A -9.00 μC point charge and +7.00 μC point charge are placed along the x-axis at x = 0.000 m and x = 0.4 m, respectively. Where must a third

Question

A -9.00 μC point charge and +7.00 μC point charge are placed along the x-axis at x = 0.000 m and x = 0.4 m, respectively. Where must a third charge be placed along the x-axis (relative to the origin) so that it does not experience any net electric force due to the other two charges? The position on the x-axis is in meters and use three significant digits.

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Thu Thủy 2 months 2021-07-24T09:29:49+00:00 1 Answers 7 views 0

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    2021-07-24T09:31:26+00:00

    Answer:

    x₃ = 0.397 m

    Explanation:

    For this exercise we use the vector sum of the forces, where the force is electric given by the Coulomb equation

               F = k \frac{q_1q_2}{r^2}

    We also use that charges of the same sign repel and charges of a different sign attract.

       In this case the force between load 1 and 3 has one direction and the force between 2 and 3 has the opposite direction, in the exercise they ask that the force on load 3 be zero

               ∑ F = F₁₃ – F₂₃ = 0

                F₁₃ = F₂₃

    we write the expressions of the forces, the distances are

             r₁₃ = (x₁-x₃) ²

             r₂₃ = (x₂ – x₃) ²

             k \frac{q_1  \ q_3}{(x_1-x_3)^2} = k \frac{q_2 \ q_3}{(x_2-x_3)^2 }

    x₁ = 0 m

             \frac{q_1}{x_3^2 } = \frac{q_2 }{(x_2 - x_3)^2 }

             q₁ (x₂ – x₃) ² = q₂ x₃²

             \sqrt{\frac{q_1}{q_2} } (x₂ -x₃) = x₃

    we substitute the values

             \sqrt\frac{9}{7} }  (0.4 -x₃) = x₃

             x₃ (1 + \sqrt{9/7}) = \sqrt{9/7}  0.4

             x₃ (1.13389) = 0.453557

             x₃ = 0.453557 / 1.13389

             x₃ = 0.397 m

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