If two such generic humans each carried 3.0 c 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?

Distance, r = 10757.05 meters

Explanation:

Charge on two generic humans, q = 3 C

They have to be for the electric attraction between them to equal their 700 N weight, F = 700 N

Electric force is given by :

, r is the distance between charges

r = 10757.05 meter

So, distance between two charges is 10757.05 meters. Hence, this is the required solution.

v=ir = 3x15 =45 v

conduction, convection, radiation