In this question i tried to apply formula $\sigma E=j$ to a portion of the medium but couldnot proceed.

2 votes

In this question i tried to apply formula $\sigma E=j$ to a portion of the medium but couldnot proceed.

Please show your calculation. Why can you not proceed?

I tried by writing down the variation of conductivity.$\sigma(x)=\sigma_1+\frac{(\sigma_2-\sigma_1)x}{d}$.I then considered a narrow area of medium of length dx at distance x from one of the plates.I couldn't get what should be substituted for j in the equation above.(Also I am not sure if the question speaks about varying conductivity in the lateral or longitudinal direction).Any ideas @sammygerbil.Thanks.

2 votes

Best answer

What you are missing is how to change the sum into an integral.

Divide the material into $n$ sheets parallel to the plates, of infinitesimal thickness $\delta=\frac{L}{n}$ and area $A$, where $L$ is the distance between the plates. The resistance of each sheet is $\frac{\rho \delta}{A}=\frac{\delta}{\sigma A}$. The sheets are in series so the resistance of the whole slab is the sum of each :

$R=\frac{\delta}{\sigma_1 A}+\frac{\delta}{(\sigma_1+d\sigma)A}+\frac{\delta}{(\sigma_1+2d\sigma)A}+$ ... $+\frac{\delta}{(\sigma_1+(n-1)d\sigma)A}$

where $d\sigma=\frac{\sigma_2-\sigma_1}{n}$ is the increment of conductivity between sheets .

This infinite sum can be converted into an integral :

$R=\Sigma_0^{\infty} \frac{\delta(n)}{\sigma(n) A}=\frac{L}{(\sigma_2 -\sigma_1)A} \int_{\sigma_1}^{\sigma_2} \frac{d\sigma}{\sigma}=\frac{L}{(\sigma_2 - \sigma_1)A} \ln (\frac{\sigma_2}{\sigma_1})$.

The rest of the question is just a matter of applying Ohm's Law.

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