Can someone show me how Green's function would apply for this simple case? I'm reading up on some stuff on basic electrostatic here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/laplace.html

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.
 A: 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. 
