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

## 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.

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?

## The Attempt at a Solution

#### Attachments

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Simon Bridge
$$\vec{F}=\frac{kQ}{r^3}q\vec{r}$$ Here r points from Q to q.