# Are Charge Imbalances Present on Resistors?

1. May 27, 2014

### Prince Rilian

I know that in electric circuits charge imbalances can build up on capacitor plates, wire bends, and antennas. But is this also the case for a resistor with a voltage applied between its terminals? In other words, which of the following diagrams would be correct?

Note that a + stands for an unbalanced, stationary positive charge; a - stands for an unbalanced, stationary negative charge; and a 0 stands for no charge imbalance present at the spot. Moving charges are depicted by the J (current density) vector.

Please note that I have depicted the resistor with electric field and current density vectors rather than using the standard voltage and current quantities as the root of my problem understanding this scenario is that I am having trouble accounting for what is giving rise to the electric field vector. I am aware that

Va-Vb = Ed,

where d is the distance across the resistor, but can't only unbalanced electric charges or changing magnetic fields give rise to electric fields? So how can you trace the charges bringing about the E vector back to the voltage applied across the resistor?

2. May 27, 2014

### UltrafastPED

The electrons in a conductor have a very rapid, random motion that is similar to that of gas molecules in the kinetic theory of gasses. Application of a voltage results in an additional drift velocity in the direction of the applied force.

Resistance is a property of conductors; some (like tungsten) have very high resistivity, while others (like copper, silver, aluminum) have quite low resistivity. The origin of resistivity (=resistance when applied to the geometry of a device) is a "resistance" to this drift velocity.

So by measuring the voltage drop at many points along a resistive material you find that there is some loss of motive force for each increment of distance. This is pretty much how Ohm derived his original law in the 1820s when he measured the voltage through different types and lengths of wires.

The loss in voltage is a measure of the reduction in the net drift velocity of the charges.

3. May 27, 2014

### Prince Rilian

Also, how would you relate my question to what gives rise to Poynting vectors? Thinking about this, I think my second figure may be the correct one. But I still need a little more explanation. It isn't all clicking yet.

http://www.furryelephant.com/conten...tric-current/surface-charges-poynting-vector/

Last edited: May 27, 2014
4. May 27, 2014

### UltrafastPED

There are some fundamental problems with this site ... the author even says "I have to admit it's quite a way outside my comfort zone but here is the gist."

Their are two kinds of electrons in a metal (which covers all conductors and resistors, but not semiconductors or insulators): electrons that are bound to the atoms that make up the crystal structure of the metal, and free electrons. These free electrons are similar to the valence electrons for covalent substances, but in the case of a metal they are very, very free!

For simplicity consider a metal with one free electron per atom. Then if there is one mole of the substance, then we have Avogadro's number of free electrons. They travel independently of each other, not even feeling the force from passing electrons - this is because the metal overall is neutral - the free electrons just balance the net charge of the bound electrons and their atomic nuclei.

Treated semi-classically (e.g., without invoking the full machinery of quantum mechanics, which was unknown at the time), Drude used Maxwell's equations plus the ideas from Maxwell's kinetic theory of gasses to describe this "electron sea" that exists inside of metals. This approach is not quite right, producing some calculations that disagree with experiment, but also some that agree with the quantum mechanical results and experiment.

It is called the Drude model, naturally, and dates from 1900.

And if you are really interested, read the attachment as well - I wrote it for the students in an introductory electrical engineering course.

You can see already that there are some problems with the statements found on your referenced web site. So let's ignore that site completely.

Now - what is your question?

5. Jun 3, 2014

### Prince Rilian

To re-ask the question bluntly, for any given circuit, where can we find out where any and all electrical charge imbalances are physically located? I know how to find many other parameters in circuits (voltage, current, resistance, capacitance, inductance, etc.) but I do not feel that I have a complete knowledge of what is going on in a circuit until I can "keep tabs on every last electron".

Last edited: Jun 3, 2014
6. Jun 3, 2014

### UltrafastPED

Why do you expect any charge imbalance? Currents are neutral.

Capacitors are the place to look.

PS: what is your level of training? Student, hobbyist, engineer, etc.

7. Jun 4, 2014

### Prince Rilian

The reason I am expecting charge imbalance is because of Ohm's Law for points in space:

Eσ = J

and only two things can ultimately make an electric field--charge imbalances or varying magnetic fields.

8. Jun 4, 2014

### Jano L.

According to the EM theory, there cannot be non-vanishing electric charge density inside the conductor, any charge artificially put inside the conductor will dissipate and move to the surface. However, as you say the electric field inside the conductor has to be due to some charges somewhere. The only place charges can sit stable is the surface of the wire and of the source of voltage. When constant current flows through the circuit, the charge redistributes itself in such a way as to create electric field necessary to drive this current.

9. Jun 4, 2014

### Jano L.

More on surface charges : http://www.ifi.unicamp.br/~assis/the-electric-force-of-a-current.pdf [Broken]

Last edited by a moderator: May 6, 2017
10. Jun 4, 2014

### Jano L.

The site explains quite well the situation from the point of view of macroscopic electromagnetic theory. What you described (Drude model) is a microscopic theory. These theories are for different scales, but they do not apparently contradict each other.

11. Jun 4, 2014

### Prince Rilian

Thank you, Jano L., for the hyperlink. That document looks like it will answer my questions.

12. Jun 4, 2014

### UltrafastPED

This version of Ohm's law applies to "current density", which is the local charge density x velocity; for individual charges use the Lorentz force law, though the concept of resistance is not really meaningful for individual charges: resistance is the result of statistics applied to large numbers of charges.

Ohm's law (in either form) makes clear that an external electric field is the electromotive force. This field is provided by the battery or generator. The field is (like any force) proportional to the gradient of the electric potential; hence the uniform change in voltage as the current passes through a region of constant resistivity means that the electric field is essentially constant there.

This is all discussed in considerable detail in Griffith's "Introduction to Electrodynamics", chapter 7: Electrodynamics.

There should not be any "surface charge on the wires" except perhaps "instantaneously" as switching occurs. Griffith's discusses this in section 7.1.2.

Last edited: Jun 4, 2014
13. Jun 4, 2014

### UltrafastPED

According to a review of this book:

"... the focus of which is refuting the charge levelled by R. Clausius, J. C. Maxwell, and others, that the alleged failure to detect a force between a current-carrying wire and a nearby stationary charge invalidates Weber's fundamental law."

Since Weber's theories are incompatible with Maxwell's equations ... I think that this is not a good resource, and certainly not for beginners!

Last edited by a moderator: May 6, 2017
14. Jun 5, 2014

### Jano L.

The vector $\mathbf E$ in the equation
$$\mathbf j = \sigma \mathbf E$$
is total electric field. It does not make sense to say it is external. Also, this electric field is not electromotive force. Electromotive force is a technical term that means "integral of electromotive intensity (Griffiths' $\mathbf f_s$) over the circuit". In situations where the current is stationary, electromotive intensity usually vanishes in the wire, and is non-zero only inside the source of voltage.

The electric field inside the wire is partially provided by the battery or generator, but usually there is contribution due to surface charges as well. This is because the wire usually does not line up with the electric field line of the battery. To maintain the wire lined up with electric field line, additional sources of electric field are necessary.

Griffiths is describing how stationary constant current arises from non-stationary charge distribution. The charges he talks about are not necessarily on the surface, but in the bulk as well. He is right in the respect that any charge inside will disappear quickly. However, he forgets to mention the charges on the surface. These stay to provide the electric field necessary to drive the constant current.

15. Jun 5, 2014

### Delta²

That statement seems quite interesting. I havent read the book but let me give this counterexample:

Suppose we have a battery with negative and positive pole. We all understand that the electric field generated by this battery will very much look alike the electric field of two point charges +q1, -q2 separated by a distance. Now suppose that we connect a piece of wire that is totally straight (with no twists and curves) to the poles of battery. I assume we all agree that the current density inside that wire will also be straight (that is the vector will always point along the straight direction of the wire). How the electric field of the battery alone produces such a current density according to Ohms law (J=sigma*E)? The only way is to assume there are surface charge density such that the total electric field inside the conductor is also a vector pointing straight along the wire. Am i mistaken somewhere here??

16. Jun 6, 2014

### vanhees71

Concerning Griffiths's book on electromagnetism I come more and more to the conclusion that it seems to be confusing sometimes and not always leads to correct ideas in the student's minds since it is well known that of course there are surface charges on conductors also for DC currents. The here critizized book by Assis is not as bad as you claim, because as far as I have looked at it, it gives correct solutions to the corresponding stationary problems. I'm not so sure concerning the didactics, because I don't see the sense of discussing old-fashioned action-at-a-distance models by Weber et al. The correct thing are the Maxwell Equations (in local form)!

So let's do the most simple example of an infinite coaxial cable. Let the inner cylinder of radius $a$ be along the [itez]z[/itex] axis and the outer conductor a cylindrical shell with inner radius $b$ and outer radius $c$.

In the stationary case the Maxwell equations separate completely into two equations for the electric and two equations for the magnetic field components. The electric part reads
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=0.$$
Further we only need the non-relativistic approximation of Ohm's law, i.e.,
$$\vec{j}=\sigma \vec{E}.$$
We make the ansatz that in the inner wire and the outer shell we have homogeneous current densities, i.e.,
$$\vec{j}=\begin{cases} \frac{I}{\pi a^2} \vec{e}_z & \text{for} \quad 0 \leq \rho <a,\\ -\frac{I}{\pi(c^2-b^2)} \vec{e}_z & \text{for} \quad b<\rho<c,\\ 0 & \text{elsewhere}. \end{cases}$$
Here $(\rho,\varphi,z)$ are the usual cylinder coordinates. Note that this ansatz takes into account global charge conservation. Local charge conservation is fulfilled anyway since $\vec{\nabla} \cdot \vec{j}=0$ everywhere, as it must be.

We shall now see that this ansatz admits an exact solution of the stationary Maxwell equations, satisfying all boundary conditions.

To that end we note that due to the first stationary Maxwell equations the electric field can be written as the gradient of a scalar potential, which due to symmetries should be of the form
$$\Phi(\vec{x})=z \phi(\rho),$$
where $\phi$ is a still to be determined continuous function.

We further know that, according to Ohm's Law we must have
$$\vec{E}=\frac{I}{\pi a^2 \sigma} \vec{e}_z \quad \text{for} \quad 0 \leq \rho<a$$
and
$$\vec{E}=-\frac{I}{\pi (c^2-b^2)\sigma} \vec{e}_z \quad \text{for} \quad b < \rho<c.$$
The field is to be determined in the other region, i.e., in the gap between the conductors (for $a<\rho<b$) and outside of the coax cable, $\rho>c$.

To that end we use the ansatz for the potential to get
$$\vec{E}=-\vec{\nabla} \Phi=-\phi(\rho) \vec{e}_z - z \phi'(\rho) \vec{e}_{\rho}.$$
Ohm's Law implies that
$$\phi(\rho)=-\frac{I}{\pi a^2 \sigma} \quad \text{for} \quad 0 \leq \rho<a$$
and
$$\phi(\rho)=+\frac{I}{\pi (c^2-b^2) \sigma} \quad \text{for} \quad b<\rho<c.$$
For the missing regions we have to solve the potential equation, following from Gauß's Law,
$$\vec{\nabla} \cdot \vec{E}=-\Delta \Phi=0 \; \Rightarrow \; \frac{\mathrm{d}}{\mathrm{d} \rho} [\rho \phi'(\rho)]=0.$$
The integration of this equation leads to
$$\phi(\rho)=A_j \ln(\rho/a) + B_j$$
with integration constants $A_{j}$, $B_{j}$ in
the regions 1 ($a<\rho<b$) and 2 ($\rho > c$). I've also written $\ln(\rho/a)$ to avoid dimensionful quantities as the argument of a logarithm. Of course, one can choose any other constant instead of $a$, but changing it just adds a constant to the potential that is physically irrelevant.

To determine the integration constants we need the boundary conditions at the surfaces of the wire and the cylindrical shell. These are given by the continuity of the potential and of the tangential component, $E_z$ of the electric field. With our ansatz the first condition already fixes also the second. So we have
$$\phi(a)=B_1=-\frac{I}{\pi a^2 \sigma}$$
and
$$\phi(b)=A_1 \ln(b/a)+B_1=+\frac{I}{\pi(c^2-b^2)\sigma} \; \Rightarrow\; A_1=\frac{I}{\sigma \pi \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2+b^2} \right ).$$
For the region outside the coax cable we must demand that the field stays finite at infinity, leading to
$$A_2=0, \quad B_2=\frac{I}{\pi(c^2-b^2) \sigma}.$$
So finally we get
$$\phi(\rho)=\begin{cases} -I/(\pi a^2 \sigma) & \text{for} \quad 0 \leq \rho<a, \\ I/[\pi \sigma \ln(b/a)] [1/a^2+1/(c^2-b^2)] \ln(\rho/a)-I/(\pi a^2 \sigma) & \text{for} \quad a \leq \rho <b,\\ I/[\pi (c^2-b^2) \sigma] & \text{for} \quad \rho \geq b. \end{cases}$$
At the surface of the wire $\rho=a$ and the inner surface of the shell of the outer conductor are surface charges of a density given by the jump of $E_{\rho}$, i.e.,
$$\sigma_{Q}=-\frac{I}{\pi \sigma \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2-b^2} \right ) \frac{z}{a}$$
and at $\rho=b$, it's
$$\sigma_{Q}=+\frac{I}{\pi \sigma \ln(b/a)} \left (\frac{1}{a^2}+\frac{1}{c^2-b^2} \right ) \frac{z}{b}.$$
The correct discussion is found in the good old book (for the case $c \rightarrow \infty$).

A. Sommerfeld, Lectures on Theoretical Physics, vol. 3 (electrodynamics)

There you also find a nice picture about the electric field lines and the energy flux (Poynting vector). One should note that the surface charges and the electric field outside the conductor are pretty small, so that we don't realize them in practice, but they are crucial for the entire conduction business and the energy flow.