According to what I was taught, if current was dispersed “uniformly,” current density would remain constant. So, in a conductor, the 'current density should be the same at all points.' But, given that electrons flow at random (their time between collisions differs from that of other electrons), how can current density be the same everywhere?

Every electron's drift velocity is different. And I am not talking about the average drift velocity here. Rather, specific drift velocities.

An electron's drift velocity differs from that of other electrons, so the j vector should differ as well, right?

Why, then, is the j vector the same everywhere?

I believe I am missing something crucial.

Please correct me and explain your reasoning.


Please review the changes.

I get that we're smoothing out the fluctuations by averaging current densities, but the actual question is, "Are there any fluctuations at all in reality?" or "Are current densities truly different at all points?" (If the electric field is consistent)

This is because the equation

Eσ = J*

(E=electric field,σ - conductivity, j=current density)

implies that if E is uniform (the same at all points of the conductor), then J is uniform at all "points." As a result, the last question is —

How is this possible that j has the same value at each position when each electron's drift velocity varies(as time between two subsequent collision varies) ?


How can q in j= q/(Δt*dA) be the same at all places for a given amount of time Δt and very small area dA(point),where q is charge crossing that point?

Please assist me in resolving my confusion.

  • 1
    $\begingroup$ The constant value of J is an approximation, here J refers to the average drift current at any location in space, and is derived from the drude model. Where the resistive force present, follows a single function, unlike the "abrupt" resistance an electron encounters when it hits a proton $\endgroup$ Commented May 22, 2022 at 19:39
  • $\begingroup$ @jensenpaul J constant can also be derived assuming the charge distribution on the surface of the wire is linear along the wire. The quesfion is then why does the charge arranges in this linear way when the power source is plugged and the circuit is closed. If you prove that it is the case then you answer OP's question. $\endgroup$ Commented May 23, 2022 at 5:30

4 Answers 4


A uniform current density $\vec{J}$ is just as much an abstraction as a uniform charge density $\rho$.

Even if a "real" charge density $\rho$ is made up of a bunch of different point charges at discrete points in space, averaging the charge over a sufficiently large volume (which is still going to be pretty small on macroscopic scales) will let us think of it as being approximately uniform so we can solve problems about it using Gauss's Law & the like.

Similarly, even if a "real" current density $\vec{J}$ is made up of many different electrons with different speeds, we can define an averaged current density by looking at the amount of charge per time that crosses a particular surface over a particular period of time, and defining the magnitude of $\vec{J}$ to be $Q/(\Delta A \Delta t)$. This averages out all the microscopic fluctuations in current that you're worried about and lets us treat $\vec{J}$ as nice and uniform on larger scales.

The situation is analogous to water in a pipe or an air current, where on a microscopic level the density is not uniform and the molecular velocities vary according to the Maxwell-Boltzmann distribution with a small additional "drift" velocity. Moreover, as with water flows and current flows, the current density $\vec{J}$ doesn't have to be uniform in space or in time; it can vary over space or time either in a "microscopic sense" or in a "macroscopic sense" (think of turbulent water flow or a storm with gusts of wind.) The only reason you usually see uniform $\vec{J}$ in introductory E&M classes is that it makes the calculations easier for beginning students.

Similarly, the equation $\vec{E} = \sigma \vec{J}$ (which applies inside a conductor) is an equation that really only applies to an "averaged" electric field. In general, when we talk about the electric field "inside a medium", we are really referring to the average of an electric field over some volume that is large compared to the size of the molecules that comprise it. This issue is discussed briefly in §4.2.3 of Griffith's Introduction to Electrodynamics, in the context of linear dielectrics.

  • $\begingroup$ Many thanks!! Your response was useful to me. $\endgroup$
    – TPL
    Commented May 23, 2022 at 11:54
  • $\begingroup$ This implies that we disregard microscopic variations since they are unimportant, correct? But how do we approximate the current flowing through the dA region (in conductor) to be the same at all points? $\endgroup$
    – TPL
    Commented May 23, 2022 at 12:01
  • $\begingroup$ @TonyPhysicslover: I'm not quite sure what you're asking. The averaged current density is not necessarily uniform over larger scales. You can certainly envision current distributions $\vec{J}$ where this is not the case; for example, imagine a beam of electrons with uniform average velocity throughout the volume of the beam but which is denser near the center. $\endgroup$ Commented May 23, 2022 at 13:04
  • $\begingroup$ (Due to the fact that current density is a point function)At each “point” in the conductor, the magnitude of the j vector equals net current/area, correct? I'm curious/puzzled as to how we can be so certain that the same number of electrons cross from every “point” (da region) in a given period of time, despite the fact that their velocity varies. $\endgroup$
    – TPL
    Commented May 23, 2022 at 13:59
  • $\begingroup$ @TonyPhysicslover: That's what the averaging takes care of. (And there's an implicit averaging over time as well as over space; I'll edit my answer to reflect this.) $\endgroup$ Commented May 23, 2022 at 14:03

"Are there any fluctuations at all in reality?"

In reality, there is Johnson-Nyquist noise: thermal current/voltage fluctuations in the conductor. This is routinely measurable with sensitive equipment, and is something that affects our ability to construct sensitive equipment.


But, given that electrons flow at random, how can current density be the same everywhere?

Electrons do not flow purely at random. There is a systematic part of their motion in addition to the random part. The microscopic random part of their motion is isotropic, so it does not contribute to the macroscopic current density. So the macroscopic current density is the same everywhere because the non-random part of the microscopic motion is the same everywhere

  • $\begingroup$ You're overlooking the fact that the time period between two collisions differs from one electron to the next in some way. As a result, each electron's particular drift velocity (systematic component of motion) or velocity owing to the Electric Field differs from the velocity of other electrons. As a result, the "non-random" component is not consistent across the board! $\endgroup$
    – TPL
    Commented May 23, 2022 at 5:06
  • $\begingroup$ @TonyPhysicslover I am not overlooking that. That is the non-random part, and the non-random part is indeed consistent, it is the random component that differs $\endgroup$
    – Dale
    Commented May 23, 2022 at 14:12
  • $\begingroup$ When you say so, do you mean that the duration between any two collisions for each electron in the conductor is the same, and that the velocity imparted by the electric field (for each electron) is also the same? Please provide proof or a supporting document. $\endgroup$
    – TPL
    Commented May 23, 2022 at 14:24
  • $\begingroup$ No, I mean that the random part is spatially isotropic and homogenous. The inhomogenous part is non-random and is consistent. You have mixed up in your mind what is random and what is non-random. The non-random part is consistent and homogenous, the random part is inconsistent and inhomogenous. The electron motion is not purely random. $\endgroup$
    – Dale
    Commented May 23, 2022 at 16:13

The average drift velocity is proportional to the electric field strength. For the potential difference between the two ends of a uniform wire to be independent of the radius, the electric field must also be not a function of the radius.


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