Finding the Net Electric Force on a Charge

[Tekmachine]

Homework Statement

Three positive point charges are arranged in a triangular pattern in a plane, as shown below.
The Coulomb constant is 8.98755*10^9 N · m^2/C^2.

http://img150.imageshack.us/img150/1052/electricforceproblemqp7.th.jpg http://g.imageshack.us/thpix.php

Find the magnitude of the net electric force on the 6 nC charge. Answer in units of N.

Homework Equations

F = \frac{K(qQ)} {r^2}}

The Attempt at a Solution

Pythagorean Theorem: \sqrt{(6)^2 + (6)^2} = 8.485

\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(2*10^{-9}C)} {(8.485m)^2}} = 1.498*10^{-9} N

\frac{(8.98755*10^9 N · m^2/C^2)(6*10^{-9}C)(9*10^{-9}C)} {(8.485m)^2}} = 6.7411*10^{-9} N

sin(45)(1.498*10^{-9} N + 6.7411*10^{-9} N) = 5.8259*10^{-9} N

Unfortunately that's not the right answer

 

[buffordboy23]

You made a good start so far. The magnitude of the force on the 6 nC particle due to the other two particles is done correctly. Your next step is wrong.

Do the following: Draw the two force vectors on the 6 nC charge, and then determine their x- and y- components. Now add the components together to get the net force components on the 6 nC charge. Apply the Pythagorean theorem to these components to get the magnitude of the net force.

 

[Tekmachine]

Attempt #142,032:
F1 = cos(45)1.498*10^{-9} N x + sin(45)1.498*10^{-9} N y
F2 = -cos(45)6.741*10^{-9} N x + sin(45)6.741*10^{-9} N y
Fnet = 3.706*10^{-9} N x + 5.825*10^{-9} N y

||Fnet|| = 6.905*10^{-9} N

Thanks Guys!

 

[lamphere]

One of the y components should be negative, not one of the x components. The magnitude still comes out the same.