1
$\begingroup$

My text poses the following question/answer:

Suppose that the potential between two point charges $q_1$ and $q_2$ separated by $r$ were actually $\frac{q_1q_2e^{-Kr}}{r}$, instead of $\frac{q_1q_2}{r}$, with $K$ very small. What would replace Poisson's equation for the electric potential?

Answer: With the assumption given, Poisson's equation is to be replaced by \begin{equation} \nabla^2\phi - K^2 \phi = -4\pi \rho \end{equation} in Gaussian units where $\rho$ is the charge density.

There is not an explanation for why Poisson's equation would be replaced by the above, so I feel like I am missing some understanding of how the $\frac{1}{r}$ potential implies that the potential is $\nabla^2\phi = -4 \pi \rho$, and how this relationship is effected by the altered potential such that we have this solution. Could someone please elaborate how we can reach the solution's conclusion a priori?

As I understand the relationship between $\frac{1}{r}$ and Poisson's equation is that it is the Green's function of the Laplacian such that it produces the correct charge density of a point charge and zero everywhere else, is this somehow to be useful in this derivation?

EDIT: The original book's solution had a typo, the book's solution implies that the Yukawa Potential is the Green's function for the Helmholtz equation when in fact it is the Green's function for the “static Klein-Gordon equation", I have corrected this sign change in the latest edit.

$\endgroup$
  • $\begingroup$ It's a little hard not knowing more about the context the question is posed. I'm going to guess that you are already supposed to know the relationship between Poisson's eqn and $1/r$, and the exercise here is to play around with the math until you find the correct modification. $\endgroup$ – garyp Aug 8 '16 at 17:25
  • $\begingroup$ The relationship between those two is that $1/r$ is the Green's function of the Laplacian such that it produces the correct charge density of a point charge and zero everywhere else, do you suppose that we make use of this? $\endgroup$ – Loonuh Aug 8 '16 at 17:29
  • $\begingroup$ It is experience that this Yukawa potential is the right Green's function for this modified equation. You can of course see it by just plugging the modified potential into the original Poisson's equation and see which additional term occurs, maybe also play around with Fourier transforms, but well ... Btw, your potential has two charges instead of just one. $\endgroup$ – Sanya Aug 8 '16 at 17:32
  • $\begingroup$ If you start with the differential equation, you can look for the Green's function, yes. But this question gives you the Green's function and asks you to find the differential equation. I haven't tried this, but if you just take the second derivative of the potential, you might end up with an expression for which it's obvious what you have to do to form a complete differential equation $\endgroup$ – garyp Aug 8 '16 at 17:32
  • 1
    $\begingroup$ @Sanya, you are indeed correct about the Yukawa potential here, actually my book has a typo! The '+' in their provided equation to solve needs to be a -' instead! $\endgroup$ – Loonuh Aug 8 '16 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.