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ludmilkaskok [199]
3 years ago
11

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 nega

tive, how far apart would they have to be for the electric attraction between them to equal their [email protected] weight?
Physics
1 answer:
bezimeni [28]3 years ago
5 0

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.

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