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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?

Respuesta :

Answer:

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 :

[tex]F=\dfrac{kq^2}{r^2}[/tex], r is the distance between charges

[tex]r=\sqrt{\dfrac{kq^2}{F}[/tex]

[tex]r=\sqrt{\dfrac{9\times 10^9\times 3^2}{700}}[/tex]

r = 10757.05 meter

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

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