Question on Calculating Coulomb force in VECTOR FORM

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
mhrob24
Messages
53
Reaction score
9
Homework Statement
Point charges q1 = 14 µC and q2 = −60 µC are fixed at r1 = (5.0î − 4.0ĵ) m and r2 = (9.0î + 7.5ĵ) m.

What is the force (in N) of q2 on q1?
(Express your answer in vector form.)
Relevant Equations
F = ke(Qq/r^2)
The only thing tripping me up here is that the answer needs to be in vector form. If the question was asking for the scalar form, then I would just find the distance between the charges (plot the charges according to their vector coordinates, then use pythagorean theorem to find the distance between the two. However, I believe (could be wrong here) that in order to get the vector form, I must calculate the force like I normally would for the scalar version, but then I would have to multiply the force by the unit vector representing the distance between the 2 charges. Here is my work done for this problem:

2ECFF85E-56D4-4911-829F-C590A3B71176 (1).jpeg
However, when I enter that vector at the bottom of my page into webassign (online HW platform), it is incorrect. I know the Force calculation is 100% correct so It has to be something with my unit vector, but I don't see the problem. The x-distance between the two charges is 4m and the y distance between the two charges is 11.5m right? So the vector representing the direct distance between the two charges is < 4.0, 11.5>. Then the unit vector is just this vector divided by its magnitude (sqrt(148.25)). I don't see where I'm going wrong...
 
on Phys.org
I haven't checked any of the arithmetic. But, without doing any calculation, you should be able to deduce the signs of the x and y components of the force.
 
  • Like
Likes   Reactions: mhrob24
Lmao. You're right. My signs of the components of the force vector were backwards. Q2 will attract Q1 so Q1 will accelerate in both the POSITIVE x and y directions. I had both components of the force vector as negative. I switched the signs and got the correct answer on Webassign. I spent like an hour trying to figure out what was going on :oldgrumpy: Thank you!
 
  • Like
Likes   Reactions: TSny
Yes. When you express the force vector in terms of a magnitude and a unit vector in the direction of the force, the magnitude part should be a positive number. When finding the magnitude of the force, just use the magnitudes (absolute values) of the charges in Coulomb's law.

1567794949413.png