# Finding the potential by Green's function

1. Apr 15, 2015

### MMS

1. The problem statement, all variables and given/known data
An infinite plane at z=0 is divided into two: the right half of it (y>0) is held at potential zero and the left half of it is held at potential \phi_0
.

Over this surface lies a point charge q at (0,y_0,z_0)
.

Use Green's function to calculate the potential at z>0.

3. The attempt at a solution

I'd appreciate it if someone could take a look at my attempt at solution as I'm not sure whether what I've gotten to is right or wrong. Mainly because I also have to calculate other things that require taking the derivative of this expression (surface charge density and force acting on the charge).
Also, I couldn't calculate the integral (Wolfram won't read it for some reason) which makes it even more suspicious. If you plug it into some program and you get an answer, please, let me know.

P.S: I've noticed earlier that someone had replied to my older post which has been deleted and I couldn't reply at the time and I can't recall what had been said.

2. Apr 15, 2015

### Telemachus

Hi, I have answered before. So here I go again. The terms in eq. (0.2) cancel. I think you have placed wrong the image charge (in the same place as the real charge), it should be located at the point $(0,-y_0,-z_0)$. You will have plus signes in the denominator for the second term in your green's function.

$\frac{q}{\sqrt{x^2+(y+y')^2+(z+z')^2}}$.

Anyway, this image charge will make the whole z=0 plane to be at potential=0. I never worked electrostatic problems with greens function, so I'm not familiar with this. The thing that is left to be done, is to set the other half plane at the potential $\phi_0$. I don't know if the surface integral in Greens theorem accounts for that (but I believe that it does). You can check Jackson's "Classical Electrodynamics" for reference.

Last edited: Apr 15, 2015
3. Apr 15, 2015

### MMS

First off thanks for the reply and the care to reply again. I appreciate it.

Second, you are completely right about the location of the image charge location in the z direction (on paper I wrote z+z' but I copied it wrong in lyx). As for the y direction, I don't think it's -y_0.
I think it needs to be the "mirror" charge to the one given from underneath the surface, hence (0,y_0,-z_0).
Don't you think so?

4. Apr 15, 2015

### Orodruin

Staff Emeritus
You are correct. Putting the image charge in -y_0 would lead to the Green's function not satisfying the correct BCs. This is an easy check. You are mirroring in the z=0 plane.

As MMS said, the y coordinate should not be mirrored. Inserting z=0 into both potential of the charge and mirror charge should give the same expression with opposite sign.

5. Apr 15, 2015

### MMS

From what I know, the potentials given on the surface simply determine your boundary conditions so I don't think there should be a problem with that. The only affect they'll have is by plugging them into the integrals with the right intervals for each coordinate.
For each problem (very specific ones) you must have some sort of symmetry (spherical/cylindrical/planar) in order to use the method of images and for it to work. Using Green, you have pretty specific patterns for each symmetry and in this case, it is the plane with the two image charges (actually, "unit" charges) for the Green function.

6. Apr 15, 2015

### MMS

Do you see where I could've possibly went wrong?

7. Apr 15, 2015

### Telemachus

Sorry for the mistake, you are right, the correct position for the image charge is $(0,y_0,-z_0)$. What you did seems ok to me now.

8. Apr 15, 2015

### Orodruin

Staff Emeritus
Why do you think you have gone wrong? (Other than copy paste errors...)

9. Apr 15, 2015

### Telemachus

10. Apr 15, 2015

### MMS

First, you expect the double improper integral to converge for it to be right. No program was able to give me a concrete answer to this (I don't know if I suck at typing functions or it simply doesn't converge to anything).

Second, after finding the potential, I'm asked to find the surface charge density sigma(x,y).
To do so, I'll have to take the derivative of the potential and equalize it to 4*pi*sigma.
This doesn't seem right to take derivatives of double improper integrals. Not to mention it is also multiplied by z, too.

Therefore, I suspect I'm wrong somewhere..

11. Apr 15, 2015

### Telemachus

12. Apr 15, 2015

### Orodruin

Staff Emeritus
Even if there exists a closed form expression for this integral (I have not checked), you should not assume that this will always be the case.

13. Apr 15, 2015

### MMS

14. Apr 15, 2015

### Orodruin

Staff Emeritus
By physical reasoning, your integrals must converge. Without the point charge, it must be between zero and $\phi_0$ everywhere.

15. Apr 15, 2015

### MMS

Yes. I completely agree with you. Yet it isn't working out for me.
Maybe I messed up some signs or something, I'll try again.

16. Apr 15, 2015

### Orodruin

Staff Emeritus
The integral you have written down is also clearly convergent for all z>0.

Switching to polar coordinates you have an integrand that goes as 1/r^3 as r goes to infinity in a two dimensional integral. With the r dr in the integration measure, the boundary term at infinity goes as 1/r.

17. Apr 15, 2015

### Telemachus

The computer isn't good for everything. Try making a change of variables. You have the primitive for the integral. Don't rely on the computer for it.

If you call u=c-y, du=-dy perhaps you can get the desired result.

I don't think its a good idea to perform the integration in polar coordinates, because you have to put one of the limits of integration in the plane y=0. But you can try that too. What is clear from what Orodruin said is that the integral converges.

18. Apr 15, 2015

### Telemachus

What I sais before was just stupid, I think that performing the integration in polar coordinates is actually the smart thing to do.

Ps. I'm not sure. You have radial symmetry around $y_0$, so I think you will have trouble at y=0 with the limits of integration, but one should try to do it to be sure.

19. Apr 15, 2015

### MMS

Well, I can tell you this. I'm not relying on computers anymore. A pen and a pad and some basic calculus with some manipulations on the integrals was all was needed for me to solve it (I think).
I sincerely apologize for this complete nonsense with the integral thingy.
The final answer which I've gotten to (the potential) after calculating the "difficult" integrals is as follows:

If any of you would like to see the complete solution, I'd be more than happy to write it down.

Thank you guys for all the help!

20. Apr 15, 2015

### MMS

The surface charge density and the force acting upon the charge: