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?

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

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