Two point charges are placed on the x axis. 4 µC at x = 0.2 m and -5 µC at x = -0.2 m. Determine the magnitude and direction of the electric

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Two point charges are placed on the x axis. 4 µC at x = 0.2 m and -5 µC at x = -0.2 m. Determine the magnitude and direction of the electric field on y axis at y = 0.15 m.

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1 year 2021-09-03T20:18:22+00:00 1 Answers 7 views 0

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    2021-09-03T20:20:20+00:00

    Answer:

    E = 10.10 10⁵ N / C,  θ = 184.75º

    Explanation:

    The electric field created by a point charge is

               E = k q / r²

    where r is the distance from the load to the desired point

    Since the electric field is a vector quantity, one of the simplest ways to calculate it is to find the components of the electric field on each axis and add them.

    charge 1  q₁ = 4 10⁻⁶ C placed at position x₁ = 0.2 m

    the distance is

               r² = (x-x₀) ² + (y-y₀) ²

               r² = (0 – 0.2) ² + (0.15 – 0) ²

               r² = 0.0625

           

              E₁ = 9 10⁹ 4 10⁻⁶ / 0.0625

              E₁ = 5.76 10⁵ N / C

    let’s find the angle of this vector

             tan θ = y / x

             θ = tan⁻¹ y / x

             θ = tan⁻¹ 0.15 / 0.2

             θ = 36.9º

    This angle is in the second quadrant since as the load is positive the electric field is salient, this angle measured from the positive side of the x axis is

              θ’= 180-  θ

              θ’= 180- 36.9

              θ’= 143.1º

    we use trigonometry

             cos 143.1 = E₁ₓ / E₁

             sin 143.1 = E_{1y} / E₁

             E₁ₓ = E₁ cos 143.1

             E_{1y} = E₁ sin 143.1

             E₁ₓ = 5.76 10⁵ cos 143.1 = -4.606 10⁵ N / C

              E_{1y} = 5.76 10⁵ sin 143.1 = 3.458 10⁵ N / C

    charge 2    q₂ = 5 10⁻⁶ at position x = -0.2 m

             r²2 = 0.2² + 0.15²

             r² = 0.0625

             E₂ = 9.10⁹  5 10⁻⁶/0.0625

             E₂ = 7.20 10⁵ N / c

    we look for the angles

             tan  θ = y / x

              θ = tan⁻¹ 0.15 / 0.2

              θ = 36.9º

    in that case the charge is negative, therefore the electric field is directed to the charge and therefore the angle is in the third quadrant

               θ’= 180 +  θ

              tea = 180 + 36.9 = 216.9º

    the components of the electric field are

              E₂ₓ = E₂ cos 216.9

              E_{2y} = E₂ sin 216.9

              E₂ₓ = 7.20 10⁵ cos 216.9 = -5.76 10⁵ N / C

              E_{2y} = 7.20 10⁵ sin 216.9 = -4.32 10⁵ N / C

    The components of the total electric field are

               Eₓ = E₁ₓ + E₂ₓ

               Eₓ = -4.606 10⁵ -5.76 10⁵

                Eₓ = – 10,366 10⁵ N / C

               E_y = E_{1y} + E_{2y}

               E_y = 3.458 10⁵ -4.32 10⁵

               E_y = -0.862 10⁵ N / C

    We can give the result in two ways

    1)        E = Ex i ^ + Ey j ^

             E = (- 10.366 i ^ -0.862 j ^) 10⁵ N / C

    2) in the form of module and angle

    let’s use the Pythagorean theorem

                 E = [tex]\sqrt{E_x^2 + E_y^2 }[/tex]Ra Ex² + Ey²

                 E = [tex]\sqrt{ (1.3666^2 + 0.862^2)}[/tex]   10⁵

                 E = 10.10 10⁵ N / C

    trigonometry

                 tan θ’= E_y / Eₓ

                 θ’= tan⁻¹ (0.862 / 10366)

                 θ’= 4.75º

    this angle is in the third quadrant, therefore measured from the positive side of the x-axis is

                 θ = 180 + θ’

                 θ = 180 + 4.75

                 θ = 184.75º

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