To practice Problem-Solving Strategy 22.1 for electric force problems. Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are loca

Question

To practice Problem-Solving Strategy 22.1 for electric force problems. Two charged particles, with charges q1=qq1=q and q2=4qq2=4q, are located on the x axis separated by a distance of 2.00cm2.00cm . A third charged particle, with charge q3=qq3=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 qqq = 2.00 nCnC .

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Thu Giang 3 years 2021-07-28T05:44:16+00:00 1 Answers 94 views 0

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  1. RuslanHeatt
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    2021-07-28T05:45:20+00:00
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    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

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