Electric Force: Charged Particles Homework Solution

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SUMMARY

The discussion focuses on solving a physics problem involving electric forces between charged particles, specifically charges q and 3q placed 90 meters apart. A third particle is positioned between them such that the total electric force on it is zero. The relevant equation used is Coulomb's Law, F(E)=(k)(q1)(q2)/(r^2). The participant initially misapplied the negative sign in the force equation but ultimately confirmed that the magnitudes of the forces should be set equal to find the correct position of the third particle.

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  • Familiarity with quadratic equations and their solutions
  • Basic knowledge of charge interactions and their directions
  • Ability to analyze and interpret physics problems involving forces
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TheExibo
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Homework Statement



Particles of charge q and 3q are placed 90m away from each other. A third particle is place in between them and the total electric force on this particle is zero. Determine the position of this particle.

Homework Equations



F(E)=(k)(q1)(q2)/(r^2)

The Attempt at a Solution



http://imgur.com/41Ngn8o
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After using the quadratic formula, I get stuck with two answers and I'm sure neigh ter of them are correct. Maybe the negative sign in the forces equation is unnecessary? Thanks
 
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TheExibo said:
After using the quadratic formula, I get stuck with two answers and I'm sure neigh ter of them are correct.
What answers did you get? Are both consistent with the constraints of the problem? Why do you think neither are correct?

TheExibo said:
Maybe the negative sign in the forces equation is unnecessary?
No, the negative sign was not correct, but you somehow managed to cancel it out. Looks to me like your final equation was correct. (Except that you left out a y in one of the terms; I assume that was just a typo.)

You know that the direction of the forces are opposite, so you just need to set the magnitudes equal.
 
Doc Al said:
What answers did you get? Are both consistent with the constraints of the problem? Why do you think neither are correct?No, the negative sign was not correct, but you somehow managed to cancel it out. Looks to me like your final equation was correct. (Except that you left out a y in one of the terms; I assume that was just a typo.)

You know that the direction of the forces are opposite, so you just need to set the magnitudes equal.

It seems the answers in the back weren't exactly correct. Thanks for checking my work!
 

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