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11 June, 11:13

Consider the electrostatic force between two particles of equal mass m = 32 * 10^-26 kg and carrying different charges q1 = 17e and q2 = 28e, where e denotes the magnitude of the charge on an electron. At what distance, in meters, is the electrostatic force between these two particles equal to the weight of one of them?

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  1. 11 June, 13:36
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    Answer: 0.187 m

    Explanation: In order to explain this problem we have to consider the Coulomb and gravitational force, which are given by;

    Electric force

    Fe = (k*q1*q2) / d^2 where k is a constant equal to 9*10^9 N/C^2*m^2 and d is the distanvce between the charges.

    gravitational force

    Fg=me*g

    equaling both forces, we have:

    (k*q1*q2) / d^2 = me*g

    d^2 = (k*q1*q2) / me*g = 9*10^9*17*28 * (1.6*10^-19) ^2 / (32*10^-26*9.8) =

    =0.187 m
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