Question

Particles q1 = +8.0 UC, 92 = +3.5 uc, and

Answer 1

Answer:

-22.3

Explanation:

Acellus

Answer 2

Answer:

   F_total = 29.4 N,    directed to the right of particle 2

Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

         F = $k \frac{q_1q_2}{r_{12}^2}$

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

         F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

         F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

        F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

        F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

        F_total = ∑ F = F₁₂ + F₂₃

        F_total = 25.2 +3.5

        F_total = 29.4 N

directed to the right of particle 2