Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that one particl

Question

Two particles, one with charge − 7.97 μC and one with charge 3.55 μC, are 6.59 cm apart. What is the magnitude of the force that one particle exerts on the other?

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Amity 2 weeks 2021-09-02T20:17:57+00:00 1 Answers 0 views 0

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    2021-09-02T20:19:26+00:00

    Answer:

    58.6 N

    Explanation:

    We are given that

    q_1=-7.97\mu C=-7.97\times 10^{-6} C

    q_2=3.55\mu C=3.55\times 10^{-6} C

    Using 1\mu C=10^{-6} C

    r=6.59 cm=6.59\times 10^{-2} m

    1 cm=10^{-2} m

    The magnitude of force that one particle exerts on the other

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

    Where k=9\times 10^9

    Substitute the values

    F=\frac{9\times 10^9\times 7.97\times 10^{-6}\times 3.55\times 10^{-6}}{(6.59\times 10^{-2})^2}

    F=58.6 N

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