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Resistance for a sphere in a half sphere containing liquid

  1. Feb 16, 2009 #1

    TFM

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    1. The problem statement, all variables and given/known data

    A small copper sphere of radius a is half immersed at the surface of a poorly conducting liquid of resistivity [tex] \rho [/tex]. The liquid is contained in a spherical copper bath of radius b, where b>>a and where the bath is concentric with the sphere. Estimate the electrical resistance between the sphere and the walls of the bath, assuming that the copper can be assumed to be a perfect electrical conductor

    2. Relevant equations

    [tex] R = \rho \frac{A}{l} [/tex]

    3. The attempt at a solution

    See I think I need to use the resistivity equation I have above, however, this is only for a wire. Is there a version for situations such as these?

    TFM
     
  2. jcsd
  3. Feb 16, 2009 #2
    It seems to me that you've sort of got a wire here, it's just that its cross-sectional area changes and it's got a weird shape. If b>>a, then doesn't that really mean that the liquid is shaped like a hemisphere? If so, then I think you can just integrate [tex]\rho dA / l [/tex]
     
  4. Feb 16, 2009 #3

    TFM

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    That makes sense. So:

    [tex] R v= \rho \frac{dA}{l} [/tex]

    But what would the limits be? Would it be from 0 to Circle(r=a) - Circle (r=a)
     
  5. Feb 16, 2009 #4
    I agree with zero as the lower limit, this would be the bottom of the pool. The disk's radius at the top would just be b; I think the problem is implying the sphere is small enough to be treated like a point. Dig out the Calculus textbook and you ought to be able to find an example like this one where you're integrating by adding a bunch of disks together.
     
  6. Feb 17, 2009 #5

    TFM

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    Could we use spherical polar coordinates, do it for a whole sphere, and then half the value?
     
  7. Feb 17, 2009 #6
    Surely this should be

    R = rho *L/A
     
  8. Feb 17, 2009 #7
    R is the integral from a to b of (rho/ 2*pi*r^2)dr

    Don't understand the relevance of b>>a though, except that you could
    make b infinity.
     
  9. Feb 17, 2009 #8

    TFM

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    So I dion't need to use:


    [tex] R = \rho \frac{l}{A} [/tex]

    Expressed in Spherical polar coordinates?
     
  10. Feb 17, 2009 #9
    r(radius) is a "polar" coordinate. 2*pi*r^2 is the area of a hemisphere.
     
  11. Feb 18, 2009 #10

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    Okay, so What should I do for the length part?
     
  12. Feb 18, 2009 #11
    Surely the length is just the difference between the radii?
     
  13. Feb 18, 2009 #12

    TFM

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    Okay, so:

    [tex] R = \rho \frac{b-a}{2\pi r^2}dr [/tex]

    Does this look okay?
     
  14. Feb 18, 2009 #13
    Edit: Sorry (had a beer) ignore what I just said (now deleted). The equation looks good in terms of units, beyond that I am unsure.

    Edit 2: Now I feel really stupid. The equation won't work because for a start you would get a negative resistance. I am going to go away now to be alone with my shame. Sorry for not being much use.
     
    Last edited: Feb 18, 2009
  15. Feb 18, 2009 #14

    TFM

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    Okay, so it should be:

    [tex] R = \rho \frac{b-a}{2\pi r dr} [/tex]

    ?
     
  16. Feb 19, 2009 #15
    Okay here is what I think is going on. You have a formula for the resistance, but this depends upon the path length and the cross-sectional area. In this problem neither of those things are constant. You have to consider an infinitesimal path length and area. If we switch to sperical polars you can see that the only factor changing the surface area is r (theta and phi are the same for whatever hemisphere you choose). I think the expression should look something like this,

    [tex]\int{RdA}=\int{\rho*dl}[/tex]

    Although if I am wrong please could someone correct me?
     
  17. Feb 19, 2009 #16
    L(or l) should be along the path that the electricity will flow. We need to break this up into infinitesimally thin sheets of thickness dr (the L) with changing area. Thus, you get [tex]R=\int_{a}^{b} \frac{\rho dr}{A}[/tex] where A (the area a hemisphere of radius r) is a function of r and thus you can integrate.
     
  18. Feb 19, 2009 #17

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    Okay, so:

    [tex] R=\int_{a}^{b} \frac{\rho dr}{A} [/tex]

    We also know the area of a hemisphere is:

    [tex] 2\pi r^2 [/tex]

    Thus giving:

    [tex] R=\int_{a}^{b} \frac{\rho dr}{2\pi r^2} [/tex]

    So now would I have to integrate this?
     
  19. Feb 20, 2009 #18
    We are making a pig's ear of solving this.

    The current I flows radially.

    dV/dr = - resistivity * current per unit area = -rho * I/(2*pi*r^2)

    Integrate from a to b to find V.

    R = V/I
     
    Last edited: Feb 20, 2009
  20. Feb 20, 2009 #19
    Bingo
    why call r tau?
     
  21. Feb 20, 2009 #20

    TFM

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    Okay, so talke the ocnstants outside:

    [tex] R = \frac{\rho}{2\pi}\int_{a}^{b} \frac{dr}{ r^2} [/tex]


    [tex] R = \frac{\rho}{2\pi}\int_{a}^{b} r^{-2 dr} dr [/tex]

    Makes:

    [tex] R = \frac{\rho}{2\pi} [r^{-1}/1]_a^b [/tex]

    Thus:

    [tex] R = \frac{\rho}{2\pi} [\frac{1}{r}]_a^b [/tex]

    [tex] R = [\frac{\rho}{2\pi r}]_a^b [/tex]

    [tex] R = \frac{\rho}{2\pi b} - \frac{\rho}{2\pi a} [/tex]

    Does this look okay?
     
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