Respuesta :
Answer:
The magnitude of the electric force of attraction between an iron nucleus is [tex]6.92\times 10^{-2}\ N[/tex].
Explanation:
Charge on an iron nucleus, [tex]q_1=q_2=26e=26\times 1.6\times 10^{-19}=4.16\times 10^{-18}\ C[/tex]
The separation between the iron nucleus is, [tex]r=1.5\times 10^{-12}\ m[/tex]
There is an electric force or repulsion between two charges. It is given by :
[tex]F=\dfrac{kq^2}{r^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (4.16\times 10^{-18})^2}{(1.5\times 10^{-12})^2}[/tex]
F = 0.0692 N
or
[tex]F=6.92\times 10^{-2}\ N[/tex]
So, the magnitude of the electric force of attraction between an iron nucleus is [tex]6.92\times 10^{-2}\ N[/tex]. Hence, this is the required solution.
