1) Gp2- Logp2--A x D Lecture o 3-Kinen x GPI A02--Electric F x c Physics questio
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1) Gp2- Logp2--A x D Lecture o 3-Kinen x GPI A02--Electric F x c Physics question X BIOL320 GeneticsL x B Courses Blackboe x Lecture 03-Kinem X C www.webassign. t/web/Student/Assignment-Responses/last?dep 15334392 Q7 Apps C Home IGradesFirst WebAssi gn LOG IN P Mastering Chemist 7. 14 points I Previous Answers OSCol Phys2016 18.7 P 049 My Notes Ask Your Teacher The point charges in the figure below are located at the corners of an equilateral triangle 10.0 cm on a side. (Assume that the +x axis is to the right and the ty-axis is up along the page (a) Find the electric field (in N/C) at the location of qa in the figure, given that qb 11.00 uC and q 7.00 HC. v N/C magnitude 8.6e+6 direction x counterclockwise from the +x-axis (b What s the force (in N) on given that q +3.00 nC? ude magn direction counterclockwise from the +x-axis 5:34 PM 4/26/2017Explanation / Answer
Ebx = -k*qb*cos60/r^2 = -9*10^9*11*10^-6*cos60/0.1^2 = -4950000 N/C
Eby = k*qb*sin60/r^2 = 9*10^9*11*10^-6*sin60/0.1^2 = 8573651.5 N/C
Ecx = -k*qc*cos60/r^2 = -9*10^9*7*10^-6*cos60/0.1^2 = -3150000N/C
Ecy = -k*qb*sin60/r^2 = -9*10^9*7*10^-6*sin60/0.1^2 = -5455960 N/C
Ex = Ebx + Ecx = -4950000 - 3150000 = -8100000
Ey = Eby + Ecy = 8573651.5-5455960 = 3117691.5
E =sqrt(Ex^2 + Ey^2 = 8.6*10^6 N/C
direction = tan^-1(Ey/Ex)
direction = 159 degrees
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F = E*q = 8.6*10^6*3*10^-9 = 0.0258 N
direction = 159
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