Bessel Function Boundary Condition on the top of a Cylinder

In summary, the author is trying to solve a problem that is very similar to one they have seen online, but their expression for the potential is not correct. They are able to use equations from their textbook to find the potential inside the cylinder.
  • #1
Othman0111
27
0
Hi everyone,

I'm working through the boundary conditions and I could not figure out what to do with the last boundary condition (when z=L)
1.png

I know that the values for K are:
2.png


How so?

1. Homework Statement


A hollow right angle cylinder of radius a and length l. The sides and bottom are grounded. The potential at the top of the cylinder
is
3.png
where X1,2 is the second zero of the first Bessel function. (Note:
the potential on the top of the cylinder is cylindrically symmetric)
a) Find the potential everywhere inside the cylinder. Find the charge
density on the surface of the cylinder.

Homework Equations


Bessel function relations, probably something have to do with the orthogonality.

The Attempt at a Solution


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  • #3
Dr Transport said:
what is [itex] r [/itex]?
It supposed to be a ρ
 
  • #4
Where did you get the original expression for the potential? My expression for the potential in Jackson is a little different.
 
  • #5
Dr Transport said:
Where did you get the original expression for the potential? My expression for the potential in Jackson is a little different.
The general expression of Φ shown here on this similar problem.
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  • #6
that doesn't look right, the last two summations are usually combined and what text quoted that expression? You cannot get a [itex] J_1(\kappa\rho)[/itex] from a sum over [itex] J_0(\kappa\rho)[/itex]'s. The entire problem should be shown, something looks fishy.
 
  • #7
Dr Transport said:
that doesn't look right, the last two summations are usually combined and what text quoted that expression? You cannot get a [itex] J_1(\kappa\rho)[/itex] from a sum over [itex] J_0(\kappa\rho)[/itex]'s. The entire problem should be shown, something looks fishy.
There is no Φ dependence → 2nd and 4th summation goes to zero.
The remaining term is after applying boundary conditions and absorbing the constants into one constant is : Φ(ρ,z) = ΣAok J0(Kρ) sinh(Kz)
After applying the last boundary condition " Φ(ρ,z=L) " is:
1.png

>
The solution that I uploaded in the last comment was not for this problem, it was for a similar problem that state:
Problem 1: A hollow right angle cylinder of radius a and length l has both
ends and the sides grounded except for a band in the middle of the side of the
cylinder. The band in the middle is held at a potential of V0 and extends from
the center of the side of the cylinder for a distance of l=4 in both directions.
a) Find the potential everywhere inside the cylinder.
 

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  • #8
A problem very similar to this I have seen online and your expression for the potential is not correct to do the problem as stated in my opinion.
 
  • #9
Dr Transport said:
A problem very similar to this I have seen online and your expression for the potential is not correct to do the problem as stated in my opinion.
Could you please give me the like to that problem. Or give me more guidance.
 
  • #10
if you do a search for potential cylindrical coordinates, you should find something, we are not supposed to give out solutions carte blanche. I'll say that your expression for the potential is not correct as written.
 
  • #11
Dr Transport said:
if you do a search for potential cylindrical coordinates, you should find something, we are not supposed to give out solutions carte blanche. I'll say that your expression for the potential is not correct as written.
I didn't find it in Jackson either. But it was what professors taght us to do. The homework is due but I just were curious. Thank you, sir.
Attached a PDF version of his notes, the equation is on page 21.

Regards
 

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  • #12
I'm not sure which edition of Jackson you have. For the 3rd edition see section 3.8, pages 117 and 118. You should be able to use equations (3.105a) and (3.105b) to determine the potential inside the cylinder. In your problem, the potential does not depend on ##\phi##. So, only one value of the index ##m## in (3.105a) contributes.

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  • #13
Dr Transport said:
You cannot get a [itex] J_1(\kappa\rho)[/itex] from a sum over [itex] J_0(\kappa\rho)[/itex]'s.
It turns out you can. I wasn't sure and I had to take some time to review this. But you can essentially expand an "arbitrary" function ##f(x)## over the interval ##0<x<a## in terms of Bessel functions of a single order, say Bessel functions of zeroth order. So, you can expand ##J_1(x)## on ##0<x<a## in terms of the infinite set of functions ## \left \{ J_0(\frac{X_{0,k}}{a}x), k = 1, 2, 3, .. \right \}##, where ##X_{0, k}## is the ##k##th zero of ##J_0(x)##.

See
https://www.math.upenn.edu/~rimmer/math241/ch12sc6frbess.pdf
 
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  • #14
TSny said:
I'm not sure which edition of Jackson you have. For the 3rd edition see section 3.8, pages 117 and 118. You should be able to use equations (3.105a) and (3.105b) to determine the potential inside the cylinder. In your problem, the potential does not depend on ##\phi##. So, only one value of the index ##m## in (3.105a) contributes.

View attachment 239655

View attachment 239656
It turns out that I don't know much about Bessel function. Thank you, sir.
 

1. What is a Bessel function?

A Bessel function is a special type of mathematical function that is used to describe oscillatory phenomena, such as sound waves or electromagnetic waves. It was first introduced by the mathematician Daniel Bernoulli in the 18th century and is named after the mathematician Friedrich Bessel.

2. What is a Bessel function boundary condition?

A Bessel function boundary condition is a mathematical condition that is applied to a Bessel function to satisfy certain physical constraints. In the case of a cylinder, the boundary condition is used to describe the behavior of a wave on the surface of the cylinder.

3. Why is the Bessel function boundary condition important in cylinder applications?

The Bessel function boundary condition is important in cylinder applications because it allows us to accurately model the behavior of waves on the surface of a cylinder. This is useful in a variety of fields, such as acoustics, electromagnetics, and fluid dynamics.

4. How is the Bessel function boundary condition applied to a cylinder?

The Bessel function boundary condition is applied to a cylinder by setting the value of the Bessel function to zero at the boundary of the cylinder. This means that the amplitude of the wave on the surface of the cylinder is zero, which is physically realistic.

5. Are there any limitations to using the Bessel function boundary condition on a cylinder?

While the Bessel function boundary condition is a useful tool for modeling waves on a cylinder, it does have some limitations. It assumes that the cylinder is infinitely long and has a perfectly smooth surface, which may not always be the case in real-world applications. Additionally, it may not accurately describe the behavior of waves at the ends of the cylinder.

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