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Coulomb's law use

  • Thread starter SupremeV
  • Start date
  • #1
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1. The problem statement, all variables given/known data

A particle of charge 6 µC is held fixed while another particle of charge 8 µC is released
from rest at a distance of 1.4 m from the first particle. If the mass of the second particle is
3x10-6 kg, what is its speed when it is very far away from the first particle?


Homework Equations


F=ma
F= kQ1Q2 / r2
V2=Vo +2ad

The Attempt at a Solution



I pretty much worked through most the problem. In the end, I used coulomb's law in combination with newtons 2nd law to get a= 73469 m/s2. However what's bugging me is the distance that the speed is "very far away". I'm under the assumption that means at a point where the force from the first particle no longer affects the second. However, when I back-tracked using the answer(454m/s) the distance ends up being 1.4( the original distance) with the velocity equations.
 

Answers and Replies

  • #2
4
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Apologies, I searched google and found a similar problem: rocket is launched straight up from the earth's surface at a speed of 1.60×10^4 m/s.
What is its speed when it is very far away from the earth?

One of the helpers suggested conservation of energy, and I applied this to this problem it worked! Sorry!
 
  • #3
gneill
Mentor
20,801
2,778
You do realize that the acceleration is not constant? It decreases as the distance between the particles grows. So your V2 = Vo + 2ad formula is not valid over the trajectory of the second particle.

Why not try a conservation of energy approach?
 

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