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19 November, 09:09

A charged particle A exerts a force of 2.65 N to the right on charged particle B when the particles are 13.8 mm apart. Particle B moves straight away from A to make the distance between them 19.0 mm. What vector force does particle B then exert on A

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  1. 19 November, 10:20
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    -1.4N

    Explanation:

    When the distance between A and B is 13.8mm (0.0138m), the force exerted by A on B is 2.65N:

    F (A, B) = k*qA*qB/r²

    => 2.65 = k*qA*qB / 0.0138²

    => k*qA*qB = 2.65 * 0.0138²

    k*qA*qB = 0.0005046

    When the distance between them increases to 19mm (0.019m), the force exerted by A on B becomes:

    F (A, B) = k*qA*qB / 0.019²

    Given that k*qA*qB = 0.0005046,

    F (A, B) = 0.0005046/0.019²

    F (A, B) = 1.4N

    The force exerted by B on A will be in the opposite direction of the force exerted by A on B, which means that the force exerted by B on A will be:

    F (B, A) = - 1.4N
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