Potential due to charge over infinite grounded plane conductor using the method of images I am reading section 3.2.1 of Griffiths 3ed which explains how to calculate potential using first uniqueness theorem.
Griffiths/3.2.1

Griffiths/First Uniqeness theorem (Its corollary actually)



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*As I understand we are taking volume $V = R^3$ , $\Phi (\vec{r})=0\ $on the boundary of $V$. But what does it mean by "boundary of $R^3$" ?

*Since $\ \rho_1(\vec{r}) \ne \rho_2(\vec{r})$ which does not satisfy the prequisites of the theorem, we cannot use the theorem. Moreover no other distribution can be used which makes this theorem useless which it is not. So what am I missing here ?

*Does not single charge distribution also satisfies the same boundary condititons ? So should not $\Phi$ also be $\frac{1}{4\pi \epsilon_0}\frac{q}{R}$ ?

 A: Question 1:
You're running into a problem because you're applying the uniqueness theorem to too large of a volume.  The bottom of an appropriate volume lies on the $z=0$ plane, where $\Phi=0$ because there's a grounded conducting plate there.  The rest of the volume's boundary is an arbitrary surface that completes the enclosure of the volume and consists of points such that $z>0$ and $x^2 + y^2 + z^2 \gg d$, the latter condition being to ensure that $\Phi \approx 0$ is a good approximation at that point of the boundary.  You can effectively think of all but the bottom of the volume's boundary as being "at infinity", loosely speaking.
Question 2:
$\rho_1(\vec{r})=\rho_2(\vec{r})$ at all points for which you need to solve for $\Phi$.  For all points for which you need to solve for $\Phi$, $z>0$, strictly.  You don't need to solve for $\Phi$ at points precisely on the boundary, because you already know that $\Phi=0$ everywhere on the $z=0$ plane, because there's a grounded conducting plate there.
Question 3:
No, a single charge doesn't satisfy the same boundary conditions, which is presumably obvious now that the bottom of the boundary is understood to be on the $z=0$ plane, instead of the volume in question being all of space.
