# Four Charges

## 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?

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.