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Electric Potential Energy, Point Charges and velocity

  • #1
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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.

U0.43m = k(-2.90x10^-6 C)(-7.2x10^-6 C)/(0.43m)

U0.43m = 0.437 J

My question at this point is why can I not just plug in my value for U0.43m (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:

U0.8m = k(-2.90x10^-6 C)(-7.20x10^-6 C)/(0.800m)
U0.8m = -.2349 J

I then subtracted U0.800m from U0.43m
= 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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Answers and Replies

  • #2
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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.
 

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