# Relationship between current and current density for a volume conductor

1. Jan 31, 2013

### Cole A.

(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) $\vec{J} = \vec{J}(x, y, z)$ is the rate of flow of charge (or the current) per unit area. (Units of $\frac{\text{C/s}}{\text{cm}^2}$.)

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 $\vec{J}$ has the form given in Figure 1 (see attachments).

The current $I$ 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

$$\begin{equation*} I = \frac{V}{R}, \end{equation*}$$

which is a scalar quantity. I believe this value is constant for each point $(x, y, z)$ 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 $\vec{J}$ should give a current. But is the current equal to $I$? I mean, does

$$\begin{equation*} \iint_S \vec{J} \cdot \vec{n} dS = I \end{equation*}$$

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

So I guess what I am asking is: is
$$\begin{equation*} I = \iint_{S_1} \vec{J} \cdot \vec{n} dS_1 = \iint_{S_3} \vec{J} \cdot \vec{n} dS_3 \neq \iint_{S_2} \vec{J} \cdot \vec{n} dS_2 \end{equation*}$$
true, where $I$ is current as found from Ohm's law? And if so, does this mean that any current $I$ 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?

Thanks

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2. Jan 31, 2013

### marcusl

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.
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.

3. Jan 31, 2013

### Cole A.

Thank you for clearing things up.