Magnitude of Electric Force Between Two Ions

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SUMMARY

The magnitude of the electric force between a Na+ ion and a Cl- ion separated by 0.5 nm can be calculated using Coulomb's Law, represented by the equation |F| = (kQq)/r^2. The correct values for Q and q should be the charges of the ions, which are +1e for Na+ and -1e for Cl-. The initial calculation of 9.23 * 10^-10 N was incorrect due to a misunderstanding of the charge contributions from electrons and protons. The final force should be calculated using the fundamental charge of an electron, 1.6 * 10^-19 C, without multiplying by the number of electrons or protons.

PREREQUISITES
  • Coulomb's Law
  • Understanding of ionic charges
  • Basic knowledge of electric forces
  • Familiarity with the concept of nanometers
NEXT STEPS
  • Review Coulomb's Law and its applications in electrostatics
  • Study the properties of ions, specifically Na+ and Cl-
  • Learn about the concept of electric charge and its quantization
  • Explore the relationship between distance and electric force in electrostatic interactions
USEFUL FOR

Students studying physics, particularly those focusing on electrostatics, as well as educators looking to clarify concepts related to ionic interactions and electric forces.

tylerc1991
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Homework Statement


Find the magnitude of the electric force between a Na+ ion and a Cl- ion separated by .5 nm.


Homework Equations


|F| = (kQq)/r^2


The Attempt at a Solution


So the problem I am having with this question is this: my first attempt at a solution was to designate |Q| = |q| = 1.6 * 10^-19 C. After which I can just plug these into the equation. But after visiting my professor he told me to multiply Q by 10 (because there are 10 electrons in the Na+ ion) and likewise to multiply q by 18 (because Cl- has 18 electrons). This makes very little sense to me and I hope that someone can help me with this problem. By the way, the answer I originally obtained was 9.23 * 10^-10 N.
 
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I think what your professor told you is incorrect.
A Na atom for example is neutral, it may have 11 electrons but it also has 11 protons.
Your Na+ ion has 10 electrons yes, but it also has 11 protons.

The only thing I can think of is maybe he wants you to do all the charges, so putting in 10 electrons and also 11 protons for Na+. Which of course is the same as just treating it as a single +e charge.
 

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