Particle 1 carrying - 4.0 μC of charge is fixed at the origin of an xy coordinate system, particle 2 carrying + 8.0 μC of charge is located on the x axis at x = 4.0 m, and particle 3, identical to particle 2, is located on the x axis at x = - 4.0 m. What is the vector sum of the electric forces exerted on particle 3? Determine the x and y components of the vector sum.

F = (2.7 10⁻² i ^ + 0j ^) N

Explanation:

The electric force is given by Coulomb's law

F = k q₁q₂ / r²

Let's apply Newton's second law in particle 3

F = F₃₁ - F₃₂

F₃₁ = k q₃q₁ / r₃₁²

F₃₂ = k q₃q₂ / r₃₂²

Let's write the values they give us

q₁ = - 4.0 10⁻⁶ C

x₁ = 0

q₂ = q₃ = + 8.0 10⁻⁶ C

x₂ = 4.0 m

x₃ = - 4.0 m

We look for the distances

r₃₁ = x₃ - x₁

r₃₁ = - 4.0 - 0 = - 4.0 m

r₃₂ = - 4.0 - 4.0 = - 8.0 m

Let's write the force equation

F = k q₃ (q₁ / r₃₁² + q₂ / r₃₂²)

Let's calculate

F = 8.99 10⁹ 8 10⁻⁶ (4 10-6/4 2 - 8 10-6 / 8 2)

F = 71.92 10³ (1/2 - 1/8) 10⁻⁶

F = 26.97 10-3 N

None of the charge are on the axis and therefore there are no forces on this axis.

F = (2.7 10⁻² i ^ + 0j ^) N