What is the Velocity of a Released Point Charge in Physics?

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The discussion centers on calculating the final velocity of a released point charge, specifically charge q2=+6nC, when it moves from a distance of 1m to 7m away from a fixed charge q1=+120nC. The correct final velocity is determined to be 0.0255 m/s. The initial approach using kinematic equations is flawed due to the variable acceleration caused by the electrostatic force, necessitating the use of integration or the Conservation of Energy principle for accurate results.

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ISITIEIW
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Hey guys, I didn't know where i could post this problem, but you guys seemed the best for the job :p

A point charge q1=+120nC, is fixed in space and can not move. A second point charge, q2=+6nC, is initially held at rest 1 m away from charge q1. Charge q2, is then released from rest. How fast will charge q2 be moving when it is 7m away from charge q1? The mass of q2 is 17g and you may ignore the effects of gravity.

Alright so the answer is 0.0255 m/s, but i can never get this,

I did that the sum of the forces in the x direction are, k being 8.85e-12, r being 1m , and m being 0.017kg (kq1q2)/(r)^2=ma

I can solve for a then substitute this into the kinematic equation vf^2 = vi^2 +2ad
where vi would be 0, and solve for vf, and d would be 6 m. But does not give the correct answer.

What am i doing wrong ?? Please help
Thanks a lot!

ISITIEIW
 
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The kinematics equation you are using to find the final velocity applies to constant acceleration only. You are going to need to integrate due to the changing acceleration with respect to position.
 
Or it might be easier to use Conservation of Energy.
 

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