Respuesta :
Answer:
The magnitude of the electrostatic force is F = 368.64 N
Explanation:
Given data,
The magnitude of charge, q₁ = 64 μ C
The magnitude of charge, q₂ = 80 μ C
The magnitude of charge, q₃ = -160 μ C
The charge q₁ is at the origin,
The position of charge q₁ in x-axis, x₁ = 0 cm
The position of charge q₂ in x-axis, x₂ = 25 cm
= 0.25 m
The position of charge q₃ in x-axis, x₃ = 50 cm
= 0.5 m
The magnitude of the electrostatic force experienced by the charge a is given by the formula,
The force acting on charge 'q₁' due to 'q₂' is,
F₁ = k q₁ q₂ / x₂²
= (9 x 10⁹ X 64 x 10⁻⁶ X 80 x 10⁻⁶) / 0.25²
= + 737.28 N
The force acting on charge 'q₁' due to 'q₃' is,
F₂ = k q₁ q₃ / x₃²
= [9 x 10⁹ X 64 x 10⁻⁶ X (-160 x 10⁻⁶)] / 0.5²
= - 368.64 N
The net electrostatic force acting on the charge, q₁ is,
F = F₁ + F₂
= + 737.28 N + (- 368.64 N)
= 368.64 N
Hence, the magnitude of the electrostatic force is F = 368.64 N
