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What is the approximate electrostatic force between two protons separated by a distance of 1.0 x 10-6 meter?
1) 2.3 x 10-16 N and repulsive
2) 9.0 x 1021 N and repulsive
3) 9.0 x 1021 N and attractive
4) 2.3 10-16 N and attractive

Sagot :

MickeyCyberkididn

Answer:

2.3×10^-16

Explanation:

Charge of proton = 1.6×10^-19

let q(a) and q(b) be the two protons.

According to the Coulumbs Laws ;

F = kq(a)×q(b) / r^2

where k = 9.0×10^9 and r is the distance of seperation.

So having a distance of 1.0×10^-6 in our case ;

F = (9.0×10^9×1.6×10^-19×1.6×10^-19)/(1.0××10^-6)^2

F = (2.304×10^-28)/(1.0×10^-12)

F = 2.304×10^-16N

Since the both charges are positively charged they will repel .

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