# Expressing force as a vector

1. Feb 3, 2009

### yoleven

1. The problem statement, all variables and given/known data
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.

3. 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?

2. Feb 3, 2009

### LowlyPion

Your vector coordinates are (-6,0) making the repulsive force -x.

I'd say then it's:

3. Feb 3, 2009

### yoleven

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. Feb 3, 2009

### LowlyPion

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