Kinetic/potential energy of charges as they approach each other

[brazen2]

Homework Statement

A charge of -4uC is fixed in place. From a horizontal distance of 55 cm, a particle of mass 0.0025kg and charge -3uC is fired with an initial speed of 15m/s directly towards the fixed charge. How far does the particle travel before it stops and begins to return back?

Homework Equations

KE=1/2mv^2
PE=qV
V=KQ/r

The Attempt at a Solution

KE=1/2mv^2
KE=1/2(0.0025kg )(15m/s)^2
KE= 0.28125
KE will equal PE when particle stops moving so...
PE=0.28125=qV=(-3e6)(9e9)(-4e6/r)
r= 0.384 = distance where particle stops moving

Is this correct? I don't want to use F= KQ1Q2/r^2 and F=ma for this question, as that would only give me the initial acceleration of the particle... Am I totally off track??
Thanks everyone

 

[Dick]

You are on the right track. But the equation you are looking for isn't KE=PE. It's KE_i+PE_i=KE_f+PE_f where the subscripts i and f refer the initial and final. Yes, KE_f=0. But neither PE_i nor PE_f is zero if you are using q*k/r for the potential.

 

[Defennder]

You should not calculate the potential at the point where it is equivalent to the initial KE of the charged particle. Otherwise this would imply that a charged particle fired at the stationary charge at that point with the same initial speed would not move at all. Rather, find the work done against the field when the fired particle moves from its origin to the point where it stops.