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A rectangular hollow conductor, open at one end

  1. Sep 30, 2011 #1
    1. The problem statement, all variables and given/known data

    A rectangular hollow conductor,open at one end of [itex]z[/itex], has a voltage impressed at [itex]z=0[/itex]. On the [itex]z=0[/itex] plane, the voltage as a function of [itex]x[/itex] and [itex]y[/itex] is simple cosines such that [itex]V=0[/itex] at [itex]x=y=+/-L/2[/itex]. All other surfaces are grounded.

    [itex]V(x,y)\approx f(cos(Ax),cos(By))[/itex]

    Calculate the potential [itex]\Phi(x,y,z)[/itex] using an "adequate" number of terms. Plot [itex]\Phi(x,y)[/itex] for at least 5 values of [itex]z[/itex] (arbitrary units)

    2. Relevant equations

    After looking at the geometry and the statement of the problem these are the boundary conditions that I came up with:

    (i) [itex]V=0[/itex] when [itex]y=+/-\frac{L}{2}[/itex]
    (ii) [itex]V=0[/itex] when [itex]x=+/-\frac{L}{2}[/itex]
    (iii)[itex]V=V_{0}(x,y)[/itex] at [itex]z=0[/itex]

    This last one is an assumption I am making, but I am not quite sure it is valid for this configuration:
    (iv)[itex]V\rightarrow\infty[/itex] as [itex]z\rightarrow\infty[/itex]

    This is rectangular geometry so the cosines make sense, but I am not quite sure how to apply boundary conditions to get my coefficients of the expansion. Heck, I may be entirely wrong with my solution.

    3. The attempt at a solution

    The generalized double sum that I come up with is

    [itex]\sum_{m,n} K_{m,n}e^{-2\pi z\sqrt(m^{2}+n^{2})/L}cos(2mx\pi/L)cos(2ny\pi/L)[/itex]

    But this seems either overcomplicated or just entirely incorrect, because whenever I try to find the coefficients of expansion I get vanishing terms, which means that there is no solution. . . . I'm lost and could really use some clarification in this problem. Thanks in advance!!
     
  2. jcsd
  3. Sep 30, 2011 #2

    G01

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    You fourth boundary condition is wrong. It should be: [itex]V\rightarrow 0[/itex] as [itex]z\rightarrow\infty[/itex].

    The periodicity of your cosines is also wrong. It should be [itex]n\pi /L[/itex], not [itex]2n\pi /L[/itex]. Do you see why?

    Now what you want to do is apply your remaining boundary condition, (iii) to find the coefficients:

    i.e. [tex]V(x,y,0)=V_0(x,y)=\sum_{n,m} C_{n,m} cos(n\pi x/L) cos(m\pi y/L)[/tex]

    You can use Fourier analysis to find the coefficients. If you've done this, and still getting zero for all of the coefficients, then I'll need to see your calculations to give more detailed advice.
     
    Last edited: Sep 30, 2011
  4. Sep 30, 2011 #3
    I do not see why the periodicity is incorrect. Could you please explain this to me?

    Also, why does the voltage go to zero as you move along z to infinity?

    I will try this method now, thank you for your help.
     
  5. Sep 30, 2011 #4
    Using this method I do get something other than zero, but I would like to see if it is indeed the correct procedure:

    The coefficients [itex]K_{m,n}[/itex] may be found by utilizing the "fourier trick"

    So,

    [itex]K_{m,n}=\frac{4V_{0}}{L^{2}} (\int cos(\frac{n\pi y}{L}) dy) (\int cos(\frac{m\pi x}{L}) dx)[/itex]

    Note: Both integrals range from 0 to L/2.

    [itex]=\frac{4V_{0}}{mn\pi^{2}} sin(\frac{m\pi}{2}) sin(\frac{n\pi}{2})[/itex]

    [itex]=\frac{4V_{0}}{mn\pi^{2}}(-1)^{n+m}[/itex]

    This is what I have come up with, please help me correct errors if there are any. Thanks a bunch!!
     
  6. Sep 30, 2011 #5

    G01

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    The potential must be zero at +- L/2 in both the x and y directions. Thus:

    [tex]cos(n\pi (L/2)/L)=cos(n\pi/2)[/tex] which is zero for all odd n.

    However,

    [tex]cos(2n\pi (L/2)/L)=cos(2n\pi/2)=cos(n\pi)[/tex] which is not zero for any integer n. So, the cosine function with this periodicity doesn't satisfy the boundary conditions.


    The V goes to zero at infinity is the standard boundary condition set in these type of problems. It makes sense physically if you think about it. We have a potential "source", V0 at z=0. As we move farther and farther away from that source, it makes sense that the potential should decrease in magnitude.
     
  7. Oct 1, 2011 #6

    G01

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    I'm short on time this morning, but I took a cursory look at your work, and you seem to have the idea. I will come back later when I'm free and double check your numbers and everything, but your method is correct.

    A few quicks points:

    1. You seem to be assuming V_o is a constant function? Is that actually the case? The problem statement seems to imply it's not.

    2. Double check the bounds on your integral. I think they should be -L/2 to L/2. However, like i said, I'm in a rush! So, I'll double check this later.


    Your welcome. Anytime! :smile:
     
  8. Oct 1, 2011 #7

    G01

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    OK. I'm back.

    I double checked and yes, the bounds on your integrals are wrong. They should go from -L/2 to L/2. Do you understand why?

    Either way, cosine is even, so it just contributes a factor of 4 (a factor of 2 for each integral).

    Also, the sum n+m will always be even, so the factor of (-1)^n+m just drops out.
     
  9. Oct 7, 2011 #8
    Well that worked out ok. Now, suppose the x and y both had boundaries at 0 to L and all sides were grounded except z=0 and [itex]z\rightarrow \infty[/itex] (which is open). The face z=0 now has a uniform surface charge density, with a total charge being +Q. Would the potential change in this case or would it be different?

    How would I go about figuring this out? Would it be about the same procedure with new boundary conditions?
     
  10. Oct 7, 2011 #9

    G01

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    Yep same procedures, but your boundaries are now at different positions. That's all. The solution should be similar to the above problem. Let me know if you have any issues.
     
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