# Conductivity of a set of layers

1. Mar 30, 2010

### KEØM

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 $$\sigma_{i}$$ (i from 1 to N) and a thickness $$m_{i}$$. 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:

$$\sigma_{x} = \frac{\Sigma_{i=1}^{N} m_{i}\sigma_{i}}{\Sigma_{i=1}^{N} m_{i}}$$

2. Relevant equations

$$\vec{j} = -\underline{\sigma} \vec{\nabla} V, \underline{\sigma} = \begin{bmatrix} \sigma_{x} & 0 \\ 0 & \sigma_{z} \end{bmatrix}}$$

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.

$$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$$

Now using the formula $$R = \frac{\rho}{g}$$

where $$\rho$$ 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 $$\rho = \frac{1}{\sigma}$$ this parallel equation, after a lot of algebra, becomes:

$$R_{eq} = \frac{1}{g_{1} \sigma_{1} + g_{2}\sigma_{2}}$$

Now, can I set $$R_{eq} = \frac{1}{\sigma_{x} g_{eq}} , g_{eq} = g_{1} + g_{2}?$$

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,

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

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?

KEØM

Last edited: Mar 30, 2010
2. Mar 30, 2010

### KEØM

I did some thinking and I think I got it. If I manipulate $$\vec{j} = -\underline{\sigma} \vec{\nabla}V$$ into V = IR by letting $$\nabla V_{x} = \frac{\Delta V}{\Delta x}$$ and

$$\nabla V_{z} = \frac{\Delta V}{\Delta z}$$

then the magnitude of my first equation will become

$$j = \sigma_{x}\frac{\Delta V}{\Delta x}$$ in the x-direction

and

$$j = \sigma_{z}\frac{\Delta V}{\Delta z}$$ in the z-direction.

Then in the x-direction:

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

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

The above equation then becomes:

$$\Delta V = \left( \frac{\rho_{x}L}{mL} \right)I$$

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

$$R = \frac{\rho_{x}}{m}}$$.

Hence m is my geometrical factor.

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

$$R = \frac{\rho_{z}}{L}}.$$

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 $$\sigma_{z}$$ similar to that of $$\sigma_{x}$$ 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?