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

1. Aug 17, 2011

### STEM2012

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.

2. Aug 17, 2011

### robert2734

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.

3. Aug 17, 2011

### STEM2012

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.

4. Aug 17, 2011

### robert2734

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).

5. Aug 17, 2011

### STEM2012

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.

Thanks