I am trying to learn some elementary EM, but I have some confusion about the basic concepts of steady current.

Suppose I have a wire of uniform cross section area. The current is always flowing from left to right.

I imagine that I can cut a segment of this wire (with the area vectors at both ends parallel to the flow of current) and compute the surface integral $\int\int_S \mathbf{J}\cdot d\mathbf{A}$, where $\mathbf{J}$ is the current density vector, $\mathbf{A}$ is the area vector, and $S$ is the boundary of the segment. I assume that $\mathbf{J}$ does not vary with time.

I believe that the magnitude of $\mathbf{J}$ can depend on its position, so lets say that its magnitude is greater on the right end, as compared to the left end. Because of that, the surface integral $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ should have a non-zero value (the dot products on both ends do not cancel).

However, $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ is precisely the net change in charge out of this segment, i.e., $-\frac{dq}{dt}$.

Now, for a steady current, $\frac{\partial \rho}{\partial t}$ is zero at every point, thus the $-\frac{dq}{dt}$ should also be zero, which contradicts my understanding that $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ can be non-zero.

This is perhaps a very stupid question, but I just cannot figure out what has gone wrong. Any help would be greatly appreciated~

  • $\begingroup$ The integral you've noted is not a closed surface it takes place over an open surface which is the cross section of the wire. As such it is not zero unless J is identically zero $\endgroup$
    – Triatticus
    Sep 11, 2018 at 0:41
  • $\begingroup$ I think your assertion "for a steady current, ∂ρ/∂t is zero at every point" is not correct. This requirement is for steady charge, not for steady current. $\endgroup$
    – verdelite
    Sep 11, 2018 at 3:11
  • $\begingroup$ @verdelite: I was attending some online (free) lectures and the professor simply said magnetostatic = steady current = ∂ρ/∂t is zero. Would you mind explaining why this assertion is in fact incorrect? $\endgroup$
    – julyfire
    Sep 11, 2018 at 3:27
  • $\begingroup$ @julyfire: That was my understanding. However, professors seem not agree with it. In Griffiths' 3rd edition, he said "By steady current I mean a continuous flow that has been going on forever, without change and without charge piling up anywhere". So "a continuous flow that has been going on forever" and ∂ρ/∂t=0 are two different things. They just require both for their definition of steady current. $\endgroup$
    – verdelite
    Sep 11, 2018 at 3:44
  • $\begingroup$ @verdelite so, am I correct to say that a "time independent current" is not necessarily equivalent to "no charge piling up anywhere", and people may define "steady current" differently? $\endgroup$
    – julyfire
    Sep 11, 2018 at 4:34

2 Answers 2


$\int\int_S \mathbf{J}\cdot d\mathbf{A}$ represents the current $(\frac {dq} {dt})$ flowing through a cross-section area. It is not "the net change in charge out of this segment".

The "net change in the charge" would be equal to the difference between the current flowing in (through the left cross-section) and the current flowing out (through the right cross-section) of the segment.

Updating the answer based on the comments:

If $S$ is the whole surface of the segment, then $\int\int_S \mathbf{J}\cdot d\mathbf{A}$ represents the net charge flow and, in a steady state, should be zero. That does not change, if the current density through the left cross-section is different from the current density through the right cross-section, since the differences in the current density would be compensated by the differences in cross-section areas, yielding the same currents.

  • $\begingroup$ Thanks for the answer. Perhaps I am unclear with my notation, but I in fact want to define S to be the boundary of a cylinder segment (including both the left and right cross sections, and the side of the cylinder that does not contribute to the dot product). $\endgroup$
    – julyfire
    Sep 11, 2018 at 1:56
  • $\begingroup$ I understand that the net change in charge is the difference between the current flowing and the current flowing out. For a steady current, this net change is zero. But why can't the current flowing in be different from the current flowing out? I mean, steady current does not depend on time, but can still be different between the left cross section and right cross section, right? $\endgroup$
    – julyfire
    Sep 11, 2018 at 1:59
  • 1
    $\begingroup$ @julyfire, what happens to the charge that flows in but does not flow out? Can that continue indefinitely or will it affect the flow after some time (so that the flow won't be steady)? $\endgroup$
    – The Photon
    Sep 11, 2018 at 5:24
  • $\begingroup$ "...since the differences in the current density would be compensated by the differences in cross-section areas". What if the cross-sectional area is constant, and there is some mechanism that causes charges to change velocity while moving through the wire segment? $\endgroup$ Sep 11, 2018 at 12:12
  • $\begingroup$ @AaronStevens If the cross-section area is the same, the density, in the steady state, should be the same. Therefore, if the charges move faster (have greater average or drift velocity) through one of the cross-sections, there must be fewer of them. $\endgroup$
    – V.F.
    Sep 11, 2018 at 12:28

$\frac{\partial}{\partial t}\rho = 0$ implies $\nabla\cdot\mathbf{J} = 0$. Then, $I = \oint_S \mathbf{J}\,d\mathbf{S} = \iiint_{\Omega} \nabla\cdot\mathbf{J}\,d\Omega = 0$

If you consider $S$ as a cylinder, you will see that the current density does not depend on the cross section you take.


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