Answer:
x=L/2 y=0 and the charge q3 is ¼ of the charge q
Explanation:
For this exercise we will use Coulomb's law.
F₁₂ = k q₁ q₂ / r₁₂²
From this expression we see that like charges repel and charges of different signs attract.
Let's apply this expression to our case, they indicate that the two charges are of equal magnitude and sign, therefore the force is repulsive, so that it is in equilibrium with a third charge (q₃) this must be of the opposite sign and be between the two charge (q)
let's apply Newton's second law to one of the charges, for example the one on the left
-F₁₂ + F₁₃ = 0
F₁₂ = F₁₃
k q₁ q₂ / r₁₂² = k q₁ q₃ / r₁₃²
q₂ / r₁₂² = q₃ / r₁₃²
q₃ = q₂ (r₁₃ / r₁₂)²
The problem indicates the charge q₁ = q₂ = 4 q and the distance between them is r₁₂ = L = 9 cm = 0.09 m, we substitute
q₃ = 4q (r₁₃ / L)²
Let's analyze the situation a bit that the charge 1 and 2 are in equilibrium with a single charge 3 this must be symmetrical between the two charge (the same force), therefore its position on the x axis must be r₁₃ = L/2 and how it is on the y axis = 0
let's substitute
q₃ = 4q (L / 2L)²
q₃ = 4q 1/4
q₃ = q
the charge q3 is ¼ of the charge q
