Question

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.

Answer 1

Answer:

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

$F_{n3} = \sqrt{(F_{n3}x)^{2} +(F_{n3}y)^{2} }$

$F_{n3} = \sqrt{( 1.955*10^{-4})^{2} +( -10.158 *10^{-4})^{2} }$

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

Answer 2

The resultant magnitude of the electrostatic force is 1.034 × 10⁻⁴ and the angle of the resultant force is 10.89⁰.

What is an electrostatic force?

The electrostatic force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

That is given as

$\rm F \propto \dfrac{q_1q_2}{r^2}\\\\F = k \dfrac{q_1q_2}{r^2}$

Where k is a constant.

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 n C and q₂ = -17.4 n C The remaining particle has a positive charge of 8.0 n C.

k = 8.99*10⁹ N*m²/C²

The force of attraction between q₁ and q₃ will be

$\rm F_{13} = \dfrac{8.99*10^9 *8.7*10^{-9}*8*10^{-9}}{0.04^2}\\\\F_{13} = 3.91*10^{-4} \ N$

Similarly, the  force of attraction between q₂ and q₃ will be

$\rm F_{23} = \dfrac{8.99*10^9 *17.4*10^{-9}*8*10^{-9}}{0.04^2}\\\\F_{23} = 7.82*10^{-4} \ N$

The component of force along x-direction will be

$\rm F_x = F_{13}*cos \ 60^o + F_{23}*cos \ 60^o\\\\F_x = - 3.91*10^{-4}*cos\ 60^o +7.82*10^{-4}*cos \ 60^o\\\\F_x = 1.955*10^{-4}$

The component of force along y-direction will be

$\rm F_y = F_{13}*sin\ 60^o + F_{23}*sin\ 60^o\\\\F_y = 3.91*10^{-4}*sin\ 60^o +7.82*10^{-4}*sin \ 60^o\\\\F_y = 10.158*10^{-4}$

The magnitude of the resultant force will be

$\rm F_R = \sqrt{(1.955*10^{-4})^2 + (10.158*10^{-4})}\\\\F_R = 1.034*10^{-3}$

And the angle will be

$\rm \theta = tan^{-1} \dfrac{F_x}{F_y}\\\\\\\theta = tan^{-1} \dfrac{1.955*10^{-4}}{10.158*10^{-4}}\\\\\\\theta = 10.89^o$

More about the electrostatic force link is given below.

https://brainly.com/question/9774180