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Conductivity of a set of layers

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Homework Statement


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}}
[/tex]


Homework Equations



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

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

[/tex]


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}}.

[/tex]

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,

KEØM
 
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Answers and Replies

  • #2
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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

and

[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.

[/tex]

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,

KEØM
 
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