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Finding resistance of cylinders with variable conductivity

  1. Dec 21, 2014 #1
    1. The problem statement, all variables and given/known data
    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.]

    2. Relevant 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##

    3. The attempt at a solution
    I began by calculating the charge density in the conductive volume: [tex]\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}[/tex] 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: [tex]E2\pi sL = \frac{1}{\epsilon_0}\left(\lambda L + \frac{J\epsilon_0L}{k}(s^2-a^2)\right)[/tex] After substituting in the above equation ##E = \frac{1}{\sigma} = \frac{s}{k}## and solving for ##J## I get: [tex]\vec{J} = \frac{k\lambda}{\pi\epsilon_0 (s^2+a^2)}\hat{s}[/tex] 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 [tex]\lambda = \frac{2\pi\epsilon_0 V}{ln[(a^2+b^2)/(2a^2)]}[/tex] Substituting, [tex]\vec{J} = \frac{2kV}{(s^2+a^2)ln[(a^2+b^2)/(2a^2)]}[/tex] 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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  3. Dec 21, 2014 #2

    ehild

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    What is s?
    How do you get the current I from the current density J?
     
  4. Dec 21, 2014 #3
    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}##
     
  5. Dec 21, 2014 #4

    ehild

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    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?
     
  6. Dec 21, 2014 #5
    Please see my edit in my previous post.
     
  7. Dec 21, 2014 #6

    ehild

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    Answer my question please. What is the surface integral ##I = \int\vec{J}\cdot\vec{da}## in terms of J and s?
     
  8. Dec 21, 2014 #7
    ##I = \int_0^{2\pi}\int_0^L J(s)sdzd\phi = J(s)sL2\pi##
     
  9. Dec 21, 2014 #8

    ehild

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    Right. Substitute J=σE. I is the same across each cross-section, it does not depend on s. What does it mean on E?
     
  10. Dec 21, 2014 #9
    ##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
     
  11. Dec 21, 2014 #10

    ehild

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

    Edit: It is zero!
     
    Last edited: Dec 21, 2014
  12. Dec 21, 2014 #11
    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##.
     
    Last edited: Dec 21, 2014
  13. Dec 21, 2014 #12

    ehild

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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.
     
    Last edited: Dec 21, 2014
  14. Dec 21, 2014 #13
    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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    Last edited: Dec 21, 2014
  15. Dec 21, 2014 #14

    ehild

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    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.
     
    Last edited: Dec 21, 2014
  16. Dec 21, 2014 #15
    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!...)
     
  17. Dec 21, 2014 #16

    ehild

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    It is very wise of you that you want to understand the problems.
    Read this, for example: http://en.wikipedia.org/wiki/Electric_displacement_field

    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.
     
    Last edited: Dec 22, 2014
  18. Dec 22, 2014 #17

    TSny

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    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.
     
  19. Dec 22, 2014 #18

    ehild

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    No, it was treated as a radial vector of magnitude dependent on the radius, but independent on the direction.
     
  20. Dec 22, 2014 #19
    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:
    [tex]\rho = \frac{J\epsilon_0}{k} = \frac{\epsilon_0}{k}\frac{k}{s}E = \frac{\epsilon_0}{s}E[/tex]
    By Gauss' law (I used ##s'## in the limit of integration to avoid confusion with ##ds##):
    [tex]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)[/tex]
    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):
    [tex]2\pi sE - \frac{\lambda}{\epsilon_0} = 2\pi(V(a)-V(s))[/tex]
    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)##) [tex]\lambda = 2\pi\epsilon_0 (bE - V)[/tex] Substituting back into the previous equation:
    [tex]2\pi sE - 2\pi(bE - V) = 2\pi(V(a)-V(s))[/tex]
    [tex](s-b)E = V(a)-V(s)-V = V(a)-V(s)-V(a)+V(b) = V(b) - V(s)[/tex]
    [tex]E = \frac{V(s)-V(b)}{(b-s)}[/tex]
    Now if we let ##s = a##,
    [tex]E = \frac{V(a)-V(b)}{b-a} = \frac{V}{b-a}[/tex]
    Finally,
    [tex] J = \sigma E = \frac{k}{s}\frac{V}{b-a} = \frac{kV}{s(b-a)}[/tex]
    and
    [tex]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[/tex]
    and hence,
    [tex]R = \frac{b-a}{2\pi Lk}[/tex]
    as before.
    Finally!!! Thanks to everyone for their input, especially ehild for his time! What a good night's sleep can do...
     
  21. Dec 22, 2014 #20

    ehild

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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 :oldconfused: ?????
    But I am pleased that you got the correct answer at the end.
     
    Last edited: Dec 22, 2014
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