Answered

Discover how IDNLearn.com can help you find the answers you need quickly and easily. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.

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 .

We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.