Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.4cm . Two of the particles have a negative charge: q1 = -8.1nC and q2 = -16.2nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Find the net force ?F? 3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude ?F3 and a direction ? measured from the positive x axis.

Answer 1

force due to q1 = [9*10^9 * 8.1*8 * 10^-18 / 0.034^2] *[ cos 60 i^ + sin 60 j^ ]

F1 = 5.044983 * 10^-4 * [ 0.5i^ + 0.866 j^ ]

so.. F1 = 2.5224915 * 10^-4 i^ + 4.369 *10^-4 j^ N

force due to q2 = [9*10^9 * 16.2*8 * 10^-18 / 0.034^2] *[ i^]

F2 = 10.089966 * 10^-4 * i^

so... total force F3 = F1 + F2 = 2.5224915 * 10^-4 i^ + 4.369 *10^-4 j^ + 10.089966 * 10^-4

so... F3 = 12.6124575 * 10^-4 i^ + 4.369 *10^-4 j^

so.. magnitude of F3 = 10^-4 * sqrt ( 12.6124575^2 + 4.369^2 ) = 13.34774 * 10^-4 N

for getting the angle I will need the figure.. please insert the figure

Answer 2

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else).

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.8cm . Two of the particles have a negative charge: q1 = -7.3nC and q2 = -14.6nC . The remaining particle has a positive charge, q3 = 8.0nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Find the net force ?F? 3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude ?F3 and a direction ? measured from the positive x axis.

vertical direction = 9*10^9 *10^-18 * 8[-21.9]/0.000784 * cos30 = -0.0017 N j

horizontal directoin = 9*10^9*10^-18 *8 [7.3]/0.000784 * cos60 = 0.000335 N i

net resultanta = 0.00173 N

direction = 78.85 degrees below x axis   

Answer 3

vertical direction = 9*10^9 *10^-18 * 8[-21.9]/0.000784 * cos30 = -0.0017 N j

horizontal directoin = 9*10^9*10^-18 *8 [7.3]/0.000784 * cos60 = 0.000335 N i

net resultanta = 0.00173 N