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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?
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