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Homework Help: Conductivity of a set of layers

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data
    We consider the following configuration. A pile of N layers of sediments. Each layer has a conductivity [tex]\sigma_{i}[/tex] (i from 1 to N) and a thickness [tex]m_{i}[/tex]. The length of the system is L. In this case, the electrical conductivity is not a scalar but a second order symmetric tensor.

    Using Ohm's Law and the idea of resistors working in parallel, demonstrate that the effective conductivity of the block along the x-direction is:

    [tex]\sigma_{x} = \frac{\Sigma_{i=1}^{N} m_{i}\sigma_{i}}{\Sigma_{i=1}^{N} m_{i}}

    2. Relevant equations

    [tex]\vec{j} = -\underline{\sigma} \vec{\nabla} V,

    \underline{\sigma} = \begin{bmatrix}
    \sigma_{x} & 0 \\
    0 & \sigma_{z}


    3. The attempt at a solution

    I am not exactly sure if this qualifies as advanced physics. This is a geophysics course and I have never seen this before (not even in the course).

    The first thing I did was I used the product over sum formula for two parallel resistors.

    [tex] R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}[/tex]

    Now using the formula [tex] R = \frac{\rho}{g}[/tex]

    where [tex]\rho[/tex] is the resistivity of the specific layer and g is the geometrical factor (I don't know what this is in this case and I don't know how to use it).

    Substituting this in for R and using [tex] \rho = \frac{1}{\sigma} [/tex] this parallel equation, after a lot of algebra, becomes:

    [tex] R_{eq} = \frac{1}{g_{1} \sigma_{1} + g_{2}\sigma_{2}}[/tex]

    Now, can I set [tex]R_{eq} = \frac{1}{\sigma_{x} g_{eq}} , g_{eq} = g_{1} + g_{2}?[/tex]

    And then can I go as far as setting g equal to m?

    If that is the case then I will have the correct answer because,

    [tex]\frac{1}{\sigma_{x}(m_{1} + m_{2})} = \frac{1}{m_{1} \sigma_{1} + m_{2}\sigma_{2}}

    \Rightarrow \sigma_{x} = \frac{m_{1} \sigma_{1} + m_{2}\sigma_{2}}{m_{1} + m_{2}}.


    Which is the formula for the specific case of N =2.

    Could someone please verify my approach or let me know how horribly wrong I am?

    Thanks in advance,

    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 30, 2010 #2
    I did some thinking and I think I got it. If I manipulate [tex]\vec{j} = -\underline{\sigma} \vec{\nabla}V[/tex] into V = IR by letting [tex] \nabla V_{x} = \frac{\Delta V}{\Delta x} [/tex] and

    [tex]\nabla V_{z} = \frac{\Delta V}{\Delta z} [/tex]

    then the magnitude of my first equation will become

    [tex] j = \sigma_{x}\frac{\Delta V}{\Delta x} [/tex] in the x-direction


    [tex] j = \sigma_{z}\frac{\Delta V}{\Delta z} [/tex] in the z-direction.

    Then in the x-direction:

    [tex] \frac{I}{A}= \sigma_{x}\frac{\Delta V}{\Delta x}

    \Rightarrow \Delta V = \left( \frac{\Delta x}{A \sigma_{x}} \right)I.


    But as is apparent in the diagram (I will post this in a bit)[tex]\Delta x = L[/tex] and [tex]A = \Delta x m = Lm[/tex]

    The above equation then becomes:

    [tex]\Delta V = \left( \frac{\rho_{x}L}{mL} \right)I[/tex]

    The L's cancel and this gives me V = IR where

    [tex]R = \frac{\rho_{x}}{m}}[/tex].

    Hence m is my geometrical factor.

    I do this same process for the z-direction (with [tex]\Delta z = m[/tex]) and I get my R to be:

    [tex]R = \frac{\rho_{z}}{L}}.[/tex]

    So then L must be my geometrical factor for the z-direction? A problem further into my worksheet asks for me to give the expression for [tex]\sigma_{z}[/tex] similar to that of [tex]\sigma_{x}[/tex] in the first problem. But using L as my geometrical factor I do not get the correct result.

    Can someone please verify my answers here and help me with the geometrical factor in the z-direction?

    Thanks in advance,

    Last edited: Mar 31, 2010
  4. Mar 31, 2010 #3
    Last edited by a moderator: Apr 24, 2017
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