Kinetic Energy of Point Charges

[cuppy]

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.

 

[chaoseverlasting]

You don't 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.

 

[m2003]

thank you.



Your approach is on the right track, but there are a few things to consider. First, you are correct in finding the electric field due to the point charge. However, since the charges are identical, they will experience the same electric field and therefore have the same electric potential energy (V=U/q). This means that the potential difference between the two charges will be zero, and therefore the change in potential energy will also be zero. This also means that the change in kinetic energy will be equal to the final kinetic energy.

Next, you are correct in using the equation V=-Ed to find the potential difference. However, this will only give us the potential difference at a single point, not the potential difference over a distance of 2.0m. To find the potential difference over a distance of 2.0m, we can use the equation V=Ed, where d is the distance between the two charges (2.0m in this case). This will give us the potential difference between the two charges, which we can then use to find the change in potential energy and the final kinetic energy.

Finally, you are correct in considering the interaction between the two charges. Since they are both positive, they will repel each other and therefore experience a force in the opposite direction of their motion. This force will do work on the charges, converting their potential energy into kinetic energy as they move towards each other. To find the force between the two charges, we can use the equation F=kq1q2/r^2, where q1 and q2 are the two charges and r is the distance between them. We can then use this force to find the work done on each charge, which will be equal to the change in kinetic energy.

To summarize, you can use the equation V=Ed to find the potential difference between the two charges over a distance of 2.0m. This potential difference can then be used to find the change in potential energy and the final kinetic energy. Additionally, you can use the equation F=kq1q2/r^2 to find the force between the two charges, which can then be used to find the work done on each charge and therefore the change in kinetic energy. Remember to take into account the fact that the charges have the same mass and therefore will have the same final kinetic energy. I hope this helps!