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# Infinitely long hollow conducting tube

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Consider n infinitely long hollow conducting tube tht is split in half with each half held at the indicated potential.

calculate B_o, B_n, A_n

what equation would plot V(rho, pi/2).

attatched copy of diagram.

https://brainmass.com/physics/system-work/infinitely-long-hollow-conducting-tube-288286

## SOLUTION This solution is FREE courtesy of BrainMass!

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The solution is attached below in two files. the files are identical in content, only differ in format. The first is in MS Word format, while the other is in Adobe pdf format. Therefore you can choose the format that is most suitable to you.

Note that I used r and theta instead of rho and phi (old habits die hard)

Laplace equation is:
(1.1)
With the following boundary conditions:
(1.2)
The potential is finite everywhere, especially at and as :
(1.3)
(1.4)
And the last condition is that the potential is periodical:
(1.5)
Since the system is completely symmetric with respect to the xy plane the potential has no dependence on the z-coordinate and equation (1.1) becomes a two dimensional problem in polar coordinates:
(1.6)
We assume a solution in the form:
(1.7)

Then:

And finally:
(1.8)
Equation (1.8) is separated and can be satisfied if and only if:
(1.9)
Where k is a constant.
Solving the angular equation we see that there are three cases we have to consider:

In this case the angular equation becomes:

Its solution is:
(1.10)
However, this solution does not comply with the periodic condition (1.5) unless and we get a trivial solution.
Case 2:
The angular equation becomes:

Its solution is:
(1.11)
Again, applying condition (1.5) we see that:

So the solution in this case is simply:
(1.12)
The third and last case is
This leads to the angular equation:

And its solution is:
(1.13)

Applying condition (1.5) we see that the solution can be periodic if:
(1.14)
So the solution for is:
(1.15)
Now we can go back to the radial equation.
We are not interested in the case since it leads to a trivial solution anyway so we will solve for the two remaining cases.
Case 1:

Therefore its solution is:
(1.16)

The second case is:

We "guess" a solution in the form and substituting it back in the equation we get:

(1.17)
The general radial solution in this case is:
(1.18)
Combining (1.18), (1.15), (1.16) and (1.12) together with (1.7), the most general solution for Laplace equation in polar coordinate is:
(1.19)

Now, in our case we divide space into two regions: outside and inside the cylinder.
The requirement that the potential be finite leads to the conclusion that inside the cylinder all the terms containing must disappear otherwise the potential will explode at r=0. By the same token, all the terms containing must vanish outside the cylinder. The term must vanish in both regions.

Therefore we can write the general solution for our problem as:
(1.20)
We also require that the potential be continuous on the cylinder itself:

(1.21)
And we are left finding the values of
On the cylinder we have:
(1.22)
We use the orthogonality relations:
(1.23)
Then:

Integrating both sides yields:

The only term that does not vanish on the left hand side is where n=m and we get:

(1.24)
But for (where ) we have:

(1.25)

By the same token:
(1.26)
In our case:
(1.27)
Then:

And the rest of the coefficients:

While for the sine coefficients:

(1.28)
Note that:
(1.29)
Therefore:
(1.30)

And the potential everywhere is:

(1.31)

For the potential on the cylinder looks like (for 20 terms in the series):

And the potential looks like:

Where the blue and red lines corresponds to repectively while the gren line corresponds to

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!