Show the potential obeys Poisson's Equation

[azupol]

Homework Statement

In class we were given the classic image problem of a point charge on the z-axis (at z=d) above an infinite grounded conducting plane, the xy plane in this case. We found the potential by getting rid of the plane and placing an image charge at a distance z= -d. Now the question is, how do we show that this potential obeys Poisson's equation in the region of interest, in this case z>0

Homework Equations

Poisson's Equation:

And the potential was found to be: V(x,y,z)=1/4piε [q/(x² + y² + (z-d)² )^1/2 ] -q/(x² + y² + (z+d)² )^1/2 ]Induced surface charge density (i.e. the induced charge in the conducting plane) is -qd/2pi(x²+y²+d²)^3/2

The Attempt at a Solution

I attempted to twice differentiate the potential, and tried to make it equal to the induced surface charge density, to check that after simplification they were both equal, but got stuck at the second partial derivative of V with respect to z. There must be a better way to solve this.

We are using Griffith's Introduction to Electrodynamics, and our prof suggested we reference the section on the Dirac delta function. I'm failing to see how the Dirac Delta would help in this case

 

[Orodruin]

The induced surface charge density is at z = 0 and not at z > 0. You have replaced all of it by a mirror charge in z = -d and should not expect it to come out of applying the Laplace operator to the potential. As long as you get a Dirac distribution in z = d, x = y = 0, you are fine.

 

[azupol]

I'm not quite sure I understand what you mean. I'm aware the induced surface charge is in the plane, at z=0. To compensate for the induced surface charge, we place an appropriate charge a distance below the xy plane such that the boundary condition of V=0 at z=0 would be satisfied. What I'm wondering is how can I show that this potential obeys Poisson's equation?

 

[BruceW]

hmm. This is a problem you can think about intuitively. Poisson equation is
\nabla^2 V = - \frac{\rho}{\varepsilon_0}
And you have two situations 1) charges on plate. 2) image charge. (and in both situations, there is also the original charge). You are trying to show that $V$ satisfies the same Poisson equation in both situations, for z>0. In other words, you want to show that
\nabla^2 V_1 = \nabla^2 V_2
for z>0. So, from this, what must be the relationship between $\rho_1$ and $\rho_2$ when z>0 ? And can you show that this relationship does hold in this case?

 

[azupol]

It seems ρ₁ = ρ₂ in z>0, otherwise the potentials would not agree right? I'm not sure what I can do to show the relationship agrees, maybe take V₃=V₂-V₁ and show that difference must result in a charge of 0 in z>0?

 

[Orodruin]

You seemed to want to get a surface charge at z = 0. This is not going to happen as it is not part of your volume. The only charge you should be getting is the original one. Think of this: What is the Laplace operator applied to the potential of the mirror charge?

Edit: Note that the solution to Poisson's equation is unique given the boundary conditions and charge density. Thus if you find one solution that gives the correct charge and boundary conditions, it is unique. The way to show this is very similar to what you proposed to do.

 

[BruceW]

It seems ρ₁ = ρ₂ in z>0, otherwise the potentials would not agree right?

right, so now you want to show that $\rho_1=\rho_2$ in the region z>0. You are given the second potential explicitly. You can get the answer just with reasoning here, but since your professor hinted at using Dirac Delta functions, you should probably apply the Laplace operator to the potential, as Orodruin says. It is not a standard kind of calculation, you need to use the properties of the Dirac Delta, like your professor hinted at.

 

[azupol]

Thanks for the pointers, I think I've figured it out. Applying the Laplace operator to the potential, I get that $\Delta V$ will go to infinity at (0, 0, d), but this is precisely calculating the divergence of 1/r², which is $4\pi \delta (r)$ and I think I can take it from there, thanks for the help!