What is the magnitude of the electric force between two electrons separated by a distance of 0.14 nm (approximately the diameter of an atom)

Respuesta :

Answer:

[tex]F = 1.174*10^{-8} N[/tex]

Explanation:

Conceptual analysis:

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

[tex]F = \frac{k*q_{1}*q_{2}}{r^2}[/tex] Formula (1)

[tex]K=8.99*10^9 \frac{N*m^2}{C^2}[/tex]: Coulomb constant

q1, q2 = charge in Coulombs (C)  

Known information :

r = 0.14 nm

[tex]1 nm = 10^{-9} m [/tex]

[tex]r=0.14*10^{-9} m [/tex]

[tex]q_{1} = -1,6*10^{-19}  C[/tex]

[tex]q_{2} = -1,6*10^{-19}  C[/tex]

Development of the problem :

Replacing the information known in formula (1):

[tex]F = \frac{8.99*10^9*1,6*10^{-19}*1,6*10^{-19}}{(0.14*10^{-9})^2}[/tex]

[tex]F = 1.174*10^{-8} N[/tex]