Question about the classic Image Problem

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Homework Help Overview

The discussion revolves around the classic image problem involving a point charge above an infinite grounded conducting plane. Participants are exploring the implications of boundary conditions on the potential in the region above the plane, particularly in relation to Laplace's equation and the presence of a singularity due to the point charge.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining the uniqueness theorem of Laplace's equation and its application to the problem. Questions arise regarding the implications of a singularity at the location of the point charge and how it affects the solution to Laplace's equation in the specified region.

Discussion Status

The discussion is actively exploring the nature of the singularity at the point charge and its compatibility with the boundary conditions. Some participants are questioning whether the region of interest can exclude the singularity without violating the conditions of Laplace's equation.

Contextual Notes

Participants note that the potential has a singularity at the location of the point charge, which raises questions about the validity of applying Laplace's equation in that region. The discussion is framed within the constraints of the problem's boundary conditions and the nature of the charge distribution.

zhuang382
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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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