1
$\begingroup$

I'm reading up on some stuff on basic electrostatic here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

enter image description here

Can someone use Green's function to show me the form of $V$?

Update:

I have made an attempt to solve for $V$ but I can't understand why it takes the form $$V = \frac{\rho}{4\pi\epsilon_o} \int \frac{1}{|r-r'|}dr^3$$

(can someone check my solution)

I can't recall this solution from anything I've previously encountered. What is the meaning of this $V$. What is the underlying geometry I'm working with? Is anyone familiar with this form of $V$?

Also, how would you derive $E$ from $V$ given the above expression? I think if I see $E$ then it would make more sense.

$\endgroup$
5
$\begingroup$

I'll try to make a simple derivation. Suppose you have a unit point charge located at position $\vec{r}'$. Then the associated charge density is $\delta(\vec{r}-\vec{r'})$, which is a Dirac distribution. The electrostatic potential produced by this charge is given by the Coloumb's law: $$G(|\vec{r}-\vec{r}'|) = \frac{1}{4\pi\varepsilon_0}\cdot\frac{1}{|\vec{r}-\vec{r}'|}.$$ Note that it obviously satisfies the Laplace equation: $$\partial^2 G(|\vec{r}-\vec{r}'|) = \frac{\delta(\vec{r}-\vec{r}')}{\varepsilon_0}.$$ Now suppose we have an arbitrary distribution of charges with density $\rho(\vec{r})$. One can write this density as a sum of point charges: $\rho(\vec{r}) = \int d^3\vec{r}' \rho(\vec{r}')\delta(\vec{r}-\vec{r}')$. Next, we multiply the above equation by $\rho(\vec{r}')$ and integrate over $\vec{r}'$, we obtain $$\partial^2\int d^3\vec{r}' G(|\vec{r}-\vec{r}'|) \rho(\vec{r}') = \frac{\rho(\vec{r})}{\varepsilon_0}.$$ Now by comparison, you find the electrostatic potential due to $\rho(\vec{r})$ is given by $$V(\vec{r}) = \int d^3\vec{r}' G(|\vec{r}-\vec{r}'|)\rho(\vec{r}').$$ This may be what you seek.

$\endgroup$
  • $\begingroup$ Do you mean 'obviously satisfies Poisson's equation'? I don't personally find this obvious --- first you must show that the Laplacian of $G$ vanishes for $r \neq r'$, and then you must show that the integral of the Laplacian of $G$ over a ball of radius $\epsilon$ centred on $r = r'$ gives you the $1/\epsilon_0$ that you seek. I would also say that the notation $\nabla^2$ is more common than $\partial^2$, and since the questioner used the former in their answer, I would favour it. As a final point, you're missing the $d^3 r'$ on one of your integrals. $\endgroup$ – gj255 Aug 28 '14 at 9:25
  • $\begingroup$ (1) By 'obviously satisfying ...', I mean it on the physics not mathematics: Coulomb's law is a solution to Possion's equation (thanks for correcting me for using Laplace equation) in the presence of a point charge. (2) I would have used the symbol by the questioner, but I do not know how to make it in Latex. (3) Thanks for pointing out that the missing $d^3\vec{r}'$. $\endgroup$ – hyd Aug 28 '14 at 10:29
  • $\begingroup$ Ah OK, I see what you're saying. $\endgroup$ – gj255 Aug 28 '14 at 11:43
  • $\begingroup$ Thanks, physically it makes sense, however, how would you go about deriving $G(|r-r'|)$ from scratch without invoking coloumb's law? $\endgroup$ – Carlos - the Mongoose - Danger Aug 29 '14 at 1:35
  • $\begingroup$ There are various ways to show that explicitly. You may take the Fourier transform of the Poisson equation. Then you see $\tilde{G}(\vec{k}) = \varepsilon^{-1}_0k^{-2}$, where $\tilde{G}$ is the Fourier transform of $G$ and $k=|\vec{k}|$. Transforming back , you obtain $G(\vec{r}) = \varepsilon^{-1} \int d^3\vec{k} \frac{e^{i\vec{k}\vec{r}}}{k^2}$. To perform this integral, you can write $d^3\vec{k} = k^2\sin(\theta)dk d\theta d\varphi$ and $\vec{k}\vec{r} = kr\cos(\theta)$, where we have chosen the $\vec{r}$ along z-axis. Now you can use standard formula to get the final result. $\endgroup$ – hyd Aug 29 '14 at 2:37

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.