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

STEM2012
Messages
7
Reaction score
0
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
 
Physics news on Phys.org
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.
 
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.
 
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).
 
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
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top