Question

The nearest neighbor interaction force is of magnitude 481 nanoNewtons, e.g., the magnitude of the force of the leftmost electron on the proton, or the magnitude of the force of any of the three on its nearest neighbor electron. Calculate the size of the net force on the leftmost proton.

Answer 1

Answer:

 F = 120.25 10⁻⁹  N

Explanation:

In this exercise, the force between the closest neighbors is indicated by f = 481 10⁻⁹ N, in general between the one-dimensional solid the distances remain the same, if the distance between the first neighbor is d, the distance between the second neighbors is 2d.

For most solids the attractive forces are electrical, therefore force is proportional to the electrical charges and the inverse of the distance squared,

             F = $k \frac{q_1 q_2}{r^2}$

if we call fo the force for the first neighbors

              F₀ = k \frac{q_1 q_2}{d^2}

the force for the second neighbors r= 2d  

              F = k \frac{q_1 q_2}{(2d)^2}

              F = F₀ / 4

let's calculate

              F = 481 10⁻⁹ / 4

              F = 120.25 10⁻⁹  N