You are asked to find the net force acting on particle 3. Centering the xy coord

Question

You are asked to find the net force acting on particle 3. Centering the xy coordinate system on particle 3 will make this easier. To practice Problem-Solving Strategy 21.1 Coulomb's Law. Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.4 cm. Two of the particles have a negative charge: q_1 = -6.1 nC and q_2 = -12.2 nC . The remaining particle has a positive charge, q_3 = 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2? Find the net force E-F3 acting on particle 3 due to the presence of the other two particles. Report you answer as a magnitude E-rs and a direction 0 measured from the positive x axis.

Explanation / Answer

F = k*Q3*Q1/0.034² = -3.8e-4 N
F = k*Q3*Q2/0.034² = -7.6e-4 N

Fx = F32cos(60°) - F31cos(60°) =-7.6e-4*0.5 +3.8e-4*0.5 =-1.9e-4 N
Fy = F32sin(60°) + F31sin(60°) = -7.6e-4*0.866 -3.8e-4*0.866 =-9.87e-4 N
60° because it's an equilateral triangle. You'll notice the signs of the forces - both F31 and F32 are negative, and only the x-component of the F31 force is positive.

Now F = sqrt(Fx²+Fy²) = 1e-3 N

while direction is =arctan(Fy/Fx). =79.1 degree

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