Answer:
The third charged particle must be placed at x = 0.458 m = 45.8 cm
Explanation:
To solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁, q₂: Charges in Coulombs (C)
d: distance between the charges in meters (m)
Equivalence
1μC= 10⁻⁶C
1m = 100 cm
Data
K = 8.99 * 10⁹ N*m²/C²
q₁ = +5.00 μC = +5.00 * 10⁻⁶ C
q₂= +7.00 μC = +7.00 * 10⁻⁶ C
d₁ = x (m)
d₂ = 1-x (m)
Problem development
Look at the attached graphic.
We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:
We use formula (1) to calculate the forces F₁₃ and F₂₃
F₁₃ = F₂₃
We eliminate k and q₃ on both sides
We eliminate 10⁻⁶ on both sides
We solve the quadratic equation:
In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .
x = 0.458m = 45.8cm
