Question

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

Answer 1

Answer:

Therefore the magnitude of required electric force is  292.864×10⁻⁵ N.

Explanation:

The number of proton of iron is 26.

$1.6\times 10^{-19}$  C is the charge of each proton.

Total charge of 26 proton is =(26×1.6×10⁻¹⁹) C

                                             =41.6×10⁻¹⁹ C

The charge of innermost electron is = - $1.6\times 10^{-19}$ C

Coulomb's Law:

The attraction or repulsion force of two charges is

• directly proportional to the product of the charges.

• inversely proportion to the square of distance    

Let $q_1 \ and \ q_2$  be two positive charge and the distance between the charges be r

Then F∝$q_1 q_2$

and  $F \propto \frac{1}{r^2}$

Therefore,

$F=\frac{kq_1q_2}{r^2}$

Here

$q_1$=   - $1.6\times 10^{-19}$C   and   $q_2$  =41.6×10⁻¹⁹ C

r = 1.5× 10⁻¹² m

K= 9.9×  10⁹C²N⁻¹m⁻²

Therefore

$F=\frac{9.9 \times 10^9\times (-1.6\times 10^{-19})\times 41.6\times 10^{-19}}{(1.5\times 10^{-12})^2}$   N

  = -292.864×10⁻⁵ N

The mins sign shows the direction of the force.

Magnitude of electric force is only the numerical value of the force.

Therefore the magnitude of required electric force is  292.864×10⁻⁵ N.