Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are placed 3.39 m a

Question

Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are placed 3.39 m apart. if the magnitude of the electric force between the charges is 2.4 × 10−6 n, what is the value of q?

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Thành Công 5 months 2021-08-02T21:48:29+00:00 1 Answers 35 views 0

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    2021-08-02T21:50:22+00:00

    Answer:

    q=3.19*10^{-8}C

    Explanation:

    The expression for the electric force between two charges

    F=k\frac{q_1q_2}{r^2}\\

    where K is the Coulomb’s constant (k=9*10^{9}Nm^2/C^2), and we have that

    q1=q

    q2=3q

    r=3.39m

    F=2.4*10^{-6}N

    By replacing in the formula we have

    F=2.4*10^{-6}N=(9*10^{9}\frac{Nm^2}{C^2})\frac{(q)(3q)}{(3.39m)^2}\\\\

    and by taking apart q we have

    q=3.19*10^{-8}C

    hope this helps!!

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