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Relationship between current and current density for a volume conductor

  1. Jan 31, 2013 #1
    (I'd like to preface this with the warning that the following question may be a very dumb one.)

    My understanding is that current density (or flux) [itex] \vec{J} = \vec{J}(x, y, z) [/itex] is the rate of flow of charge (or the current) per unit area. (Units of [itex] \frac{\text{C/s}}{\text{cm}^2} [/itex].)

    Say we know that in an irregularly shaped volume conductor (e.g. a nerve fiber, where the irregularity is due to some sort of biological obstruction), the current density [itex] \vec{J} [/itex] has the form given in Figure 1 (see attachments).

    The current [itex] I [/itex] is --- I think --- given from Ohm's law: specifically, if the potential difference across the ends of the fiber and the total resistance of the fiber are known, then

    I = \frac{V}{R},

    which is a scalar quantity. I believe this value is constant for each point [itex] (x, y, z) [/itex] in the fiber, assuming no build-up of current anywhere (by the law of conservation of charge).

    Now I know that the surface integral of a flux gives a flow rate, so the surface integral of [itex] \vec{J} [/itex] should give a current. But is the current equal to [itex] I [/itex]? I mean, does

    \iint_S \vec{J} \cdot \vec{n} dS = I

    for every [itex] S [/itex], or only for [itex] S [/itex] = cross-sectional area of the conductor? The reason I ask is because, if I draw three example surfaces [itex] S_1, S_2, S_3 [/itex] (Figure 2), the surface integral of [itex] \vec{J} [/itex] over [itex] S_1 [/itex] is obviously not equal to that over [itex] S_2 [/itex].

    So I guess what I am asking is: is
    I = \iint_{S_1} \vec{J} \cdot \vec{n} dS_1 = \iint_{S_3} \vec{J} \cdot \vec{n} dS_3 \neq \iint_{S_2} \vec{J} \cdot \vec{n} dS_2
    true, where [itex] I [/itex] is current as found from Ohm's law? And if so, does this mean that any current [itex] I [/itex] through a conductor is not simply the rate of flow of charge, but a rate of flow of charge implicitly with respect to the cross-sectional area?


    Attached Files:

    • Fig1.png
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    • Fig2.png
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  2. jcsd
  3. Jan 31, 2013 #2


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    Fine up to here.
    No, this is wrong. Current I is a total amount, that is, it's the integral of J across a surface as you've written below. It doesn't make sense to talk about I at an infinitesimal point (x,y,z).
    In your example, it is the second option.
    Yes. Usually it's obvious what that surface is (current in a wire refers to the area of the wire), but if the surface isn't obvious then you need to specify it.
  4. Jan 31, 2013 #3
    Thank you for clearing things up.
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