Calculating Net Force on a Charge at the Center of a Square

[Liketothink]

Homework Statement

Four charges Aq,Bq,Cq, and Dq (q = 1.00 × 10-7C) sit in a plane at the corners of a square whose sides have length d = 85.0 cm, as shown in the diagram below. A charge, Eq, is placed at the origin at the center of the square.

DATA: A = 6, B = 2, C = 3, D = 6, E = -1. Consider the charge at the center of the square. What is the net force, in the x-direction, on this charge?

Homework Equations

F=kq1q2/r^2

The Attempt at a Solution

F1=(9*10^9)(6*10^-7(-1*10^-7)/(.601^2)-because it is a square, I took half the distance of the square and found the hypotenuse. Then I used that for the distance between the charges.
So,
F2=(9*10^9)(6*10^-7(-1*10^-7)/(.601^2)-
F3=(9*10^9)(3*10^-7(-1*10^-7)/(.601^2)-
F4=(9*10^9)(3*10^-7(-1*10^-7)/(.601^2)-
Then I multiplied each one by cos(45) and found the x-components
-.000528, -.000352, -.001057, -.001057
Then I added all of them together
so total Frx=-.002995 N
But that's wrong. I don't know what I did wrong. Thank you for helping me.

 

[mustang1988]

Im having the same problem can anyone help except the problem wants you to find the total charge on a corner and can't figure out how to find the diagonal

 

[fantispug]

Liketothink: Think about the geometry of the problem. Two charges are on the left of the centre and two are on the right, moreover all the charges are attracting the centre charge. Draw a picture and think carefully about the signs/directions of each force.

(Or think this way: is cos(135)=cos(45)?)

mustang1988: I don't understand what you're asking. Find the total CHARGE on a corner? You're given all the charges. Find what about the diagonal? It's length can be found by Pythagoras' theorem.