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 650-N weight?

Answer 1

r = 3721.04 m

Step-by-step explanation:

Given that,

Weight of human, F = 650 N

Charge on two humans,
`q_1=q_2=1\ C`

We need to find the distance between charges if the electric attraction between them to equal their 650 N weight. It is given by :

`F=(kq^2)/(r^2)`

`r=\sqrt{(kq^2)/(F)}`

`r=\sqrt{(9* 10^9* 1^2)/(650)}`

r = 3721.04 m

So, the distance between charges is 3721.04 m if the electric attraction between them to equal their 650 N weight. Hence, this is the required solution.