What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original magnitude?

Question

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original
magnitude?

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Thành Công 2 months 2021-09-05T14:18:53+00:00 1 Answers 4 views 0

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    2021-09-05T14:20:07+00:00

    Answer:

    F’ = 64 F

    Explanation:

    The electric force between charges is given by :

    F=\dfrac{kq_1q_2}{r^2}

    Where

    q₁ and q₂ are charges

    r is the distance between charges

    When  each charge is doubled and the distance between them is 1/4 its original magnitude such that,

    q₁’ = 2q₁, q₂’ = 2q₂ and r’ = (r/4)

    New force,

    F'=\dfrac{kq_1'q_2'}{r'^2}

    Apply new values,

    F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F

    So, the new force becomes 64 times the initial force.

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