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 d = 1.25 m (Fig. 16-53). The charges are Q1 = +4.0 µC, Q2 = -5.0 µC, and Q3 = -6.0 µC. Calculate the magnitude and direction of the net force on each due to the other two.

Force on Q1: ______N at ________° counterclockwise from +x axis (to the right) (Force on Q1 = 0.2197, the angle wasn't 266.99)

Force on Q2:_______ N at _______° counterclockwise from +x axis (to the right) (Force on Q2 wasn't 0.1151, the angle wasn't 120.01)

Force on Q3:_______ N at _______° counterclockwise from +x axis (to the right) (Force on Q3 wasn't 0.0155, the angle wasn't 33.71)

Appreciate the help

Explanation / Answer

Magnitude of force on Q1 due to Q3 = F = k*Q1*Q3/1.25²

F =9*10^9*4.0*10^-6* -6.0*10^-6/1.25^2 = 0.1382

Magnitude of force on Q1 due to Q2 = F = k*Q2*Q1/1.25²

F =9*10^9*  -5.0*10^-6*4.0*10^-6/1.25^2 = 0.1152

Now you must add up the components in vector form:
Fx = Fcos(60°) - Fcos(60°)

Fx = 0.115cos(60°) - 0.138cos(60°) =-0.0115

Fy = Fsin(60°) + Fsin(60°)

Fy = 0.0199

magnitude of the net force on a particle,

F = sqrt(Fx²+Fy²),

0.0229N

direction is

=arctan(Fy/Fx). = 59.97degree

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*Q1/1.25²

F =9*10^9*-5.0*10^-6*4.0*10^-6*/1.25^2 = 0.115

F = k*Q2*Q3/1.25²

F =9*10^9*-5.0*10^-6*-6.0*10^-6/1.25^2 = 0.172

Fx = -Fcos(60°) - F

-0.2295
Fy = -Fsin(60°)

  0.115 sin(60°) =-0.0995

magnitude of the net force on a particle,

F = sqrt(Fx²+Fy²),

0.249N

direction is

=arctan(Fy/Fx).= 23.43

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*Q1/1.25²

F =9*10^9*-6.0*10^-6*4.0*10^-6/1.25^2 = 0.138

F = k*Q3*Q2/1.25²

F =9*10^9*-6.0*10^-6*-5.0*10^-6/1.25^2 = 0.172

Fx = Fcos(60°) + F

0.241
Fy = -Fsin(60°)

0.119

magnitude of the net force on a particle,

F = sqrt(Fx²+Fy²),

0.268N

direction is

=arctan(Fy/Fx).= 26.27

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.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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