Electrostatic Force?

Question

Here is the image of the picture given below:

http://img262.imageshack.us/img262/110/52459164bn7.jpg

The problem then wants to know the magnitude of the resulting force on the charge at the origin. It wants the answer in nN.

I worked through the problem and found that the force due to the top charge to be a magnitude of 5.86942 nN and the magnitude of the force from the right charge to be 6.99032 nN, resulting in a net force of 9.12769 nN. I was just wondering if these numbers looked right?

The next part of the question wants the know the direction of the resultant force. Here are the image choices and question.

http://img209.imageshack.us/img209/1862/60591657xs8.jpg

Just looking at it i thought it would be answer choice 1 since the top and the origin are attracted making a force pointing down, and the origin and the right repel making a force pointing to the right, so I thought the net force would be pointing down to the right. But I didn't seem to work.

I am confused, I thought since the top and the origin have opposite charges charges, the sign would be negative, since q+ * q- = -, thus j pointing down, and since the right is the same charge q- * q- = + so i would be pointing right?

Answer 1

For the first question, regarding the magnitude of the force:
Fy=k*(8 nC)*(-4 nC)/(49 m^2)
Fy=5.87 nN

Fx=k*(-7nC)*(-4nC)/(36m^2)
Fx=6.99 nN

Then F=sqrt(Fx^2 + Fy^2)
F=9.13 nN

So, you're correct.

For the second part, use the same technique, but with vectors, which preserve directions as well as magnitudes.
You will get the same magnitudes as in the first part, with directions in -i and +j.