How Do You Convert a Scalar Force into Vector Notation in Physics?

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Homework Help Overview

The discussion revolves around converting a scalar force into vector notation in the context of electrostatics. The original poster presents a scenario involving a charge and a test charge, seeking to express the resulting force as a vector.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the force between two charges and seeks guidance on expressing this force in unit vector notation. Some participants provide insights on the direction of the force based on the positions of the charges, while others question the implications of changing the position of the test charge.

Discussion Status

The discussion is active, with participants exploring different interpretations of the force's direction based on the positions of the charges. There is an exchange of ideas regarding the implications of relocating the test charge and how it affects the calculation of the force vector.

Contextual Notes

Participants are considering the effects of different charge positions and the resulting vector directions, which may lead to confusion regarding the representation of forces in vector notation.

yoleven
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Homework Statement


I have a charge Q of +1.5*10^-7C located at co-ordinates (x,Y) (-6cm,0cm)
what is the force due to Q on a test charge of +1 C at the origin. Expressed as a vector.




The Attempt at a Solution


F= [tex]\frac{\kappa*Q1*Q2}{r^2}[/tex]

F= [tex]\frac{8.99*10^{9}*1.5*10^{-7}*1}{.06^{2}}[/tex]

F= 3.746*10 [tex]^{5}[/tex] N repulsive force.

How do I express this in [tex]\widehat{i}[/tex], [tex]\widehat{j}[/tex] unit vector notation?
 
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Your vector coordinates are (-6,0) making the repulsive force -x.

I'd say then it's:
F = - (your answer) i + 0 j
 
Okay, thanks. What if the test charge wasn't located at the origin but at (0,7) so that it made a right triangle with Q.
Could I "move" the test charge to the origin and calculate the x coordinate and then "move" Q to the origin and calculate the y coordinate?
Or is this wrong?
 
The Force at the origin (0,0) is what you wanted isn't it.

It's repulsive from the (-6,0) charge, so then at the origin it will be +x directed, not -x directed as it is on the charge at (-6,0).

So in that case the repulsive force is F = +(your answer) i + 0 j
 

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