Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the magnitude and d

Question

Two positive point charges, each of which has a charge of 2.5 × 10−9 C, are located at y = + 0.50m and y = − 0.50m. Find the magnitude and direction of the resultant electric force acting on a charge of 3.0 × 10−9 C located at x = 0.70m.

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niczorrrr 1 week 2021-07-21T20:52:35+00:00 1 Answers 0 views 0

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    2021-07-21T20:54:30+00:00

    Answer:

    F = 147,78*10⁻⁹ [N]

    Explanation:

    By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .

    The angle β  ( angle between the line running through one of the charges in y axis and the charge in x axis) is

    tan β  =  0,5/0,7

    tan β  = 0,7142    then   β = arctan 0,7142      ⇒   β = 35 ⁰

    cos  β = 0,81

    d = √ (0,5)²  +  (0,7)²       d1stance between charges

    d = √0,25 + 0,49

    d =  √0,74  m

    d = 0,86  m

    Now Foce between two charges  is:

    F  = K* q₁*q₂/ d²      (1)

    Where  K  = 9*10⁹ N*m²/C²

    q₁  =  2,5* 10⁻⁹C

    q₂  = 3,0*10⁻⁹C

    d²  = 0,74 m²

    Plugging these values in (1)

    F  =  9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74       [N*m²/C²]*C*C/m²

    F = 91,21 * 10⁻⁹   [N]

    And  Fx  =  F*cos  β

    Fx  =  91,21 * 10⁻⁹ *0,81

    Fx =73,89*10⁻⁹  [N]

    Then total force acting on charge located at x = 0,7 m is:

    F = 2* Fx

    F = 2*73,89*10⁻⁹  [N]

    F = 147,78*10⁻⁹ [N]

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