Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating the electrostatic force

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.

    3. 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!
  2. jcsd
  3. Jan 10, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
  4. Jan 10, 2010 #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?
  5. Jan 12, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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 .
  6. Jan 12, 2010 #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).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook