Calculating the electrostatic force

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Homework Help Overview

The discussion revolves around calculating the electrostatic force between a Ca2+ ion and an O2– ion, with a specified separation distance of 1.25 nm. The original poster references Coulomb's Law and expresses confusion regarding the distinction between attractive force and electrostatic force.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the difference between attractive force and electrostatic force, questioning whether the calculation should consider valence electrons. Other participants clarify that Coulomb's Law provides the electrostatic force, which can be attractive or repulsive based on charge signs.

Discussion Status

Participants are exploring the nuances of force directionality and the implications of charge signs on the electrostatic force. Some guidance has been offered regarding the interpretation of force values, but there remains uncertainty about the sign of the force in relation to attraction and repulsion.

Contextual Notes

The discussion includes considerations about the definition of positive and negative forces in different dimensional contexts and the implications of Newton's third law on force interactions.

Hockey07
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Homework Statement



Calculate the (electrostatic) force between a Ca2+ and an O2– ion the centers of which are separated by a distance of 1.25 nm.

Homework Equations



I know that:

Coulomb's Law is F = (k q1 q2)/r2

Attractive Force is F = (k z1e z2e)/r2
where z1 and z2 are the valence electrons, and e is the charge of an electron.

The Attempt at a Solution



The textbook problem asks for the attractive force, so:

F = [ (9 x 109)(2)(2)(1.6 x 10-19)2 ] / (1.25 x 10-9)2

Which is 5.89 x 10-10 N

That's the correct answer to the attractive force, but the question my instructor has posed is to find the electrostatic force.

I'm unsure what the difference between the attractive force and electrostatic force is. Would I just calculate it using Coulomb's Law without taking into account the valence electrons?

Thanks in advance!
 
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Coulomb's Law gives the electrostatic force, and you have correctly calculated it.

The electrostatic force can be either attractive or repulsive, depending on the signs of the charges involved. In this case it happens to be attractive.

p.s. Welcome to Physics Forums.
 
Thank you. :)

One more question:

Would the value of the electrostatic force be negative because it's attractive? And positive if it's repulsive?
 
Hockey07 said:
Thank you. :)

One more question:

Would the value of the electrostatic force be negative because it's attractive? And positive if it's repulsive?
No.
Saying that a force is positive vs. negative is meaningful only in a 1-dimensional situation. In that case, you would need to define which direction is positive and which is negative, or at least make it clear from the context. Furthermore, the force on one charge would be positive while the force on the other would be negative, in accordance with Newton's third law .
 
The answer to this problem was actually negative. I know that a force vector being positive or negative would indicate direction (which is what I think you were saying).

However, if there were a positive charge and a negative charge (which attract), Coulomb's law would be negative. That's why I thought it would be negative in this case (because it's an attractive force).
 

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