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6 February, 08:09

Two charges that are 1 meter apart repel each other with a force of 2 N. If the distance between the charges is increased to 2 meters, the force of repulsion will be:

a) 1 N b) 0.5 N c) 8 N d) 4 N

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  1. 6 February, 10:29
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    b) 0.5 N

    Explanation:

    From coulomb's law,

    F = kq'q/r² ... Equation 1

    Where F = force of repulsion between the charges, q' = first charge, q = second charge, r = distance between the charges, k = proportionality constant.

    q'q = Fr²/k ... Equation 2

    Given: F = 2 N, r = 1 m, k = 9.0*10⁹ Nm²/C²

    Substituting into equation 2

    q'q = 2 (1) ² / (9.0*10⁹)

    q'q = 2/9.0*10⁹ C².

    If the distance between the charges is increased to 2 meters,

    r = 2 m, q'q = 2/9.0*10⁹ C².

    Substitute into equation 1

    F = 9.0*10⁹ (2/9.0*10⁹) / 2²

    F = 2/4

    F = 1/2 = 0.5 N.

    The right option is b) 0.5 N
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