Question about the classic Image Problem

In summary, the conversation discusses the potential in a region above an infinite grounded conducting plane with a point charge held at a distance above it. The method used is to create another system with the same boundary conditions using the uniqueness theorem of Laplace's equation. However, there is a singularity at the point where the charge is located, which violates the equation. The region of interest is the space above the plane, excluding the point where the charge is located.
  • #1
zhuang382
10
2
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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  • #2
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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  • #3
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##?
 
  • #4
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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1. What is the classic Image Problem?

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