Question

A if two such generic humans each carried 2.5 coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 700 n weight?

Answer 1

Using Coulomb's Law 

F = {k(Q1)(Q2)}/r² 

Where F =force in Newtons, k = Coulomb constant, Q1 = - Q2= charge of the objects in Coulumbs, and r is the distance between the objects. 

Solving for r: 

r = sqrt[{k(Q1)(Q2)}/F] 

where k = 8.897 x 10^9 N•m²/C², Q1 = 2.5 C, Q2 = -2.5 C, and F = -700 N 

Note: Negative magnitude of Force indicates attraction. 

r = 8,912.77 m