An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the object?

Question

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the object?

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Nem 2 weeks 2021-09-05T13:27:21+00:00 1 Answers 0 views 0

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    2021-09-05T13:28:42+00:00

    Answer:

    the force of attraction between the two charges is 3.55 N.

    Explanation:

    Given;

    first charge carried by the object, q₁ = 15.5 µC

    second charge carried by the q₂ = -7.25 µC

    distance between the two charges, r = 0.525 m

    The force of attraction between the two charges is calculated as;

    F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N

    Therefore, the force of attraction between the two charges is 3.55 N.

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