A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

Question

A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge

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Thu Thảo 3 years 2021-08-17T15:37:51+00:00 2 Answers 9 views 0

Answers ( )

    0
    2021-08-17T15:39:22+00:00

    Answer:

    F = 2.49 x 10⁻⁹ N

    Explanation:

    The electrostatic force between two charged bodies is given by Colomb’s Law:

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

    where,

    F = Electrostatic Force = ?

    k = colomb’s constant = 9 x 10⁹ N.m²/C²

    q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

    q₂ = second charge = 1.4 C

    r = distace between charges = 0.9 m

    Therefore,

    F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

    F = 2.49 x 10⁻⁹ N

    0
    2021-08-17T15:39:33+00:00

    31.5N

    I have confirmed it is the right answer.

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