Method of image with Green function

hubrisos
Messages
3
Reaction score
0
Homework Statement
A conducting sphere of radius a is made up of 3 different shells, upper part a/2 \leq z \leq -a/2, the middle part -a/2 \leq z\leq a, the lower part -a \leq z \leq -a/2 where the center of the sphere is at the origin. The upper part V and the lower part has -V potentials, while the mid part is grounded. Find the electric potential at x up to a 4 T order in the expansion of a/x << 1.



Basically the shell has 3 regions, upper, middle, lower. the lower/upper parts are not hemispherical, middle part is relatively broad, and the upper part has V, the lower part has -V potential.
Relevant Equations
green functions
A conducting sphere of radius a is made up of 3 different shells, upper part a/2
91fe1446-dc3f-4e80-a323-984eb2d97416.png
z
91fe1446-dc3f-4e80-a323-984eb2d97416.png
-a/2, the middle part -a/2
91fe1446-dc3f-4e80-a323-984eb2d97416.png
z
91fe1446-dc3f-4e80-a323-984eb2d97416.png
a, the lower part -a
91fe1446-dc3f-4e80-a323-984eb2d97416.png
z
91fe1446-dc3f-4e80-a323-984eb2d97416.png
-a/2 where the center of the sphere is at the origin. The upper part V and the lower part has -V potentials, while the mid part is grounded. Find the electric potential at x up to
323afa11-ef34-4ae4-907c-ae5e2581007f.png
order in the expansion of a/x << 1.

Basically the shell has 3 regions, upper, middle, lower. the lower/upper parts are not hemispherical, middle part is relatively broad, and the upper part has V, the lower part has -V potential.

I don't know where to start, I don't ask you to solve the entire thing, but give me a few tips/hints. I am lost due to the fact that the mid part is huge.
 

Attachments

  • hem.jpg
    hem.jpg
    16.1 KB · Views: 165
Physics news on Phys.org
hubrisos said:
I don't know where to start, I don't ask you to solve the entire thing, but give me a few tips/hints. I am lost due to the fact that the mid part is huge.
Can you be more specific? Why is that throwing you? Your thread title is "Method of image with Green function." That seems to suggest the approach you're supposed to take.
 
Gee why did not I think of that? Wow such an insight.
Thank you.
What a toxic community...
 
Well, the rules of the forum are that you need to show some effort, that you actually thought about the problem. "I have no idea where to start" doesn't cut it.
 
I did not ask you to solve the whole problem, did I?

I was a member of this website 5-6 years ago, and the same smug attitude was flying around everywhere. it seems it is still the same.

Enjoy your "I know everything" attitude. I am out.
 
hubrisos said:
I was a member of this website 5-6 years ago, and the same smug attitude was flying around everywhere. it seems it is still the same.
Our rules are still the same. As @vela said, you need to have shown some effort, at least to the point of thinking about an approach.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top