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CrzyMunky
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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
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