Question

An average human weighs about 650 N. If each of two average humans could carry 1.0 C 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 [email protected] weight?

Answer 1

Answer:

Distance, r = 3721.04 meters

Explanation:

It is given that,

Charges on both humans, $q_1=q_2=1\ C$

Electric force of attraction between them, F = 650 N

We need to find the separation between two humans. It can be solved using the following formula as :

$F=\dfrac{kq^2}{r^2}$

$r=\sqrt{\dfrac{kq^2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times (1)^2}{650}}$

r = 3721.04 meters

So, the distance between the humans is 3721.04 meters. Hence, this is the required solution.