# Homework Help: Method of Images - Need a hint!

1. Feb 22, 2009

### Sonolum

1. The problem statement, all variables and given/known data

A conducting sphere of charge Q and radius A is located at a distance D from the sphere's center to an infinite, grounded, conducting plane. Ultimately, I would like to find the capacitance, but that is simple once I find the potential. So the question isow does one determine the potential of this system using the method of images? The hint given is that the solution is a very nasty infinite series.

2. Relevant equations

Miscellaneous electrostatic equations, Method of Images

3. The attempt at a solution

Initially we have an infinite grounded conducting plane with no net charge and a potential = 0 everywhere. We place this plane in the YZ plane.

Then we bring in our charged conducting sphere, placing it at distance D from the plane. Obviously the charges on the sphere will no longer be homogeneously distributed.

The method of images states that any stationary charge distribution near a grounded conducting plane can be treated by introducing an oppositely charged mirror image (sphere).

The charged sphere behaves as though all of its charge is concentrated at the center, as does its mirror sphere... This is where I take a step that I am somewhat unsure of...

If this is the case, then the situation is similar to that of a charged sphere and a point charge, except the boundary conditions are that V=0 at the plane (or when rvector = dvector = |d|ihat, or more simply, when rvector dot dvector= |d|)

In this scenario, the mirror sphere behaves like a point charge, which I call q' = -Q. There is an image charge induced in the real sphere of q'' by the image sphere/charge, at a distance d' from the center of the real sphere.

So, I know that
[[EQ 1]]V = Vground + (4*pi*epsilon0)-1 (Q - q'')(|rvector|-1,

But this is for a charged sphere and a point charge! Can I make the assumption that the image sphere will beave like a point charge? Not quite, I'm sure!

So I churn through copious amounts of vector algebra and the like, and I discover that from the solution for the grounded sphere & point charge is gives us:

q'' = (AQ)/(2D), and d' = (A^2)/(2D)

And then I can substitute into [[EQ 1]]:

V(rvector) = k( -Q/(|rvector - 2d*ihat|) + (aQ)/(2d(|rvector - (a^2)/(2d)*ihat| + (Q(1-(a/(2d))))/(|rvector|)

(where k = (4 * pi * epsilon0)-1

This is only a valid solution when V(rvector dot dvector = d ) (because that would represent the surface of the grounded plane)

Mucking about with some more math, I discover I am missing a term:

(-Q(a^3))/(4(d^3)|rvector - (a^2)/(2d) * ihat|)

I'm can see a pattern *trying* to emerge, but I'm not 100% sure how to morph this into an infinite series... Any pointers? Can I treat this situation the way that I have? Are my Boundary Conditions correct?

Also, is there an equation editor I can use to make expressing these less cumbersome?

2. Feb 22, 2009

### gabbagabbahey

Are you required to use the method of images for this problem?

Separation of variables is much easier.

3. Feb 22, 2009

### Sonolum

It's supposed to be solved with the method of images, and we're supposed to get an infinite series solutions.

I was thinking about multipole expansion, but I'm not sure about that... It would give me legendre polynomials, or so I've read, but I'm pretty weak in those....

4. Feb 22, 2009

### Sonolum

Is this a bad question? It's got VERY few views, yet other questions posted after mine have gotten a lot of attention...

5. Feb 22, 2009

### gabbagabbahey

Well, the problem is solved easiest using Laplace's equation, but if you are supposed to use MoI, then by all means, use MoI!

The first step is to determine what region you want to find the potential in, and hence what region(s) you are allowed to place image charges in. So where do you need to find the potential in order to calculate capacitance? which regions are you allowed to place image charges in then?

6. Feb 22, 2009

### Sonolum

Yeah, the prof is pretty adamant about us doing our problems just so (which is uaully the most tedious, difficult method possible)... And he wants method of images here.

I only need potential in the region outside the real charged sphere, and outside th infinite plane. I can place image charges within the plane, and inside the real charged sphere. That step is simple.

What I am mot struggling with is whether or not I can treat the image sphere as a point charge, or essentially treat is as a charged sphere/point charge problem with the potential 0 at r(dot)d (or at the surface of the plane)... Is this a legal move? I feel that, by the Uniqueness Theorem, if I can satisfy my boundary conditions I should have a valid solution... and since r(dot)d = |r|^2 + |d|^2 + |r||d| cos T, where T is the angle between r and d, then I could somehow finagle this into a legendre polynomial, which is an infinite series or Pl(cos T)?

7. Feb 22, 2009

### Sonolum

Also, to solve with laplace's equation (so I can double check my work), I'd say (del^2)V = 0 inside the charged sphere, and inside the plane, then use these BCs to solve for V, right? With a little separation of variables thrown in there for good measure, of course.

8. Feb 22, 2009

### gabbagabbahey

Close, you need to find the potential outside the sphere and above the plane. Which means you can place image charges inside the sphere and below the plane.

Certainly, by the uniqueness theorem, if you can find a potential that satisfies Laplace's equation in the region outside the sphere and above the plane, and also satisfies the boundary conditions then you are guaranteed that it is the potential in that region.

However, you seem to be neglecting one of your boundaries. The conducting plane (which I would place at z=0 for simplicity) is one boundary, and z-->infinity is another, but there is also one more boundary to the region you are interested in. Can you think of where that boundary is?

9. Feb 22, 2009

### gabbagabbahey

Right, but be sure to use the correct boundary conditions!

I'd do the MoI method first, and then see if you can solve Laplace's equation after to check your answer.

10. Feb 22, 2009

### Sonolum

Well, I've set up my problem with the charged sphere to the left of the plane, so really it's the area outside the sphere and to the left of my plane, but that's just a metter of setting up the vectors.

I DID choose my origin at the center of the charged sphere, though, so the plane is at x=d, not x=0 (or z=0 if you prefer to set things up that way).

The only other boundary condition I am neglecting is tht the potential is some constant at the surface of the real charged sphere, right?

11. Feb 22, 2009

### gabbagabbahey

Okay, that should work too!

Yes, exactly.

(1)Let's forget about all the rest of the charges for a second and just place a point charge $Q$ at $x=0$....Which boundary conditions are satisfied for this, and which one isn't?

(2)Where would you need to put an image charge in order for the potential to satisfy that boundary condition?

(3) Does the potential still satisfy the other boundary conditions? If not, where would you nee to place an image charge, and what is its necessary magnitude to satisfy that BC?

(4)Does the potential now satisfy the other BC? etc....

12. Feb 22, 2009

### Sonolum

Easy: With Q at x=0, there needs to be -Q at x=2d in order to satisfy V(d)=0. But then V(a) != B.

The same place you'd put it for a sphere/point charge problem - But then wouldn't you have to change the charge you placed at the very beginning to (Q-q')?

In order to make V(a) = B, you have to place a second image charge at x' of q'', where q'' = (a/(2d))*Q, and d' = (a^2)/(2d)...

Welllllllll... Maybe? Because, see, when you have a charged sphere, it's the grounded sphere solution, plus the solution with (Q-q') at the center, right?

I don't know until I know whether or not I'm changing the charge I put at the center... I'm concerned that the inhomogeneous charge distribution is going to change what I started with....

13. Feb 22, 2009

### gabbagabbahey

Good

Don't worry about changing the original charge, what is important here is not how much imaginary charge ends up inside the sphere, but only the potential that the resulting distribution produces....at the end of the derivation, I'll help you show that this does indeed produce the correct potential.

Just leave the original charge where it is and proceed:

Yes, great

So now to recap, we have charges (1)Q at x=0, (2)-Q at x=2d and (3)(a/(2d))Q at x=a^2/(2d)

Pu t that concern aside for the moment, and continue on adding image charges....

14. Feb 22, 2009

### Sonolum

Sorry - Had to take a break and work on something else for awhile, I was getting uber frustrated.

I think I see what you mean... I talked to another classmate, and they said there are 16 image charges necessary in order to solve the problem to 1% accuracy like requested.... I have a feeling that I'm not going to get it done in time for the deadline, but I DO think that I have a clearer understanding of the method - even if I don't make it all the way to 16, perhaps I'll be able to spot the pattern and generalize the series.

Thank you SO much for your insight - I think I see what's going on now and am going to make a final attempt... Are there any other pointers for me when I get there - any final considerations I should take into account (counting self energy or something)?

Thanks again!