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 ï½˜10⁻³ N) 



Given:

∴  Charge on electron = e = 1.6 ï½˜ 10⁻¹⁹ C 

Charge on nucleus = q = 26 e = 26 ï½˜ 1.6 ï½˜ 10⁻¹⁹ C = 41.6 ï½˜ 10⁻¹⁹ C  = 4.16 ï½˜ 10⁻¹⁸ C 

Separation  distance = r = 1 ï½˜ 10⁻¹² m

∴  Coulomb's Constant = K = `\frac {1}{4ㄫε_0}`= 9 ï½˜ 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 = k`\frac {eq}{r^2}`

by putting values 

F = 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {1.6 ｘ 10⁻¹⁹ C ｘ 4.16 ｘ 10⁻¹⁸ C}{(1 ｘ 10⁻¹² m)^2}`

F = 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {6.656 ｘ 10⁻¹⁹⁻¹⁸ C}{1 ｘ 10⁻²⁴ m^2}`

F = 9 ｘ 10⁹ Nm²C⁻² ｘ `\frac {6.656 ｘ 10⁻³⁷ C²}{1 ｘ 10⁻²⁴ m^2}`

F = `\frac {59.9 ｘ 10⁹⁻³⁷ Nm²}{1 ｘ 10⁻²⁴ m^2}`

F = `\frac {59.9 ｘ 10⁻²⁸ Nm²}{1 ｘ 10⁻²⁴ m^2}`

F = 59.9 ï½˜ 10⁻²⁸⁺²⁴N

F = 59.9 ï½˜ 10⁻⁴N

F = 5.99 ï½˜ 10⁻³N ---------------Ans.



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