Calculating the electrostatic force

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The discussion revolves around calculating the electrostatic force between a Ca2+ ion and an O2– ion separated by 1.25 nm using Coulomb's Law. The correct attractive force calculation yields 5.89 x 10^-10 N, but confusion arises regarding the distinction between attractive force and electrostatic force. It is clarified that the electrostatic force can be either attractive or repulsive, depending on the charge signs, and in this case, it is indeed attractive. The notion of force being positive or negative is explained as context-dependent, with directionality being significant in one-dimensional scenarios. Ultimately, the electrostatic force is recognized as negative due to the attractive nature between the oppositely charged ions.
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).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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