Question

What is the magnitude of the electric force of attraction between an iron nucleus (q=+26e)(q=+26e) and its innermost electron if the distance between them is 1.5×10−12m1.5×10−12m?

Answer 1

The magnitude of the electric force of attraction between an iron nucleus is
`6.92* 10^(-2)\ N`.

Step-by-step explanation:

Charge on an iron nucleus,
`q_1=q_2=26e=26* 1.6* 10^(-19)=4.16* 10^(-18)\ C`

The separation between the iron nucleus is,
`r=1.5* 10^(-12)\ m`

There is an electric force or repulsion between two charges. It is given by :

`F=(kq^2)/(r^2)`

`F=(9* 10^9* (4.16* 10^(-18))^2)/((1.5* 10^(-12))^2)`

F = 0.0692 N

or

`F=6.92* 10^(-2)\ N`

So, the magnitude of the electric force of attraction between an iron nucleus is
`6.92* 10^(-2)\ N`. Hence, this is the required solution.