Kinetic Energy of Point Charges

  • Thread starter cuppy
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  • #1
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Identical 8-μc point charges are positioned on the x axis at x=+/-1.0m and released from rest simultaneously. What is the kinetic energy of either of the charges after it has moved 2.0m?

some relevant formulas are:
E=kq/r^2
V=-Ed
V=U/q

my attempt at the question:
i started by finding the electric field due to the point charge, E=kq/r^2=71920V/m??? since i took 1.0m as the distance from the origin.
i'm confused about the next few steps. would it be right to use V=-Ed (with d as 2.0m) to find the potential difference then use V=U/q to get the change in potential energy and finally equate this with the change in kinetic energy?

since the charges were initially at rest the initial KE was zero, the change in kinetic energy would just be the final kinetic energy. but i don't know how to to approach the question since there are two identical charges. would that mean i have to take into account their interaction and actually find the force between them and somehow relate that to kinetic energy?

i think i'm completely off the mark so any hints would be great.
 

Answers and Replies

  • #2
You dont need to find the electric field here. You are right, you can find the change in potential energy and equate it to change in kinetic energy here.

The potential due to the charges is a scalar quantity and so is potential energy, so, you can just add them up without worrying about the vector field.
 

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