Electric Potential Energy, Point Charges and velocity

[Northbysouth]

Homework Statement

A small metal sphere, carrying a net charge of q₁ = -2.90 μC , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = -7.20 μC and mass 1.70g , is projected toward q₁. When the two spheres are 0.800 m apart, q₂ is moving toward q₁ with speed 22.0 m/s as in figure 1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.

I have attached an image of the question

Homework Equations

1/2mv^2 = kinetic energy
u = kq₁q₂/r

The Attempt at a Solution

As my attached image shows, I have gotten the correct answer fort his question but I'm not sure how I did it.

So I began by calculating the electric potential energy (U) when q₂ was 0.43m away from q₁.

U_0.43m = k(-2.90x10^-6 C)(-7.2x10^-6 C)/(0.43m)

U_0.43m = 0.437 J

My question at this point is why can I not just plug in my value for U_0.43m (0.437 J) into the kinetic energy formula?

If I do this I get:

0.437 J = 1/2(1.70/1000)(v^2)
v = 22.67 m/s

Which, I'll admit, does not make sense because this says that the velocity of q₂ has increased when it should decrease because q₁ and q₂ have same charges, thus they should repel each other, thereby slowing down q₂.

I then calculated the electric potential energy (U) when q₂ is 0.8 m from q₁:

U_0.8m = k(-2.90x10^-6 C)(-7.20x10^-6 C)/(0.800m)
U_0.8m = -.2349 J

I then subtracted U_0.800m from U_0.43m
= 0.437 J - 0.2349J
= 0.2026 J

I then plugged this ΔU into the kinetic energy formula
1/2mv^2 = 0.2026J
v = 15.7 m/s

Why was it necessary to find the electric potential energy of the two distances?

 

[voko]

This is because the total energy is conserved. PE + KE = constant. So if the system has PE1 at some point, and PE2 at some other point, then PE1 + KE1 = PE2 + KE2, hence PE1 - PE2 = KE2 - KE1.