Calculating the electrostatic force

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 10K views
Hockey07
Messages
6
Reaction score
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!
 
Physics news on Phys.org
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).