Three charged particles are placed at the corners of an equilateral triangle of side d = 1.30 m (Fig. 16-53).

Three charged particles are placed at the corners of an equilateral triangle of side d = 1.30 m (Fig. 16-53). The charges are Q1 = +4.0 µC, Q2 = -9.0 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 _____degrees counterclockwise from +x axis (to the right)
Force on Q2: ________ N at _____degrees counterclockwise from +xaxis (to the right)
Force on Q3: ________ N at _____degrees counterclockwise from +xaxis (to the right)

Figure 16-53 just looks like an equalateral triangle with Q1 at the top, Q2 on the bottom left, and Q3 at the bottom right. all sides have length d = 1.30m

Q1)
Magnitude of force on Q1 due to Q3 = F₁₃ = k*Q1*Q3/1.3²
Magnitude of force on Q1 due to Q2 = F₁₂ = k*Q2*Q1/1.3²

Now you must add up the components in vector form:
Fx = F₁₂cos(60°) - F₁₃cos(60°)
Fy = F₁₂sin(60°) + F₁₃sin(60°)
60° because it's an equilateral triangle. You'll notice the signs of the forces - both F₁₂ and F₁₃ are negative, and only the x-component of the F₁₃ force is positive.

Q2)
F₂₁ = k*Q2*Q1/1.3²
F₂₃ = k*Q2*Q3/1.3²

Fx = -F₂₁cos(60°) - F₂₃
Fy = -F₂₁sin(60°)
F₂₁ is negative, which means it's an attractive force, so Q1 is pulling Q2 in the positive x-direction and the positive y-direction. F₂₃ is positive, which means it's repulsive, so Q3 is pushing Q2 in the negative x-direction.

Q3)
F₃₁ = k*Q3*Q1/1.3²
F₃₂ = k*Q3*Q2/1.3²

Fx = F₃₁cos(60°) + F₃₂
Fy = -F₃₁sin(60°)
F₃₁ is negative, and Q1 pulls Q3 in the negative x-direction, but the positive y-direction. F₃₂ 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 θ=arctan(Fy/Fx).

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