Conductivity Tensor: Problem 4.7 Purcell Analysis

[psholtz]

Homework Statement

I'm working through Problem 4.7 in Purcell, on conductivity tensors.

We build a conductor by alternately stacking plates of tin and silver (differing conductivities). When we pass a current through this "composite" conductor, we will get a different value for the conductivity depending on whether we apply the current (a) parallel; or (b) perpendicular to the plates.

Let $$\sigma_1$$ designate the conductivity of the silver plate, and $$\sigma_2$$ designate the conductivity of the tin.

The silver plate is 100 angstrom thick, and the tin plate is 200 angstrom thick.

We are seeking the "ratio" of $$\sigma_{\perp} / \sigma_{\parallel}$$

Homework Equations

The important equation is the "microscopic" version of Ohm's Law:

$$\mathbf{J} = \sigma \mathbf{E}$$

The Attempt at a Solution

First let's derive a value for $$\sigma_{\perp}$$, which is to say, the conductivity we get when passing the current perpendicular to the plates. The current density must be the same in both regions (silver and tin). Because the conductivities of the metals differ, we must have a charge layer that builds up at the junction between the metal plates. In other words:

$$J_1 = J_2$$

$$\sigma_1 E_1 = \sigma_2 E_2$$

and since $$\sigma_1 > \sigma_2$$, we must have $$E_1 < E_2$$.

Let $$E_0$$ be the 'net" electric field driving the (uniform) current density in both regions (i.e., $$J_{net} = \sigma_{net} E_0 = \sigma_{\perp} E_0$$). We have:

$$E_1 = E_0 - E_{charge layer}$$

$$E_2 = E_0 + E_{charge layer}$$

from which we obtain:

$$\large{ E_0 = \frac{1}{2}(E_1 + E_2)}$$

$$\large{E_0 = \frac{J}{2}(\frac{1}{\sigma_1} + \frac{1}{\sigma_2})}$$

$$\large{E_0 = \frac{J}{2}( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} )}$$

$$\large{J = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2} E_0 }$$

From which we conclude:

$$\sigma_{\perp} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$$

Note that if $$\sigma_1 = \sigma_2 = \sigma$$, we have:

$$\sigma_{\perp} = \frac{2 \sigma^2}{2 \sigma} = \sigma$$

which is the result we expect.

Now let's derive the "parallel" conductivity. Again, we will have a net "applied" electric field $$E_0$$ but this time, the applied field will produce two distinct current denstities in the respective metal layers. The "net" current density, the one we will use to calculate the "parallel" conductivity of the material, will be the vector sum of these two (in other words, there will be no net charge layer). We have:

$$J_1 = \sigma_1 E_0$$

$$J_2 = \sigma_2 E_0$$

Bearing in mind that the silver layer (sigma 1) is 100 ang thick, and the tin layer (sigma 2) is 200 ang thick, we have:

$$J_{total} = J_1 + J_2 = \frac{\sigma_1 + 2\sigma_2}{3} E_0$$

So that:

$$\sigma_{\parallel} = \frac{\sigma_1 + 2\sigma_2}{3}$$

Again, if $$\sigma_1 = \sigma_2 = \sigma$$, we have:

$$\sigma_{\parallel} = \frac{\sigma + 2\sigma}{3} = \sigma$$

which is the expected result.

The ratio we initially set out to calculate is thus:

$$\sigma_{\perp} / \sigma_{\parallel} = \frac{6 \sigma_1 \sigma_2}{(\sigma_1 + \sigma_2)(\sigma_1 + 2\sigma_2)}$$

If we suppose that $$\sigma_1 = k\sigma_2$$ this reduces to:

$$\sigma_{\perp} / \sigma_{\parallel} = \frac{6 k \sigma_2^2}{(k\sigma_2 + \sigma_2)(k\sigma_2 + 2\sigma_2)} = \frac{6 k \sigma_2^2}{(k + 1)(k + 2)\sigma_2^2}$$

$$\sigma_{\perp} / \sigma_{\parallel} = \frac{6k}{(k+1)(k+2)}$$

Taking k=7.2, we have:

$$\sigma_{\perp} / \sigma_{\parallel} = 0.573$$

However, the answer in the book is 0.457.

What am I doing wrong?

 

[pleu]

Your $\sigma_\perp$ equation is wrong. It should be $\sigma_\perp = \frac{3 \sigma_Ag \sigma_sn}{\sigma_Sn + 2 \sigma_Ag}$.