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

[Violagirl]

Homework Statement

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

Homework Equations

F = k(q)(Q)/r²r-hat

The Attempt at a Solution

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

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

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

F = F_x + F_y = -kQ²2/(a)² x-hat + -kQ₁²/a² y-hat

Is this right? Or can more be done to solve it?

 

[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.