Calculating force on the charges placed in the corners of a triangle

[mmoadi]

Homework Statement

We place three charges in the corners of the equilateral triangle; the size of the sides is 1 m. Charges are of sizes: e1, e2 = −2e1, and e3 = 3e1, where e1 = 10 ^-6 As. What is the force on the first (second, third) charge and in which direction are they showing?

http://item.slide.com/r/1/20/i/uNrFtWB4tT_sS7zkuV1ExJS_gf3980kJ/
http://item.slide.com/r/1/130/i/1G12EjLq5D8Lo4lU8yYKxmOQfVTQfoSl/
http://item.slide.com/r/1/142/i/dpHAZtcg7j-o-U_uDNFiH60YH-Do8Dwr/

Homework Equations

F= k[q(1)q(2)/ d²]

The Attempt at a Solution

Calculating for e(1):
d= 1 m

F(1,2)= k[q(1)q(2)/ d²] → F(1,2)= 1.8 N

F(1,2-y)= -F(1,2) sin 60º= - 1.56 N
F(1,2-x)= -F(1,2) cos 60º= - 0.4 N

F(1,3)= k[q(1)q(3)/ d²] → F(1,3)= 2.7 N

F(1,3-y)= f(1,3) sin 60º= 2.34 N
F(1,3-x)= -F(1,3) cos 60º= - 1.35 N

F(1-x)= -2.25 N
F(1-y)= F(1,2-y) + F(1,3-y)= 0.78 N

F(1)= sqrt[F(1-x)² + F(1-y)²]= 2.39 N

Are my calculations for e(1) correct?
Thank you for helping!

 

[denverdoc]

Have't looked at them all but cos 60=0.5 so not sure where F(1,2-x) came from.

 

[mmoadi]

Homework Statement

F(1,2-x)= -F(1,2) cos 60º= - 0.4 N

<sub>My mistake, it was a typo. here is the correct calculation as I made it:

F(1,2-x)= -F(1,2) cos 60º= - 0.9 N

Now are my calculations correct for e(1)?
Thank you for helping!</sub>

 

[denverdoc]

My results are different:

Here is how I did force of particle 2 on 1: k= 9.0 E9

total force (2 on 1) = K (1E-6)(2E-6)/1m^2=18 E-3. Since this is attractive force, it is directed upward and rightward.

Fx = F(2-1) cos 60= 9 E-3N
Fy = F(2-1) sin 60 = 0.87 * 18E-3= 15.6 E-3

Do the same for 3 on 1, and then add x components, then Y components and use Pythags formula for total. Direction can be found using tan(angle)=sum y/ sum x

 

[mmoadi]

_{Thank you for helping, I think I understand now}