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Homework Help: Grounding resistance

  1. Nov 14, 2006 #1
    I'm not sure of the proper term here, but I'd guess it is either earth, ground or grounding resistance.

    The task is to compute this grounding resistance for a sphere electrode (radius a) in a ground (conductivity sigma). The definition given for grounding resistance "between the surface of the electrode and a distant point in the ground" is the ratio of voltage between these two points (let say a and b) and current to the electrode, U_ab / I.

    But I don't really know how I should use the definition in this computation. All I can come up with is this:

    sigma = 1/rho => rho = 1/sigma

    [tex]\Large R=\frac{\rho L}{A},L=dr,A=2 \pi r^2[/tex]

    [tex]\Large dR = \frac {\rho dr} {2 \pi r^2}[/tex]

    [tex]\Large R=\int_a^b dR[/tex]

    But this gives the resistance of the whole sphere of ground, right?

    So should I just compute using the first equation, where L=distance between a and b, A=pi*a^2 (cross section of an imaginary tube just big enough to swallow the electrode)? Doesn't make sense either.
    Last edited: Nov 14, 2006
  2. jcsd
  3. Nov 14, 2006 #2


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    I think you are on the right track, but the area of a sphere is [itex] A=4 \pi r^2[/itex]. I would take the upper limit of integration to infinity. That is as distant as you can get.
  4. Nov 14, 2006 #3
    Oh, of course. Many thanks.

    However, I fail to understand the solution. Why is the area of the whole sphere around the electrode taken into the computation? I mean: if an electron tries to get from point a on the surface of the electrode to point b somewhere in the ground, what does this have to do with the part of the sphere that's on the opposite side of the electrode than b?

    Like this:

    (-b) ------------- (-a) --- (a) ------------- (b)

    Q tries to get from (a) to (b), but in the integral the whole area is taken into account, also the area in (-b).
  5. Nov 14, 2006 #4


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    Imagine a few million electrons all trying to make their escape from the sphere at the same time. Their mutual repulsion would spread them uniformly around the sphere travelling in all directions. At a fundametal level, the resistance to the movement of a single charge is very complicated. It really only makes sense to talk about conductivity and resistivity in an average sense involving many charge carriers. That would be true even for a wire of uniform diameter. The simple conductivity relationship that is the basis for your caclulation is consistent with the observed average behavior when many charges are migrating through a conductor.
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