## Relationship between current and current density for a volume conductor

(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
Attached Thumbnails

 PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus

Recognitions:
Science Advisor
 Quote by 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.
Fine up to here.
 Quote by Cole A. 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).
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).
 Quote by Cole A. 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?
In your example, it is the second option.
 Quote by Cole A. 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?
Yes.
 Quote by Cole A. 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
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.

 Thank you for clearing things up.
 Thread Tools

 Similar Threads for: Relationship between current and current density for a volume conductor Thread Forum Replies Engineering, Comp Sci, & Technology Homework 8 Advanced Physics Homework 1 Advanced Physics Homework 0 Quantum Physics 8 Classical Physics 4