Expressing force as a vector

  • Thread starter yoleven
  • Start date
  • #1
78
0

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?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,097
5
Your vector coordinates are (-6,0) making the repulsive force -x.

I'd say then it's:
F = - (your answer) i + 0 j
 
  • #3
78
0
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?
 
  • #4
LowlyPion
Homework Helper
3,097
5
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
 

Related Threads on Expressing force as a vector

Replies
11
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
2
Views
1K
Replies
2
Views
927
  • Last Post
Replies
4
Views
17K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
8K
Replies
6
Views
3K
  • Last Post
Replies
3
Views
578
Top