Question

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 600 weight? *An average human weighs about 600 r= .............. km

Answer 1

They would be
`r=9.7\,km` apart

Step-by-step explanation:

Electric force between two charged objects is:

`F_(e)=k(\mid q_(1)q_(2)\mid)/(r^(2))` (1)

With q1, q2 the charges of the humans, r the distance between them and k the constant
`k=9.0*10^(9)\,(Nm^(2))/(C^(2))`, so if we want the electric force between them will be equal to their 600 N weight, we should make W=Fe=600 N on (1):

`600=k(\mid q_(1)q_(2)\mid)/(r^(2))`

solving for r:

`r^(2)=k(\mid q_(1)q_(2)\mid)/(600)`

`r=\sqrt{k(\mid q_(1)q_(2)\mid)/(600)}`

`r=\sqrt{(9.0*10^(9))(\mid(-2.5)(2.5)\mid)/(600)} =9682 m`

`r=9.7\,km`