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The resistance in water between two bouys

  1. Jun 27, 2013 #1
    1. The problem statement, all variables and given/known data
    Two spherical buoys float half submerged in a calm deep sea. Their radius a is very much smaller than their separation b. Calculate the resistance between them if the resistivity of the water is ρ.


    2. Relevant equations
    [tex]R=V/I[/tex]
    [tex]V=\int Edl[/tex]
    [tex]I=\sigma \int_S \textbf{E} \cdot d \textbf{S}[/tex]
    [tex]\rho = \sigma^{-1}[/tex]


    3. The attempt at a solution
    The field between the bouys is
    [tex]E=\frac{Q}{\varepsilon 4 \pi} \Big(\frac{1}{x^2}-\frac{1}{(b-x)^2} \Big) =E_1+E_2[/tex]
    I already know how to show that
    [tex]V \simeq \frac{Q}{2 \pi \varepsilon_0} \Big( \frac{1}{a}-\frac{1}{b-a} \Big) \simeq \frac{Q}{a 2 \pi \varepsilon_0}[/tex]
    Now here comes the tricky part and I am not sure what I am doing but it feels like the current between the two bouys should pass a theoretical earth before reaching each other. The reason for this is that when the two bouys are a distant apart half of their field lines should still join. Is this reasonable? So reasoning like that
    [tex]I= \sigma \int_{S_{between}} \textbf{E} \cdot d \textbf{S}=2 \sigma \int_{earth} \textbf{E}_1 \cdot d \textbf{S} = \sigma \int_{sphere} \textbf{E}_1 \cdot d \textbf{S} = \frac{Q}{ \rho \varepsilon_0} [/tex]

    And finally
    [tex]R=V/I=\frac{\rho}{2 \pi a}[/tex]
    But the correct answer is ##R=\frac{\rho}{\pi a}## so I guess I overestimated the current by a factor 2. (Where did it go wrong? I always tend to miss a factor 2 in electromagnetism.)
     
  2. jcsd
  3. Jun 27, 2013 #2

    gneill

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    Could it be because the spheres are only half submerged? How does that affect the current?
     
  4. Jun 27, 2013 #3

    rude man

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    The OP calculated R = half the alleged correct answer but if the spheres are only half submerged then the resistance should be doubled, not halved (since air resistance >> salt water).

    Anyway, I wouldn't offhand know how to approach this problem. Interesting that the answer is not a function of b.
     
  5. Jun 27, 2013 #4
    Ok, I'm glad there was nothing wrong with my calculations or reasoning, but the problem was that I thought the bouys were submerged very Deep in the ocean! (yes the current becomes halved and the resistance doubled.)
     
  6. Jun 27, 2013 #5

    haruspex

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    Yes - as Order noted, you can treat each sphere separately. Model the ocean as a semi-infinite space (z < 0, say) minus a hemispherical portion, r < a. The resistance of a hemispherical shell radius r > a and thickness dr is ρdr/(2πr2), so the total resistance to infinity is ρ/(2πa). Getting from one sphere to the other involves (in the limit) going out to infinity and back again, giving a total ρ/(πa).
     
  7. Jun 28, 2013 #6
    Yes. This is a correct assessment.

    For a single isolated source of radius a in an infinite ocean,
    [tex]V=4\pi \rho a I[/tex]
    and for a single isolated sink of radius a in an infinite ocean,
    [tex]V=-4\pi \rho a I[/tex]
    So, for the combination of a source and a sink, both with the same current I,
    [tex]ΔV=\pi \rho a (2I)[/tex]

    Because of symmetry, the half-submerged (semi-infinite) problem is the same as the fully submerged problem with the 2I replaced by I.
     
    Last edited: Jun 28, 2013
  8. Jun 28, 2013 #7

    rude man

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    Chet, I don't understand: V/I is resistance but your expression is not ohms:
    [tex]ΔV=\pi \rho a (2I)[/tex] → ΔV/I = 2πρa which has units of (ohm-m)*(m) = ohm-m^2.
     
  9. Jun 29, 2013 #8
    Oops. Thanks rude man. I realized I had made a mistake and was about to correct it when I saw your posting. So, here goes again:

    For a single isolated current source of radius a in an infinite ocean,
    [tex]V=\frac{ρI}{4\pi a}[/tex]
    and for a single isolated current sink of radius a in an infinite ocean,
    [tex]V=-\frac{ρI}{4\pi a}[/tex]
    So, for the combination of a source and a sink in an infinite ocean, both with the same current I,
    [tex]ΔV=\frac{ρI}{2\pi a}[/tex]
    Because of symmetry, the half-submerged (semi-infinite) problem is the same as the fully submerged problem with the I replaced by 2I, since, in the infinite ocean problem, half the current goes to the top half of the spheres and half the current goes to the bottom half of the spheres. So, for the half-submerged problem,
    [tex]ΔV=\frac{ρI}{\pi a}[/tex]
    Sorry for my previous error.

    Chet
     
    Last edited: Jun 29, 2013
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