Answer:
x₂ = 0.01336
Explanation:
In this exercise we use the translational equilibrium equation
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
at the point where charge 3 is placed the two electric forces have the same magnitude
let us use the expression of Coulomb's law for the electric force
F₁₃ = ka q₁ q₃ / r₁₃²
F₂₃ = ka q₂ q₃ / r₂₃²
we substitute
k q₁ q₃ / r₁₃² = k q₂ q₃ / r₂₃²
now imprescriptibility suppose that particle 1 is at the origin of the coordinate system, particle 2 is at a distance d = 2.00cm = 2 10-2 m, therefore let's call the distance from particle 1 to particle 3 as x
r₁₃ = x
R₂₃ = d-x
In the exercise we are given the charges for the particle1 q1 = q, for the particle 2 the charge is q2 = 4q the distance between them is d = 2.00cm = 0.0200 m, the value of q = 2.00 nC = 2.00 10⁻⁹ C
let's substitute these values
q₁ / x₂ = q₂ / (d-x)²
let's clear x
(d-x)² = q₂ / q₁ x²
d² - 2dx + x₂ = q₂ /q₁ x²
x² (1-q₂ / q₁) - 2d x + d² = 0
let's substitute the values and solve the quadratic equation
x² (1 - 4q / q) - 2 0.02 x + 0.02² = 0
-3 x² - 0.04 x + 0.0004 = 0
x² + 0.0133 x - 0.0001333 = 0
x = [-0.0133 ±√(0.0133² + 4 0.00013333)] / 2
x = [-0.0133 + - 0.026664] /2
x₁ = -0.01998 m
x₂ = 0.01336 m
Since load 3 must be between charged 1 and 2 the correct answer is
x₂ = 0.01336
