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:
[tex]F = \frac{k*q_1*q_2}{d^2}[/tex] 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
[tex]F_{13} = \frac{k*q_1*q_3}{d_1^2}[/tex]
[tex]F_{23} = \frac{k*q_2*q_3}{d_2^2}[/tex]
F₁₃ = F₂₃
[tex]\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}[/tex] We eliminate k and q₃ on both sides
[tex]\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}[/tex]
[tex]\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}[/tex]
[tex]\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}[/tex] We eliminate 10⁻⁶ on both sides
[tex](1-x)^2 = \frac{7}{5} x^2[/tex]
[tex]1-2x+x^2=\frac{7}{5} x^2[/tex]
[tex]5-10x+5x^2=7 x^2[/tex]
[tex]2x^2+10x-5=0[/tex]
We solve the quadratic equation:
[tex]x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m[/tex]
[tex]x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m[/tex]
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
