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9 February, 13:52

Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of their electrical charges q1 and q2 and inversely as the square of the distance d between the particles. Find the effect on F of doubling q1 and q2 and halving the distance between them.

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  1. 9 February, 14:55
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    Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

    Explanation:

    Let the charges be q1 and q2 and the distance between the charges be 'd'

    Mathematical representation of coulombs law will be;

    F1=kq1q2/d² ... (1)

    Where k is the electrostatic constant.

    If q1 and q2 is doubled and the distance halved, we will have;

    F2 = k (2q1) (2q2) / (d/2) ²

    F2 = 4kq1q2 / (d²/4)

    F2 = 16kq1q2/d² ... (2)

    Dividing equation 1 by 2

    F1/F2 = kq1q2/d² : 16kq1q2/d²

    F1/F2 = kq1q2/d² * d²/16kq1q2

    F1/F2 = 1/16

    F1 = 1/16F2

    This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved
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