Attraction of -Q to +Q in x and y directions

1. Oct 14, 2013

Violagirl

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

Find the magnitude and direction of the force on the charge -Q on the figure in the attached document.

2. Relevant equations

F = k(q)(Q)/r2r-hat

3. The attempt at a solution

So I have a force in the x direction and a force in the y direction.

Fx = k(Q1)(-Q)/(a) x-hat

Fy = (k)(Q2)(-Q)/(a) y-hat

F = Fx + Fy = -kQ22/(a)2 x-hat + -kQ12/a2 y-hat

Is this right? Or can more be done to solve it?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Attached Files:

• Attraction of -Q to +Q in +x and +y directions.jpg
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2. Oct 14, 2013

Simon Bridge

Almost there - check the directions of the forces.
Your formula has the -Q charge being pushed in the -x direction.
Is that what would happen?

In general - the force experienced by charge q due to charge Q is:
$$\vec{F}=\frac{kQ}{r^3}q\vec{r}$$ Here r points from Q to q.

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