Calculate Magnitude of q & Electric Field Strength for Na+ & Cl- Ions: 0.24nm

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The separation between Na+ and Cl- ions is 0.24nm, with an attractive force of 4.0x10^-19N. To calculate the magnitude of the charge (q), Coulomb's law is used, yielding a value of approximately 1.03x10^-14C. For the electric field strength at the midpoint, it is noted that the fields from the two charges do not cancel due to their opposite signs. The discussion emphasizes the need to apply the concept of electric fields from point charges to find the solution. Understanding these principles is crucial for accurately determining the electric field strength in this context.
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Question The separation between Na+ and Cl- ions in a molecule is 0.24nm. They are point charges of +q and -q and force of attraction between them is 4.0x10^-19N.

a) Get the magnitude of the q
Possible Answer: Do I just use Coloumbs law? I got 1.03x10^-14C

b) What is the electrical field strength (in magnitude and direction) at the mid-point of the molecule?
Possible Answer: Not sure what to do here? I'm thinking maybe its 0 but that would just be too easy.

Thanks in advance!? :smile:
 
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sci0x said:
a) Get the magnitude of the q
Possible Answer: Do I just use Coloumbs law? I got 1.03x10^-14C
I didn't check your answer, but using Coulomb's law is all you need to do.

b) What is the electrical field strength (in magnitude and direction) at the mid-point of the molecule?
Possible Answer: Not sure what to do here? I'm thinking maybe its 0 but that would just be too easy.
You just found the charges in part a, so now find the field strength. What's the field at a distance from a point charge? (Hint: the charges have opposite signs, so the fields don't cancel.)
 

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