Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 4.0 cm. two of the particles have a negative charge: q1 =-8.7 nc and g2 =-17.4 nc the remaining particle has a positive charge, 93 8.0 nc. what is the net electric force acting on particle 3 due to particle 1 and particle 2? find the net force f3 acting on particle 3 due to the presence of the other two particles. report you answer as a magnitude and a direction θ measured from the positive x axis express the magnitude in newtons and the direction in degrees to three significant figures.

F3 = 1.03 * 10⁻³ N : net force F3 acting on particle 3 due to the presence of the other two particles.

θ= -79.1°  :  direction of the net force F3

θ= 79.1° measured from the positive x axis ,clockwise

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces.

Equivalences

1nC= 10⁻⁹ C

Known data

k = 8.99*10⁹ N*m²/C²

q₁ = -8.7 nC = -8.7 * 10⁻⁹ C

q₂=-17.4 nC =-17.4 * 10⁻⁹ C  

q₃= +8 nC= +8 * 10⁻⁹ C  

d₁₃= 0.04 m

d₂₃= 0.04 m

d₁₃: distance from q₁ to q3

d₂₃: distance from q₂ to q3

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure:

The force F₁₃ of q₁ on q₃ is attractive because the charges have opposite signs.  

The force F₂₃ of q₂ on q₃ is attractive because the charges have opposite signs.  

Calculation of the net force exerted for q₁ and q₂ on the charge q₃

The net force exerted for q₁ and q₂ on the charge q₃ is the algebraic sum of the forces in x-y of F₁₃ and F₂₃  

Fn₃x =F₁₃x+ F₂₃x

Fn₃y = F₁₃y+F₂₃y

Calculation of the magnitudes of F₁₃ and F₂₃

To calculate the magnitudes of the forces exerted by the charges q₁, and q₂ on q₃ we apply Coulomb's law:

F₁₃=(k*q₁*q₃)/d₁₃²  

F₁₃=(8.99*10⁹*8.7*10⁻⁹*8*10⁻⁹)/(0.04)² = 3.91*10⁻⁴ N

F₂₃=(k*q₂*q₃)/d₂₃²  

F₂₃=(8.99*10⁹*17.4*10⁻⁹*8*10⁻⁹)/(0.04)² = 7.82*10⁻⁴ N

Calculation of the x-y components of F₁₃ and F₂₃

F₁₃x=F₁₃*cos60°= 3.91*10⁻⁴ N *cos60°= 1.955*10⁻⁴ N

F₁₃y=F₁₃*sin60°= 3.91*10⁻⁴ N *sin60°= 3.386*10⁻⁴ N

F₂₃x=F₂₃*cos60°= 7.82*10⁻⁴ *cos60°= 3.91*10⁻⁴ N

F₂₃y=F₂₃*sin60°= 7.82*10⁻⁴ *sin60°= 6.772*10⁻⁴ N

Calculation of the x-y components of Fn₃

Fn₃x = F₁₃x+F₂₃x= - 1.955*10⁻⁴ N + 3.91*10⁻⁴ N = 1.955*10⁻⁴ N

Fn₃y = F₁₃y+F₂₃y=  - 3.386*10⁻⁴ N - 6.772*10⁻⁴ N = - 10.158 *10⁻⁴ N

Calculation of the magnitude of Fn₃

F3=Fn₃= 1.034 * 10⁻³ N : net force F3 acting on particle 3 due to the presence of the other two particles.

Calculation of the direction (θ) of Fn₃

θ= tan⁻¹ (Fn₃y/Fn₃x)

θ= tan⁻¹  (- 10.158 *10⁻⁴ / 1.955*10⁻⁴ N)

θ= -79.1°  

θ= 79.1° measured from the positive x axis, clockwise

- measure reactants precisely. ( first choice)

- stir to the reaction take place. ( second choice)

- wait patiently until the reaction completes. ( fourth choice)

- make sure to recover as much copper as possible. ( last choice)

the answer is b

explanation: