Where Should a +2.5Q Charge Be Placed to Nullify the Force?

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SUMMARY

The discussion centers on determining the position of a +2.5Q charge necessary to nullify the force exerted on another charge in a physics problem involving electrostatics. The initial force on the charge is given as \(\vec{F_{res}} = Qkq (-0.222 \vec{i} - 0.250 \vec{j})\) N. Participants suggest using Coulomb's law and vector analysis to find the appropriate position vector for the +2.5Q charge, ultimately leading to the conclusion that the charge should be placed at a distance of approximately 2.7345 meters along the vector direction of \(-1.8157 \vec{i} - 2.0446 \vec{j}\).

PREREQUISITES
  • Understanding of Coulomb's Law and electrostatic forces
  • Familiarity with vector notation and operations
  • Knowledge of unit vectors and their properties
  • Basic skills in algebra and solving equations
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  • Study the application of Coulomb's Law in electrostatics problems
  • Learn how to decompose vectors into their components
  • Explore the concept of unit vectors and their significance in physics
  • Practice solving electrostatic force problems involving multiple charges
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This discussion is beneficial for physics students, particularly those studying electrostatics, as well as educators seeking to clarify concepts related to forces and charge interactions.

  • #31
Jsut take what you have, and split it up:
( 0,222 \vec{i} + 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} x \vec{i} + y \vec{j}

becomes
( 0,222 \vec{i}) = \frac{2,5}{ x^2 + y^2} x \vec{i}
and
( 0,250\vec{j}) = \frac{2,5}{ x^2 + y^2} y \vec{j}

Then its basically 2 equations with 2 unknowns.

I would then say solve the top one for y^2 and then plug it into the bottom. Its a lot of algebra from there but it would work. I honestly think splitting up vectors into the two parts gets rid of the point of even having vectors. But it DOES make things easier to computer in classes with dynamics and forces acting on objects. Sum the forces in each direction, set equal to m*a, youll see that a lot.

But you should always make sure you feel comfortable using something like full vector notation before you toss it aside for an easier way.
 
Last edited:
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  • #32
When I solve for x and y , I get x= (+-) 3,16 and y= 1,27 ...
Anyway, I already understood your first method, it's a lot more easier.. but I don't know why it never works like you have written.
Thanks a lot for all your time and help, Healey01 :)
 

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