position dependence.

ma = cqQ/x^2

mvdv = cqQ(dx/x^2)
.5mv^2 = -cqQ/x

and then ?

i got nothing. any suggestions, clarifications or anything of the like would be much appreciated.

 

[maverick280857]

The distance of closest approach is attained when the kinetic energy of the particle wholly equals the electrostatic potential energy of the system constituted by the nucleus and particle. If you are unclear about the calculations and the sense they ought to make, look this up: Distance of Closest Approach.

 

[photon_mass]

oh geez. i read the question wrong.
thank you.