Calculating the electrostatic force

In summary, using Coulomb's Law, the electrostatic force between a Ca2+ and an O2– ion separated by a distance of 1.25 nm is calculated to be -5.89 x 10-10 N. The electrostatic force can be either attractive or repulsive, depending on the signs of the charges involved. In this case, it is attractive because the charges are opposite in sign. The value of the electrostatic force being negative indicates the direction of the force on the Ca2+ ion, while the force on the O2– ion would be positive.
  • #1
Hockey07
6
0

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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  • #2
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.
 
  • #3
Thank you. :)

One more question:

Would the value of the electrostatic force be negative because it's attractive? And positive if it's repulsive?
 
  • #4
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 .
 
  • #5
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).
 

FAQ: Calculating the electrostatic force

How is the electrostatic force calculated?

The electrostatic force is calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

What is the unit of measurement for electrostatic force?

The unit of measurement for electrostatic force is the Newton (N). This is the same unit used to measure other types of forces, such as gravity.

Can the electrostatic force between two particles be negative?

Yes, the electrostatic force can be negative if the two particles have opposite charges. In this case, the force will be attractive, pulling the particles towards each other.

How does distance affect the electrostatic force?

According to Coulomb's law, the electrostatic force is inversely proportional to the square of the distance between two charged particles. This means that as the distance between the particles increases, the force decreases.

What happens to the electrostatic force if the charges of the particles are doubled?

If the charges of the particles are doubled, the electrostatic force between them will also double. This is because the force is directly proportional to the product of the charges.

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