Question

An average human weighs about If two such generic humans each carried 1.0 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 weight?

Answer 1

Answer:

Explanation:

excess charge, q1 = 1 C, q2 = - 1 C

Let the distance is r.

let your mass is, m = 50 kg

Weight, W = mass x gravity = 50 x 9.8 = 490 N

Use the Coulomb's law

$F=k\frac{q_{1}q_{2}}{r^{2}}$

$490 = \frac{9\times10^{9}\times 1\times 1}{r^{2}}$

r = 4285.7 m

r = 4.285 km