Four point charges, each of magnitude 5.5[tex]\mu[/tex]C, are placed at the corners of a square 76.5 cm on a side. The value of Coulomb's constant is 8.9875 x 10^9 Nm^2/C^2. If three of the charges are positive and one is negative, find the magnitude of the force experienced by the negative charge. Answer in units of N. First I drew a free body diagram and found that the negative charge would be attracted to all of the other charges. So then I used Coulomb's Law. F_14= [tex] 1/4\pi\epsilon_o * (5.5 x 10^-6 *5.5 x 10^-6)/ .765^2. [/tex] I know that E_o = 8.85 x 10^-12, by setting 8.9875 x 10^9= [tex] 1/4\pi \epsilon_o. [/tex] Solving this gave me 4.58 x 10^-22. This would be the force for 2 of the charges, since all of the charges and distances are equal, except one force is in the x-direction, and the other is in the y-direction. Next I solved for the other force, which would be the diagonal from the negative charge. I found the distance between the two charges to be 1.08 m by using the pythagorean theorem. I split it up into x and y components and used Coulomb's Law. In the x-direction: F= [tex] 1/4\pi \epsilon_0 * (5.5 x 10^-6 * 5.5 x 10^-6) / 1.08^2 * (.765/1.08). [/tex] and I got 1.83 x 10^-11. Since it is a square, I got the y-direction to be the exact same thing. Then I added together the components for both parts to get 1.83 x 10^-11 in both directions. Then I did the [tex] \sqrt (1.83 x 10^-11)^2 +(1.83 x 10^-11)^2 [/tex] for my final force and found it to be 2.59 x 10^-11, which isn't right. Can someone help me? thanks in advance.