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31 May, 22:48

Two men each weigh 650 N. One man carries + 1.0 C of excess charge, the other - 1.0 C of excess charge. How far apart must they be for their electric attraction to equal 650 N?

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Answers (2)
  1. 1 June, 00:32
    0
    The answer is 3721.04 m

    Explanation:

    To solve the exercise, we will use the expression of force of attraction or repulsion between two electric charges:

    F = (k*q1*q2) / d^2, where

    F = force of attraction = 650 N

    q1 and q2 are the charge

    k = electric proportionality constant = 9 x 10^9 N x m^2

    Clearing d, we have:

    d = ((k*q1*q2) / F) ^1/2 = ((9x10^9*1*1) / (650)) ^1/2 = 3721.04 m
  2. 1 June, 02:48
    0
    Coulomb's law:

    F = k*q₁*q₂/r² where k ≈ 9.00*10⁹NC⁻²m²

    Given values:

    q₁ = + 1.0C

    q₂ = - 1.0C

    F = 650N

    Substitute the terms in Coulomb's law with our given values. We will have to use the absolute value of q₂ to so the algebra works out. Solve for r:

    650 = 9.00*10⁹*1.0*1.0/r²

    r = 3721m

    Taking significant figures into account:

    r = 3700m
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