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Modifying Coulomb's Law for Use in Particle Energy (chemistry)

  1. Aug 17, 2011 #1
    New to PF.

    How is the "original" form of coulomb's law F=kQ1Q2/D^2 derived into the modified chemistry form used to predict the energy released when bonds form (or the inverse), E=kQ1Q2/r?

    Please describe your mathematical steps. Feel free to just post links explaining this. I've searched everywhere...I'm only a high school student, trying to write a book but do not have the greatest resources.

    Thanks in Advance
  2. jcsd
  3. Aug 17, 2011 #2
    E equals the integral of Force*dx. So that's how you get from force to energy (and vica versa). There's actually a minus sign E=-kQ1Q2/r. There's also a constant of integration but we conventionally say the two particles have zero potential energy when they are infinite distance apart.
  4. Aug 17, 2011 #3
    Okay, now that the relationship between E and F is established, I can say that
    dE/dx=-kQ1Q2/d^2. Bu from here, how do I get to the modified form, where r is the denominator.
  5. Aug 17, 2011 #4
    dE/dr=(kQ1Q2)r^-2. No minus sign here. We'll ignore the denominator and just say r is raised to the negative two power.

    The general form, when df/dr=r^k has solution f(r)=(1/(k+1))r^(k+1).

    Ok. So we plug in k=-2 and get (-1)r^(-1). So restoring the constants E=kQ1Q2(-1)r^(-1).
  6. Aug 17, 2011 #5
    Ok. I understand how you integrated from dF/dr=r^k to find a solution, but how the heck did you get that differential to begin with? Also, I don't understand where did the k=-2 come from?

    Keep in mind, I'm only in high school and my highest level of math education is AP calc AB (which is calc I and half of calc II) so you probably have to be more thorough then when you're usually talking to mathemeticians, physicists, etc.

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