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Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged particle, with charge q3=q, is placed on the x-axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 2.00 nC . Assuming charge 1 is located at the origin of the x-axis and the positive x-axis points to the right, find the two possible values x3,r, and x3,l for the position of charge 3.

Respuesta :

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

lhmarianateixeira

In this exercise we have to use the knowledge of charged particles to calculate the distance between them, this corresponds to:

[tex]X_{31} = 0.58 m \\ X_{32} = -1.38 m[/tex]

For this exercise we use Newton's second law where the force is the Coulomb force, we can say that is:

[tex]F_{13} - F_{23} = 0\\ F_{13} = k q_1 q_3 / r_{13}^2\\ F_{23} = k q_2 q_3 / r^2_{23}\\ 4- 4x_3 + x_3^2 = 4 x_3^2\\ 5x_3^2 + 4 x_3 - 4 = 0\\ x_3 = [-4 +\sqrt{(16 - 4 5 (-4))} ] / 2 5\\ X_{31} = 0.58 m\\X_{32} = -1.38 m [/tex]

See more about charged particles at brainly.com/question/6903736

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