6
$\begingroup$

If we take a (lets imagine cylindrical) resistor in DC (steady state), we have that the electric field follows Ohm’s law:

$\mathbf J_f=\sigma \mathbf E$. where $\mathbf J_f$ is the free electron current density.

Since it is in steady state it also follows from the continuity equation that $\nabla \cdot \mathbf J_f=0$

Putting the first equation inside the second we get for an homogeneous medium that: $\nabla \cdot (\sigma \mathbf E)=0$, hence $\nabla \cdot \mathbf E=0$, hence according to Gauss’ law $\nabla \cdot \mathbf E= \rho_f/\epsilon =0$.

In other words in steady state the free charge inside a resistor is zero.

The big problem is that according to any book I have read (although not a mathematical reason have been given) charge density and electric field are spatially uniform inside a resistor in DC.

Yet, $\mathbf J=\rho_f \mathbf V$ (where $\rho_f $ is the free charge density), and since $\rho_f=0$ , $\mathbf J$ and $\mathbf E$ should be zero

How can $\mathbf E$ and $\mathbf J$ be nonzero and uniform in steady state, if the above equations indicate they should be zero?

$\endgroup$

4 Answers 4

5
$\begingroup$

Yet, $\mathbf J=\rho_f \mathbf V$ (where $\rho_f $ is the free charge density), and since $\rho_f=0$ , $\mathbf J$ and $\mathbf E$ should be zero

The problem that you are running into is that this statement is not really correct. It should actually be: $\mathbf J_f= \Sigma \rho_i \mathbf V_i$ (where $\rho_i$ is the charge density of the i’th type of free charge and $\mathbf V_i$ is its drift velocity).

In a typical metallic conductor you will have a very large negative $\rho_{electron}$ with a very small $\mathbf V_{electron}$. You will have an equally large positive $\rho_{proton}$ with $\mathbf V_{proton}=0$. Note that even though the protons are fixed in place they are still considered free charges. This is because they do not form dipoles with an overall neutral charge. Bound charges are overall neutral but with a dipole moment that can be polarized.

In an electrolyte like a sodium chloride solution you will have a negative $\rho_{Cl^-}$ and an equal magnitude but positive $\rho_{Na^+}$ each with their own velocities pointed in opposite directions.

Since different types of free charge will have different velocities you cannot simply lump them all together as your expression tried to do.

$\endgroup$
4
  • $\begingroup$ Thanks, this one is the answer I was looking for. Indeed my mistake was to not consider the inmovable (zero velocity) holes the electrons in the conductor leave as "free charge". I didnt use to think of them as free charge since the can not move. Most books only mention electrons as free charge and hence the origin of my mistake. I always thought of them as bounded charges. I actually realized my mistake by reading the book "Electromagnetic Fields and Energy" by Haus, I can not recommend it enough. It can be found online in the following link: web.mit.edu/6.013_book/www/body.html $\endgroup$ Commented Oct 9, 2020 at 18:35
  • 1
    $\begingroup$ @Daniel Rodriguez said “I always thought of them as bounded charges.” Yes, I think that is common. I will add a paragraph explaining that. $\endgroup$
    – Dale
    Commented Oct 9, 2020 at 19:04
  • 1
    $\begingroup$ Nitpick: the drift velocity of an ion species in an electrolyte is determined by its electrical mobility $\mu$. Specifically, $\vec{v}_d = \mu \vec{E}$. Electrical mobility varies according to the mass & physical size of the ion. This means that the drift speeds of different ion species in a solution are not generally the same, as your penultimate paragraph implies. $\endgroup$ Commented Oct 14, 2020 at 13:42
  • $\begingroup$ @MichaelSeifert you are absolutely correct. Thanks for the catch. I will correct that immediately. $\endgroup$
    – Dale
    Commented Oct 14, 2020 at 13:52
1
$\begingroup$

$\rho$ is zero inside a resistor, because the positive and negative charges cancel each other. $J$ is non-zero since only electrons do the moving.

$\endgroup$
0
$\begingroup$

You have mis-stated Gauss's Law.

Gauss's Law can be written as either

$${\bf\nabla}\cdot{\bf E}=\frac{\rho}{\epsilon_0}$$ where $\rho$ is the total charge (not the free charge); or as $${\bf\nabla}\cdot{\bf D}=\rho_f$$ where $\bf D$ is the electric displacement field and $\rho_f$ is the free charge.

So we don't have 0 free charge (if we did, $\sigma$ would be zero), we have 0 total charge, including both free (current carriers) and fixed charges (nuclear protons).

$\endgroup$
2
  • $\begingroup$ You are only partially correct, indeed the total charge is also zero, but your second equation of gauss law is wrong. The correct form is $div(D)=\rho_f$. Now for a linear homogeneous material $D=\epsilon E$, and taking epsilon out of the divergence you can write gauss law as $div(E)=\rho_f /\epsilon$. $\endgroup$ Commented Oct 9, 2020 at 6:45
  • $\begingroup$ After I have finally understood the answer to my question, I wanted to clarify something in your comment for any future readers with the same question. Indeed the total charge is zero as you put. However, also the NET free charge is zero. As pointed above this means that the electron and hole density cancel each other. Yet the current density is only proportional to the electron density so we can indeed have a zero net free charge and a finite current density. $\endgroup$ Commented Oct 9, 2020 at 18:52
-1
$\begingroup$

In a current carrying conductor (with any amount of resistance), the free electron density and the total charge density is not constant. A power supply takes electrons from the positive end and puts them into the negative end. In a uniform conductor a uniform current requires a uniform electric field which implies a uniform gradient in the charge density. Gauss's law tells us that a uniform field can be produce by one or more infinite sheets of charge. Clearly, we don't have that in electric circuit. So the question becomes; what kind of charge distribution can maintain a nearly uniform field within a length of wire which can have multiple bends, curves or loops? Each short segment of wire must see more charge beyond one end than beyond the other. This requires a non-uniform distribution of charge along the wire. Then each such segment (except one near the center) must contain a net charge. But the flux coming in from one end must equal that leaving at the other. This means that a segment with a net + charge must have flux leaving through the side of the wire, and one with a negative charge will have flux entering through the side. These components will participate in the divergence of the field, and will cause a radial shift in the distribution of free electrons.

$\endgroup$
7
  • $\begingroup$ This answer is about diffusion currents, but the current in a typical resistor is drift current. $\endgroup$
    – The Photon
    Commented Oct 10, 2020 at 21:23
  • $\begingroup$ A drift current requires a an electric field, and a uniform electric field in a conductor requires a charge density gradient. $\endgroup$
    – R.W. Bird
    Commented Oct 11, 2020 at 19:53
  • $\begingroup$ A uniform charge density gradient would produce a quadratic variation of electric field, according to Gauss's law. $\endgroup$
    – The Photon
    Commented Oct 11, 2020 at 19:58
  • $\begingroup$ Gauss's law is as valid in a conductor as anywhere else. The only way to get a uniform electric field is with zero net charge density (meaning also zero gradient of charge density). In your scenario (charge density varying along the length of the wire) how do you resolve Kirchhoff's current law? Is the charge velocity also varying along the length of the wire to compensate for the changing charge density? $\endgroup$
    – The Photon
    Commented Oct 12, 2020 at 16:43
  • $\begingroup$ Gauss's law is always valid, the problem is choosing a closed surface for which you can calculate the flux. Keep in mind that there are electric fields leaving (or entering) from the sides of a current carrying wire. Your final comment does seem to be a good point. The field and drift velocity will need to be larger near the positive end of the wire where the number of free electrons is lower, in order to maintain continuity of current flow. This affect is apparently quite small. $\endgroup$
    – R.W. Bird
    Commented Oct 13, 2020 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.