# Finding resistance of cylinders with variable conductivity

## Homework Statement

Suppose the conductivity of the material separating two coaxial cylinders, of radius a and b (a < b) and held at a potential difference ##V##, would not be uniform. Specifically, ##\sigma(s) = k/s##, for some constant ##k##. Find the resistance between the cylinders. [Hint: Because ##\sigma## is a function of position, ##\nabla\cdot\vec{E}≠0##, the charge density is not zero in the resistive medium, and ##\vec{E}## does not go like ##1/s##. But we do know that for steady currents ##I## is the same across each cylindrical surface. Take it from there.]
[Image of the problem is attached below.]

## Homework Equations

Ohm's law: ##\vec{J} = \sigma\vec{E}## where ##\sigma## is the conductivity.
Gauss' law: ##\nabla\cdot\vec{E} = \frac{\rho}{\epsilon_0}## where ##\rho## is the volume charge density.
Definition of the electric potential: ##V(a) - v(b) = -\int_b^a \vec{E}\cdot\vec{dl}##
A handy product rule I used: ##\nabla\cdot(f\vec{A}) = f(\nabla\cdot\vec{A}) + \vec{A}\cdot(\nabla f)## where ##f## is a scalar function and ##\vec{A}## is a vector function.
For steady currents: ##\nabla\cdot\vec{J} = 0##

## The Attempt at a Solution

I began by calculating the charge density in the conductive volume: $$\nabla\cdot\vec{E} = \nabla\cdot\vec{\frac{1}{\sigma}\vec{J}} = \frac{1}{\sigma}(\nabla\cdot\vec{J}) + \vec{J}\cdot(\nabla \frac{1}{\sigma}) = \vec{J}\cdot(\nabla \frac{s}{k}) = \frac{J}{k} = \frac{\rho}{\epsilon_0}$$ From this, ##\rho = \frac{J\epsilon_0}{k}##. Now, by symmetry, both ##\vec{J}## and ##\vec{E}## point only in the ##\hat{s}## direction - i.e. perpendicular to the axis of the cylinders. Thus, I thought about using Gauss' law with ##\lambda## being the line density in the inner cylinder: $$E2\pi sL = \frac{1}{\epsilon_0}\left(\lambda L + \frac{J\epsilon_0L}{k}(s^2-a^2)\right)$$ After substituting in the above equation ##E = \frac{1}{\sigma} = \frac{s}{k}## and solving for ##J## I get: $$\vec{J} = \frac{k\lambda}{\pi\epsilon_0 (s^2+a^2)}\hat{s}$$ By definition: ##V = -\int_b^a \vec{E}\cdot\vec{dl} = \int_a^b \vec{\frac{s}{k}\vec{J}}\cdot\vec{dl}##. From this, we can get $$\lambda = \frac{2\pi\epsilon_0 V}{ln[(a^2+b^2)/(2a^2)]}$$ Substituting, $$\vec{J} = \frac{2kV}{(s^2+a^2)ln[(a^2+b^2)/(2a^2)]}$$ I could go further and find ##I## from its definition via ##J## but I already see a problem: we assumed ##\nabla\cdot\vec{J} = 0## but it is clearly not the case for the above formula... Something clearly went wrong and I can't figure out what, the argument seems sound to me (obviously...).

Sorry if this is a bit long but I got utterly confused here... Any suggestion/help/comment will be greatly appreciated on this one!

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ehild
Homework Helper
What is s?
How do you get the current I from the current density J?

ELB27
What is s?
How do you get the current I from the current density J?
Sorry, I should have stated this in the OP. ##s## is the distance from the axis of the cylinders (I define coordinates such that the axis is the z-axis). As for the current, by definition: ##I = \int\vec{J}\cdot\vec{da}## where ##da## is the surface area element perpendicular to the flow. In this case, it would be ##\vec{da} = \hat{s}sd\phi dz##. If I do the integration I get ##I = \frac{4\pi Lks}{(s^2+a^2)ln[(a^2+b^2)/(2a^2)]} V## and thus, apparently, ##R = \frac{(s^2+a^2)ln[(a^2+b^2)/(2a^2)]}{4\pi Lks}##

ehild
Homework Helper
Sorry, I should have stated this in the OP. ##s## is the distance from the axis of the cylinders (I define coordinates such that the axis is the z-axis). As for the current, by definition: ##I = \int\vec{J}\cdot\vec{da}## where ##da## is the surface area element perpendicular to the flow. In this case, it would be ##\vec{da} = \hat{s}sd\phi dz##.
J depends on s only. You can pull it out from the integral with respect to the surface element. So what is I at a certain s?

J depends on s only. You can pull it out from the integral with respect to the surface element. So what is I at a certain s?
Please see my edit in my previous post.

ehild
Homework Helper
Please see my edit in my previous post.
Answer my question please. What is the surface integral ##I = \int\vec{J}\cdot\vec{da}## in terms of J and s?

Answer my question please. What is the surface integral ##I = \int\vec{J}\cdot\vec{da}## in terms of J and s?
##I = \int_0^{2\pi}\int_0^L J(s)sdzd\phi = J(s)sL2\pi##

ehild
Homework Helper
##I = \int_0^{2\pi}\int_0^L J(s)sdzd\phi = J(s)sL2\pi##
Right. Substitute J=σE. I is the same across each cross-section, it does not depend on s. What does it mean on E?

ELB27
Right. Substitute J=σE. I is the same across each cross-section, it does not depend on s. What does it mean on E?
##I = \sigma E 2\pi Ls = \frac{k}{s}s2\pi LE = kLE2\pi##. If ##I## is constant than ##E = \frac{I}{2\pi kL}## is too. From this, ##V = -\int_b^a E ds = E(b-a) = \frac{b-a}{2\pi kL} I## and thus ##R = \frac{b-a}{2\pi kL}##. Is this the correct answer? And if so, what was the boggy step in my first attempt? (it seemed more intuitive to me..). In any case, it is already the 2nd time my attempts to approach a problem in electrodynamics with methods from electrostatics fails (the first was with discharge of a capacitor) and I very much want to know what is wrong with it

ehild
Homework Helper
(Why did you assume that div J =0? )

Edit: It is zero!

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Why did you assume that div J =0? J is parallel with the radial vector, s. In two dimension div s = 2
Well, the continuity equation states that ##\nabla\cdot\vec{J} = -\frac{∂\rho}{∂t}## where ##\rho## is the charge density and ##t## is time. Since the current is steady in the sense that it is the same for every ##s##, there cannot be any charge piling up anywhere and thus, the aforementioned time derivative must be zero (my book just states that for steady currents, ##\nabla\cdot\vec{J} = 0##).
EDIT: Also, it does hold for our solution, where ##J ∝ 1/s##.

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ehild
Homework Helper
You are right div J=0. But Gauss' Law is valid for Electrostatics. And in a medium, the charge density is equal to the divergence of the electric displacement.
I also do not understand what you mean on line density on the inner cylinder.

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You are right div J=0. But Gauss' Law is valid for Electrostatics. And in a medium, the charge density is equal to the divergence of the electric displacement.
I also do not understand what you mean on line density on the inner cylinder.
First of all, I think that Gauss' law is valid in electrodynamics too if the E-field is constant in time (my book uses it extensively and it is one of Maxwell's equations written in general form - see Wikipedia). As for the line charge density, I tried to replicate the argument of a simplified worked-out example of the problem where the conductivity is uniform. Picture attached (I used Griffiths' "Introduction to Electrodynamics"). I understand that it is the line charge density along the smaller cylinder which is used just as a convenient constant related to the total charge enclosed by the Gaussian surface. It is simply used to later find and it is convenient as it causes the ##L## to cancel. If I were to choose a surface charge ##\sigma## I would just add a constant ##2\pi a## to the equation involving Gauss' law which would then get canceled.

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ehild
Homework Helper
I meant the Maxwell equation in vacuum div E = ρ/ε0 is not valid in media. It has the form div D=ρ. (Gauss Law means the relation between volume integral and surface integral which is valid).
I did not learn Electrodynamics from Griffiths. For stationary currents, I use the Maxwell equations in the form curl H = J, curl E = 0 together with the material equation for the current density JE . As curl E = 0, E= - grad U. These equations are enough to find the current or potential difference without using the relation between the surface integral of E and enclosed charge in case of a current carrying conductor.

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ELB27
I meant the Maxwell equation in vacuum div E = ρ/ε0 is not valid in media. It has the form div D=ρ. (Gauss Law means the relation between volume integral and surface integral which is valid).
I did not learn Electrodynamics from Griffiths. For stationary currents, I use the Maxwell equations in the form curl H = J, curl E = 0 together with the material equation for the current density JE . As curl E = 0, E= - grad U. These equations are enough to find the current or potential difference without using the relation between the surface integral of E and enclosed charge in case of a current carrying conductor.
Thank you very much for your input. I will try to remember these methods next time. I will think a bit more about my original approach - it must be valid too after all - and will post if I figure it out (I'm not obsessed, I can stop whenever I want!...)

ehild
Homework Helper
Thank you very much for your input. I will try to remember these methods next time. I will think a bit more about my original approach - it must be valid too after all - and will post if I figure it out (I'm not obsessed, I can stop whenever I want!...)
It is very wise of you that you want to understand the problems.

In your original approach, you said the div E = q/ ε0. Then you applied Gauss Law to get E again, determining E(s) from the charge enclosed inside a cylinder of radius s. Meanwhile you introduced a non-existent volume charge density between the cylinders.

div E = ρ/ ε0 is valid for the microscopic field in the vacuum and ρ is the whole charge density of the charged particles the material consists of (Bound charge) + the excess charge (free charge) . But we can measure only the average of the electric field in such volumes where the bound charge density averages to zero.

The volume between the cylinders has no excess charge. It is electrically neutral. When connecting a battery across them or adding opposite charges to them, so as a potential difference and electric field is set up, no free charges are created. But the constituents of the material between the cylinders get polarized or aligned and dipole chains appear. The bound charge is inside those dipoles. It appears on the inner surfaces of the metal cylinders as surface charge.

The free charge density is equal to the divergence of the electric displacement vector, D, defined as D0E+P where P is the polarization, the dipole moment of unit volume. When you derived div E, it was 1/ε0(div D-divP)=1/ε0(ρ(free) +ρ(bound)). The free charge density is zero, and the average bound charge is also zero in the volume, it appears on the cylinders only as surface charge..

There is no need to work with Gauss Law in the conducting medium. You know the conductivity, σ. You can use JE, and find the current as integral over a the cross section. And you get the voltage U as negative integral of the electric field. From these, R=U/I

In case of inhomogeneity in a material, it is possible that space charge is created locally. At a pn junction, between p type and n type semiconductors, some of the free electrons of the p-type side go over the p type region and some of the free holes go over to the n type side.
At the junction between two different metals some electrons go over from one metal to the other, establishing potential difference, dependent on the temperature.
There is a space-charge region also at the interface between an electrode and electrolyte. But inhomogeneity of the conductance does not create free charge density.

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ELB27
TSny
Homework Helper
Gold Member
##R = \frac{b-a}{2\pi kL}##. Is this the correct answer?
Looks correct to me.
And if so, what was the boggy step in my first attempt?
When you used Gauss's law in your first post, it appears that you treated J as a constant when finding the total charge enclosed by the Gaussian surface.

ELB27
ehild
Homework Helper
Looks correct to me.

When you used Gauss's law in your first post, it appears that you treated J as a constant when finding the total charge enclosed by the Gaussian surface.
No, it was treated as a radial vector of magnitude dependent on the radius, but independent on the direction.

ELB27
No, it was treated as a radial vector of magnitude dependent on the radius, but independent on the direction.
I figured it out!
My mistake was that I simply multiplied ##\rho## by the volume to get the charge that it corresponds to, but this assumes ##\rho## and hence ##J## to be constant! Instead, I should have integrated ##\rho## by the volume. Here is the final (long, cumbersome but satisfying) solution using Gauss' law:
$$\rho = \frac{J\epsilon_0}{k} = \frac{\epsilon_0}{k}\frac{k}{s}E = \frac{\epsilon_0}{s}E$$
By Gauss' law (I used ##s'## in the limit of integration to avoid confusion with ##ds##):
$$E2\pi sL = \frac{1}{\epsilon_0}\left(\lambda L + \int_0^L \int_0 ^{2\pi} \int_a^{s'} \rho sdsd\phi dz\right) = \frac{1}{\epsilon_0}\left(\lambda L + 2\pi L \int_a^{s'}\frac{\epsilon_0}{s}E sds\right) = \frac{1}{\epsilon_0}\left(\lambda L + 2\pi L\epsilon_0 \int_a^{s'}E ds\right)$$
But the latter integral is just the potential difference between ##s## and ##a## since by definition ##V(a)-V(b) = -\int_b^a Edl##. Thus (cancelling ##L## and rearranging):
$$2\pi sE - \frac{\lambda}{\epsilon_0} = 2\pi(V(a)-V(s))$$
If we let ##s=b## this turns gives us ##\lambda## in terms of the given potential difference ##V = V(a)-V(b)## (since the current flows from the inner cylinder a to the outer one - ##V(a)>V(b)##) $$\lambda = 2\pi\epsilon_0 (bE - V)$$ Substituting back into the previous equation:
$$2\pi sE - 2\pi(bE - V) = 2\pi(V(a)-V(s))$$
$$(s-b)E = V(a)-V(s)-V = V(a)-V(s)-V(a)+V(b) = V(b) - V(s)$$
$$E = \frac{V(s)-V(b)}{(b-s)}$$
Now if we let ##s = a##,
$$E = \frac{V(a)-V(b)}{b-a} = \frac{V}{b-a}$$
Finally,
$$J = \sigma E = \frac{k}{s}\frac{V}{b-a} = \frac{kV}{s(b-a)}$$
and
$$I = \int Jda = \int_0^{2\pi} \int_0^L \frac{kV}{s(b-a)}sdzd\phi = 2\pi L \frac{kV}{(b-a)} = \frac{2\pi Lk}{b-a}V$$
and hence,
$$R = \frac{b-a}{2\pi Lk}$$
as before.
Finally!!! Thanks to everyone for their input, especially ehild for his time! What a good night's sleep can do...

ehild
Homework Helper
Finally!!! What a good night's sleep can do...
It was difficult to follow what you did in the OP. Did you really took J independent on the radius??? But whenever you showed J, it depended on s ?????
But I am pleased that you got the correct answer at the end.

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It was difficult to follow what you did in the OP. Did you really took J independent on the radius??? But whenever you showed J, it depended on s ?????
I still do not follow what you did, and the long integral can not be read.
But I am pleased that you got the correct answer at the end.
I didn't took J to be independent of the radius explicitly. When I used Gauss' law, I needed to express the total charge enclosed by the Gaussian surface via ##\rho##. Namely - ##Q_{volume} = \rho v## where ##v## is the enclosed volume with the charge. In the OP, I simply multiplied ##\rho (s) = \frac{\epsilon_0 J(s)}{k}## by the volume of the cylindrical shell that holds the volume charge (the shell between the two cylinders ##a## and ##b##), i.e. ##v = L\pi (s^2-a^2)##. But this assumes ##\rho## and hence ##J## to be constant over the volume when it is clearly not the case. So instead of multiplying I should have integrated ##Q_{volume} = \int \rho d\tau## where ##d\tau## is the volume element. (In other words, I subconsciously pulled ##\rho## out of this integral).
As for the long integral - which one are you referring to? In my browser everything appears fine.

ehild
Homework Helper
I didn't took J to be independent of the radius explicitly. When I used Gauss' law, I needed to express the total charge enclosed by the Gaussian surface via ##\rho##. Namely - ##Q_{volume} = \rho v## where ##v## is the enclosed volume with the charge. In the OP, I simply multiplied ##\rho (s) = \frac{\epsilon_0 J(s)}{k}## by the volume of the cylindrical shell that holds the volume charge (the shell between the two cylinders ##a## and ##b##), i.e. ##v = L\pi (s^2-a^2)##. But this assumes ##\rho## and hence ##J## to be constant over the volume when it is clearly not the case. So instead of multiplying I should have integrated ##Q_{volume} = \int \rho d\tau## where ##d\tau## is the volume element. (In other words, I subconsciously pulled ##\rho## out of this integral).

That was what I could not find out...
As for the long integral - which one are you referring to? In my browser everything appears fine.
I managed to see it at the end...

That was what I could not find out...
I managed to see it at the end...
I have just one last question related to your previous suggestion to use ##\vec{D}##. Is there any relation between the conductivity of the material ##\sigma## and its electric susceptibility ##\chi_e##? My book doesn't elaborate on this categorization and I'm a bit confused - previously we had only conductors and insulators, now we have various conductors. What properties do they share? Is everything I learned with ##\vec{H}## and ##\vec{D}## applies to "weak" conductors too?

ehild
Homework Helper
To sum up what was really needed for the solution:
The problem gave the conductivity between two concentric cylinder with radii a and b and length L as σ=k/r (r is the distance form the axis)
It is known that the current density is JE. Because of symmetry, E and J are radial vectors.
The current across a cross-section dA perpendicular to J is I = JdA.
Assuming stationary current and time-independent electric field, div J = 0 and the current is the same across any cylindrical surface of radius r:
I = 2πrL J, that is J(r) =σE(r) =I/(2πrL)---> E(r)=(r/k)I/(2πrL)=I/(2kπL). The potential difference U(a)-U(b) = ∫Edr from a to b: I/(2kπL)(b-a).
The resistivity is R=ΔU/I.
There was no need for Gauss Law.

ELB27
ehild
Homework Helper
I have just one last question related to your previous suggestion to use ##\vec{D}##. Is there any relation between the conductivity of the material ##\sigma## and its electric susceptibility ##\chi_e##? My book doesn't elaborate on this categorization and I'm a bit confused - previously we had only conductors and insulators, now we have various conductors. What properties do they share? Is everything I learned with ##\vec{H}## and ##\vec{D}## applies to "weak" conductors too?
In case of stationary fields and currents, you can treat conduction and Electrostatic behaviour separately, σ and χ are not related.
In case of time-varying fields, the fields can be written with their Fourier components, of angular frequency ω. There is not only conductive current, but also displacement current, proportional to ∂D/∂t, and the two together are considered current density, a complex quantity. The permittivity, conductivity, susceptibility are considered complex quantities and dependent of ω, and can be expressed by each other.
The simplest model describing the behaviour of materials in time-varying electric field considers the electrons, atoms, molecules as oscillators vibrating about their equilibrium position and driven by the applied electric field.The vibration is not necessarily in phase with the applied field. The displacement from equilibrium creates dipole moments. The polarization varies harmonically and its amplitude is proportional to that of the electric field, but there can be a phase difference, the susceptibility is complex.
The motion of charges that creates dipole moments, can be also considered as currents. That current is proportional to the time derivative of the polarization, and related to the electric field by a complex conductivity.
The complex conductivity and susceptibility can be derived from the same model, so they are related.

"Conductors" and "insulators" are ideal models for the materials. Even the best insulator conducts electricity at some extent and all conductors have finite resistivity.
Electrodynamics and electric characteristics of materials is quite a difficult topics! But Maxwell equations with the appropriate "material equations" are valid for a very broad range of materials and phenomena.

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ELB27