Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC i

Question

Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?

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Philomena 6 months 2021-07-29T09:21:41+00:00 1 Answers 81 views 0

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    2021-07-29T09:23:09+00:00

    Answer:

     F = 0.111015 N

    Explanation:

    For this exercise the force is given by Coulomb’s law

            F = k q₁q₂ / r₂₁²

    we calculate the electric force of the other two particles on the charge q1

    Charges q₁ and q₂

    the distance between them is

              r₁₂ = y₁ -y₂

              r₁₂ = 0.30 + 0.30

              r₁₂ = 0.60 m

    let’s calculate

              F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2

              F₁₂ = 1 10⁻¹ N

    directed towards the positive side of the y-axis

    Charges 1 and 3

    Let’s find the distance using the Pythagorean Theorem

                 r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]

                 r₁₃ = 0.50 m

                F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²

                F₁₃ = 1.697 10⁻² N

    The direction of this force is on the line that joins the two charges (1 and 3), let’s use trigonometry to find the components of this force

               tan θ = y / x

               θ = tan⁻¹ y / x

              θ = tan⁻¹ 0.3 / 0.4

               tea = 36.87º

        The angle from the positive side of the x-axis is

             θ ’= 180 – θ

            θ ’= 180 – 36.87

            θ ’= 143.13º

           sin143.13 = F_13y / F₁₃

               F_13y = F₁₃ sin 143.13

               F{13y} = 1.697 10⁻² sin 143.13

               F_13y = 1.0183 10⁻² N

                cos 143.13 = F_13x / F₁₃

               F₁₃ₓ = F₁₃ cos 143.13

               F₁₃ₓ = 1.697 10⁻² cos 143.13

               F₁₃ₓ = -1.357 10-2 N

    Now we can find the components of the resultant force

              Fx = F13x + F12x

              Fx = -1,357 10-2 +0

              Fx = -1.357 10-2 N

              Fy = F13y + F12y

             Fy = 1.0183 10-2 + ​​1 10-1

              Fy = 0.110183 N

    We use the Pythagorean theorem to find the modulus

             F = Ra (Fx2 + Fy2)

             F = RA [(1.357 10-2) 2 + 0.110183 2]

             F = 0.111015 N

    Let’s use trigonometry for the angles

             tan tea = Fy / Fx

              tea = tan-1 (0.110183 / -0.01357)

              tea = 1,448 rad

    to find the angle about the positive side of the + x axis

               tea ‘= pi – 1,448

               Tea = 1.6936 rad

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