Question

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

Answer 1

F_net = 9.87 10⁻⁴ N

Step-by-step 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 =
`k (q_1q_3)/(r_(13)^2 ) - k (q_2q_3)/(r_(23)^2 )`

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