Electric Forces in a Triangle (math check?)

[CrzyMunky]

Hi Guys I've tried to do this problem but the answers are not being accepted is the math wrong? or did i forget a concept? Thanks for taking a look

The triangle is set by Q1 on the top of the triangle Q2 on the bottom left and Q3 on the bottom right



Calculate the magnitude of the net force on each due to the other two. Three charged particles are placed at the corners of an equilateral triangle of side 1.20 (see the figure ). The charges are Q1= 7.0 , Q2= -8.6 , and Q3= -5.4 .

F(e)= k (Q1*Q2)/ r^2

F(12) =(8.98e9*7e-6*8.6e-6)/ (1.2)^2 F(23) = 8.98e9*8.6e-6*5.4e-6/ (1.2)^2

F(31) =(8.98e9*5.4e-6*7e-6) / (1.2)^2

F(12) = .3754N
F(23) = .2896N
F(31) = .2357N

Fx1 = -Cos(60)(.3754)+Cos(60)(.2357)= -.069
Fy1 = -Sin (60) (.3754)-Sin(60)(.2357)=-.121

F1= sqrt((-.069)^2+(-.121)^2)= .139

Fx2 = -Cos(60)(.3754)-.2896=-.1025
Fy2 = Sin(60)(.3754)=.3247

F2= sqrt((-.1025)^2+(.3247))=.340

Fx3 = Cos(60)(.2357)+.2896=.4075
Fy3 = Sin(60) (.2357)=.204

F3= sqrt((.4075)^2+(.204)^2)=.226

 

[anathema]

Hi there! It looks like you have correctly calculated the magnitudes of the net forces on each particle. However, it's important to also consider the direction of the force, which is why you calculated the x and y components of the force. In this case, the net force on each particle is directed towards the center of the equilateral triangle, which is why you calculated the x and y components using the cosine and sine functions. It looks like you may have made a small error in your calculations for Fx1 and Fy1, as the correct values should be -0.074 and -0.217, respectively. But overall, your calculations seem to be correct. Keep up the good work!