Respuesta :
Answer:
F = 8.98 j N
Explanation:
Coulomb's strength is given by
[tex]F_{e}[/tex] = k q₁ q₂ / r₁₂²
this force applied between each pair of electric charges and the total force is the sum of all the forces
F = F₁₄ + F₂₄ + F₃₄
where charge 1 is at position x = 1.0, charge 2 at x = 0 and charge 3 at x = -1.0 m, the cahrge 4 at point x = 0 y = 2.0 m
The force between charge 2 and 4 is attractive because of different signs
F₂₄ = k q₂q₄ / r₂₄²
r₂₄ = 0-2
F₂₄ = 9 10⁹ 3 10⁻⁴ 3 10⁻⁵ / 2²
F₂₄ = 20 N in y axis
The force between charges 1 and 4 is
F₁₄ = k q₁ q₄ / r₁₄²
r₁₄² = (1-0)² + (0-2)²
r₁₄² = 5
F₁₄ = 9 10⁹ 3 10⁻⁴ 3 10⁻⁵ / 5
F₁₄ = 16.2 N
this force is repulsive since the two charges have the same sign.
Let's use trigonometry to find its components
tan θ = y / x
θ = tan⁻¹ 2/1
θ = 63.4º
let's break down the force
sin 63.4 = [tex]F_{14y}[/tex] / F
cos 63.4 = F_{14x} / F
F_{14y} = F sin 63.4 = 16.2 sin 63.4
F_{14y} = 14.49 N
F_{14x} = F cos 63.4 = 16.2 cos 63.4
F_{14x} = -15.69 N
The force between charges 3 and 4 has the same magnitude and angles, so its value is
F₃₄ = 16.2 N
[tex]F_{34y}[/tex] = 14.49 N
F_{34x} = 15.69 N
to find the total force we add each component
Fₓ = F_{14x} + F_{24x} + F_{34x}
Fₓ = -15.69 + 0 + 15.69
Fₓ = 0
F_{y} = F_{14y} + F_{24y} + F_{34y}
F_{y} = 14.49 - 20 + 14.49
F_{y} = 8.98 N
the total force is
F = 8.98 j N
