Question about the classic Image Problem

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The discussion centers on the classic image problem involving a point charge above an infinite grounded conducting plane. The potential in the region above the plane must satisfy Laplace's equation, but the presence of the point charge introduces a singularity at the charge's location. The participants clarify that the region of interest excludes the singularity, focusing on the area where z > 0 and (x, y, z) ≠ (0, 0, d). They emphasize that the singularity does not violate the conditions of Laplace's equation, as the governing equation accounts for the charge's presence through a delta function representation. Ultimately, the discussion highlights the importance of correctly defining the region of interest to analyze the potential effectively.
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Homework Statement
Question about Laplace' equation and the classic Image Problem
Relevant Equations
##\nabla^2 V = 0##
##\nabla^2 V = \frac{-p}{\epsilon_0}##
I am studying the classic image problem (griffins, p. 124)

Suppose a point charge ##q## is held a distance ##d## above an infinite grounded conducting plane. Question: What is the potential in the region above the plane?

boundary conditions:

1. V = 0 when z = 0 (since the conducting plane is grounded)
2. V ~ 0 far from the charge ( that is ## x^2 +y^2 + z^2 >> d##)

The method is basically to use uniqueness theorem of Laplace's equation to create another system with the same boundary conditions. Therefore the potential function will be uniquely determined.

My question is:
since the region we are interested in is all the region of ##z > 0##, the Laplace's equation has to be satisfied i.e. ##\nabla^2 V = 0##
But the problem is that there is a point charge within this region, at ##z = d##. How to understand this?
 
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At that point the potential has a singularity. No wonder: ##\rho = \infty## there.

Your problem exists whether there is a grounded plane present or not
 
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BvU said:
At that point the potential has a singularity. No wonder: ##\rho = \infty## there.

Your problem exists whether there is a grounded plane present or not
So we are actually interested in the region that ##\{(x,y,z)| z > 0, (x,y,z) \neq (0,0,d)\}##?
Or a singularity at ##z = d## do not violate ##\nabla^2 V = 0##?
 
Of course it does. The equation you want to solve is
$$\Delta V(\vec{r})=-\frac{1}{\epsilon_0} \rho(\vec{r})=-\frac{q}{\epsilon_0} \delta^{(3)}(\vec{r}-\vec{r}_0),$$
where ##\vec{r}_0=(0,0,d)## with the boundary condition mentioned in #1.
 
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