What is the magnitude of the force of attraction between an iron nucleus bearing charge q = 26e and its innermost electron, if the distance between them is 1ｘ10⁻¹² m? (Answer 6 ｘ10⁻³ N) 



Given:

∴  Charge on electron = e = 1.6 ｘ 10⁻¹⁹ C 

Charge on nucleus = q = 26 e = 26 ｘ 1.6 ｘ 10⁻¹⁹ C = 41.6 ｘ 10⁻¹⁹ C  = 4.16 ｘ 10⁻¹⁸ C 

Separation  distance = r = 1 ｘ 10⁻¹² m

∴  Coulomb's Constant = K = 14ㄫε0= 9 ｘ 10⁹ Nm²C⁻²

To Find:

Coulomb's force of attraction  = F = ?

Solution:

Using Coulomb's law, the force of attraction between the electron of charge "e" and iron nucleus of charge "q" is given by:

F = keqr2

by putting values 

F = 9 ｘ 10⁹ Nm²C⁻² ｘ 1.6ｘ10⁻¹⁹Cｘ4.16ｘ10⁻¹⁸C(1ｘ10⁻¹²m)2

F = 9 ｘ 10⁹ Nm²C⁻² ｘ 6.656ｘ10⁻¹⁹⁻¹⁸C1ｘ10⁻²⁴m2

F = 9 ｘ 10⁹ Nm²C⁻² ｘ 6.656ｘ10⁻³⁷C²1ｘ10⁻²⁴m2

F = 59.9ｘ10⁹⁻³⁷Nm²1ｘ10⁻²⁴m2

F = 59.9ｘ10⁻²⁸Nm²1ｘ10⁻²⁴m2

F = 59.9 ｘ 10⁻²⁸⁺²⁴N

F = 59.9 ｘ 10⁻⁴N

F = 5.99 ｘ 10⁻³N ---------------Ans.



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