Method of image with Green function

In summary: As it is, we can't really give much more help with what you've asked. In summary, a conducting sphere of radius a has three different shells: an upper part with potential V, a middle part that is grounded, and a lower part with potential -V. The center of the sphere is at the origin. To find the electric potential at x, we can use the method of images with Green's function. The middle part, although relatively broad, can be treated as a thin shell. The non-hemispherical shape of the upper and lower parts may complicate the calculations, but the approach remains the same. When asking for help, it's important to show some effort and thought in order to receive valuable assistance.
  • #1
hubrisos
3
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.
 

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  • #2
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.
 
  • #3
Gee why did not I think of that? Wow such an insight.
Thank you.
What a toxic community...
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

Related to Method of image with Green function

What is the method of image with Green function?

The method of image with Green function is a mathematical technique used to solve boundary value problems in electrostatics and other areas of physics. It involves creating a set of "imaginary" charges or sources outside of the problem domain to mimic the behavior of the actual charges or sources inside the domain. The Green function is a mathematical function that describes the potential or field at a point due to a point source. By combining the method of images with the Green function, we can find the solution to the boundary value problem.

How does the method of image with Green function work?

The method of image with Green function works by using the principle of superposition. This means that the total potential or field at a point is the sum of the potentials or fields due to each individual charge or source. By creating a set of "imaginary" charges or sources outside of the problem domain, we can mimic the behavior of the actual charges or sources inside the domain and use the Green function to find the potential or field at any point.

What are the advantages of using the method of image with Green function?

One of the main advantages of using the method of image with Green function is that it simplifies the problem by reducing it to a set of point sources. This makes it easier to solve compared to other techniques such as using the Laplace or Poisson equations. Additionally, the method of images is often used to solve problems with complex geometries that would be difficult to solve using other methods.

What are the limitations of the method of image with Green function?

One limitation of the method of image with Green function is that it can only be used to solve problems in electrostatics or other areas where the potential or field is governed by the Laplace or Poisson equations. It also assumes that the medium is linear and isotropic, which may not always be the case in real-world situations. Additionally, the method of images may not always provide an exact solution, and some approximation may be necessary.

Can the method of image with Green function be applied to problems in other areas of physics?

Yes, the method of image with Green function can be applied to problems in other areas of physics where the potential or field is governed by the Laplace or Poisson equations. This includes problems in fluid mechanics, heat transfer, and elasticity. However, the specific form of the Green function may differ depending on the problem at hand.

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