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

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To convert a scalar force into vector notation, the force due to charge Q at coordinates (-6 cm, 0 cm) on a test charge at the origin is calculated using Coulomb's law, yielding a repulsive force of approximately 3.746 x 10^5 N. This force can be expressed in unit vector notation as F = - (3.746 x 10^5) i + 0 j, indicating a direction towards the positive x-axis. If the test charge is moved to (0, 7), the approach involves calculating the force components separately for the x and y coordinates. The repulsive nature of the force means it will direct away from the charge at (-6, 0), resulting in a positive x-direction at the origin. Thus, the correct vector notation for the force at the origin remains F = + (3.746 x 10^5) i + 0 j.
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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= \frac{\kappa*Q1*Q2}{r^2}

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

F= 3.746*10 ^{5} N repulsive force.

How do I express this in \widehat{i}, \widehat{j} 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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