Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m

Sagot :

Answer:

 F_net = 9.87 10⁻⁴ N

Explanation:

Let's use that force is a vector magnitude

         ∑ F = F₁₃ + F₂₃

De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract

 

the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m

       ∑ F = F₁₃ - F₂₃

       F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]

in this case q₁ = q₂ = q

       F_net = k q q₃  (  )

 

      let's look for the distance

      r₂₃ = y₂ - y₃

      r₂₃ = -7 -16

      r₂₃ = - 23 m

       

      r₁₃ = 38 - 16

      r₁₃ = 22 m

let's calculate

      F_net = 9 10⁹ 24 26 10⁻¹² ( )

      F_net = 5.616 ( 1.758 10⁻⁴ )

      F_net = 9.87 10⁻⁴ N

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.