Three charged particles are placed at each of three corners of an equilateral tr

Question

Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -8.4 nC and q2 = -16.8 nC. The remaining particle has a positive charge, q3 = 8.0 nC . 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.

Explanation / Answer

Q1)
Magnitude of force on Q1 due to Q3 = F = k*Q1*Q3/2.7²
Magnitude of force on Q1 due to Q2 = F = k*Q2*Q1/2.7²

Now you must add up the components in vector form:
Fx = Fcos(60°) - Fcos(60°)
Fy = Fsin(60°) + Fsin(60°)
60° because it's an equilateral triangle. You'll notice the signs of the forces - both F and F are negative, and only the x-component of the F force is positive.

Q2)
F = k*Q2*Q1/2.7²
F = k*Q2*Q3/2.7²

Fx = -Fcos(60°) - F
Fy = -Fsin(60°)
F is negative, which means it's an attractive force, so Q1 is pulling Q2 in the positive x-direction and the positive y-direction. F is positive, which means it's repulsive, so Q3 is pushing Q2 in the negative x-direction.

Q3)
F = k*Q3*Q1/2.7²
F = k*Q3*Q2/2.7²

Fx = Fcos(60°) + F
Fy = -Fsin(60°)
F is negative, and Q1 pulls Q3 in the negative x-direction, but the positive y-direction. F is positive, and Q2 pushes Q3 in the positive x-direction.

To determine magnitude of the net force on a particle, F = sqrt(Fx²+Fy²), while direction is =arctan(Fy/Fx).

Make sure you convert µC into Coulombs before you get started.

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