Three charged particles are placed at the corners of an equilateral triangle of

Question

Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m (see (Figure 1) ). The charges are Q1 = 7.0 ?C , Q2 = -9.2 ?C , andQ3 = -6.2 ?C

Part A

Calculate the magnitude of the net force on particle 1 due to the other two.

Express your answers using two significant figures.

Part B

Calculate the direction of the net force on particle 1 due to the other two.

Express your answer as positive angle using two significant figures.

Part C

Calculate the magnitude of the net force on particle 2 due to the other two.

Express your answers using two significant figures.

Part D

Calculate the direction of the net force on particle 2 due to the other two.

Express your answer as positive angle using two significant figures.

Part E

Calculate the magnitude of the net force on particle 3 due to the other two.

Express your answers using two significant figures.

Part F

Calculate the direction of the net force on particle 3 due to the other two.

Express your answer as positive angle using two significant figures.

Explanation / Answer

Magnitude of force on Q1 due to Q3 = F = k*Q1*Q3/1.2² =0.271 N
Magnitude of force on Q1 due to Q2 = F = k*Q2*Q1/1.2² =0.403 N

Now you must add up the components in vector form:
Fx = Fcos(60°) - Fcos(60°) = 0.202- 0.136 = 0.066 N
Fy = Fsin(60°) + Fsin(60°) = 0.349 +0.235 = 0.584 N
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.

magnitude of the net force on a particle 1, F = sqrt(Fx²+Fy²) = 0.59 N while direction is =arctan(Fy/Fx)= arctan(0.584/0.066)= 83.55 deg

Q2)
F = k*Q2*Q1/1.2² =0.403 N
F = k*Q2*Q3/1.2² =0.357 N

Fx = -Fcos(60°) - F = -0.202 - 0.357 = -0.559 N
Fy = -Fsin(60°) = -0.349 N
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.

magnitude of the net force on a particle 1, F = sqrt(Fx²+Fy²) = 0.66 N while direction is =arctan(Fy/Fx)= arctan(0.559/0.349)= 58.02 deg

Q3)
F = k*Q3*Q1/1.32² = 0.271 N
F = k*Q3*Q2/1.2² =0.357 N

Fx = Fcos(60°) + F =0.136 +0.357 =0.493 N
Fy = -Fsin(60°)= -0.235 N
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)= 0.55 N while direction is =arctan(Fy/Fx) = 25.49 deg

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.