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26 August, 05:16

A 2 microcoulomb charge is placed at a distance of 0.25 m away from a 3.6 microcoulomb charge. Describe the type of electrostatic force between the charges and calculate the magnitude of the electrostatic force between them.

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  1. 26 August, 06:34
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    The force between the charge is a repulsive force

    The magnitude of the electrostatic force between them = 1.0368 N

    Explanation:

    Since both charges are positive charge, and they have the same sign, from the law of electrostatics, The type of force between the charges is Repulsive force.

    Using

    F = kqq'/r² ... Equation 1

    Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulombs constant.

    Given: q = 2 μC = 2*10⁻⁶ C, q' = 3.6 μC = 3.6*10⁻⁶ C, r = 0.25 m, k = 9*10⁹ Nm²/C²

    Substitute into equation 1

    F = 9*10⁹ (2*10⁻⁶) (3.6*10⁻⁶) / 0.25²

    F = 64.8*10⁻³/0.0625

    F = 1036.8*10⁻³

    F = 1.0368 N

    Hence the magnitude of the electrostatic force between them = 1.0368 N
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