- #1

- 249

- 2

## Homework Statement

A small metal sphere, carrying a net charge of q

_{1}= -2.90 μC , is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q

_{2}= -7.20 μC and mass 1.70g , is projected toward q

_{1}. When the two spheres are 0.800 m apart, q

_{2}is moving toward q

_{1}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 = kq

_{1}q

_{2}/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 q

_{2}was 0.43m away from q

_{1}.

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 q

_{2}has increased when it should decrease because q

_{1}and q

_{2}have same charges, thus they should repel each other, thereby slowing down q

_{2}.

I then calculated the electric potential energy (U) when q

_{2}is 0.8 m from q

_{1}:

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?