What is the strength of the interaction of two protons at a distance of 1.6 * 10 ^ (- 14) m?

To calculate the force of electrostatic repulsion of two protons located at a distance R, we use the Coulomb law: F = k ∙ q₁ ∙ q₂ / R ^ 2, where q₁ and q₂ are the charges of interacting bodies, the coefficient k = 9 ∙ 10 ^ 9 (N ∙ m ^ 2) / Cl ^ 2. Since the proton charge is equal to the elementary charge: q₁ = q₂ = e, then:

F = k ∙ e ^ 2 / R ^ 2, where the charge e = 1.6 ∙ 10 ^ (- 19) Cl.

From the condition of the problem it is known that the distance between the protons is R = 1.6 ∙ 10 ^ (- 14) m. Substitute the values ​​of the quantities into the calculation formula:

F = 9 ∙ 10 ^ 9 (N ∙ m ^ 2) / Kl ^ 2 ∙ (1.6 ∙ 10 ^ (- 19) Kl) ^ 2 / (1.6 ∙ 10 ^ (- 14) m) ^ 2 ;

F = 0.9 N.

Answer: the force of interaction of two protons is 0.9 N.




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