Irodov 3.163: Thin Layers of a Graded Conductor – Amit & Deepali Sinha Foundation

Problem Irodov 3.163

The space between the plates of a parallel-plate capacitor is filled with a weakly conducting medium whose conductivity changes linearly across the gap, from $\sigma_1 = 1.0$ pS/m at one plate to $\sigma_2 = 2.0$ pS/m at the other. Each plate has area $S = 230$ cm $^2$ and the plates are $d = 2.0$ mm apart. A voltage $V = 300$ V is applied. What current flows through the capacitor?

The capacitor edge on: the gap d is filled with a medium whose conductivity rises linearly from σ_1 at one plate to σ_2 at the other. — The capacitor edge on: the gap $d$ is filled with a medium whose conductivity rises linearly from $\sigma_1$ at one plate to $\sigma_2$ at the other.

Given

$\sigma_1 = 1.0$ pS/m
$\sigma_2 = 2.0$ pS/m
$S = 230$ cm $^2$
$d = 2.0$ mm
$V = 300$ V
edge effects neglected (assumed)

Find

the current

I

Solution

Step 1The conductivity across the gap

Measure $x$ from the plate where the conductivity is $\sigma_1$ . Rising linearly to $\sigma_2$ at $x = d$ , it is

\sigma(x) = \sigma_1 + (\sigma_2 - \sigma_1)\frac{x}{d} \tag{1}

Step 2The current density is the same at every depth

Neglecting the edges, everything depends on $x$ alone and the current runs straight across the gap. In the steady state $\nabla\cdot\vect j = 0$ , which here reads $dj/dx = 0$ : the current density $j$ is the same in every layer, while the field $E = j/\sigma(x)$ is strongest where the medium conducts worst.

Step 3The voltage as a sum over the layers

A thin layer dx at the height x above the plate of conductivity σ_1. All such layers carry the same current density. — A thin layer $dx$ at the height $x$ above the plate of conductivity $\sigma_1$ . All such layers carry the same current density.

The voltage across the gap is the field integrated across it, with (1) for $\sigma(x)$ :

\begin{aligned} V &= \int_0^d \frac{j}{\sigma(x)}\,dx = j\int_0^d \frac{dx}{\sigma_1 + (\sigma_2 - \sigma_1)x/d}\\ &= \frac{j\,d}{\sigma_2 - \sigma_1}\ln\frac{\sigma_2}{\sigma_1} \end{aligned}

Step 4The current

Solving the last line for the current density:

j = \frac{V(\sigma_2 - \sigma_1)}{d\ln(\sigma_2/\sigma_1)}

The current is this times the plate area, $I = jS$ . Numerically, with $S = 2.3\times10^{-2}$ m $^2$ and $\ln 2 = 0.693$ ,

I = \frac{300\cdot 2.3\times10^{-2}\cdot 1.0\times10^{-12}}{2.0\times10^{-3}\cdot 0.693}\ \text{A}

I = \frac{V S(\sigma_2 - \sigma_1)}{d\ln(\sigma_2/\sigma_1)} \approx 5.0\ \text{nA}

(2)

Check

Uniform medium: as $\sigma_2 \to \sigma_1 = \sigma$ , with $\sigma_2 = \sigma(1 + \delta)$ and $\ln(1 + \delta) \approx \delta$ ,

\frac{\sigma_2 - \sigma_1}{\ln(\sigma_2/\sigma_1)} \approx \frac{\sigma\delta}{\delta} = \sigma

and the current becomes $\sigma S V/d$ , Ohm’s law for a uniform slab. Here that effective conductivity is $1.44$ pS/m, between the two conductivities, as an effective value must.

Remark

The effective conductivity is the logarithmic mean of $\sigma_1$ and $\sigma_2$ , smaller than their plain average $1.5$ pS/m: the poorly conducting side carries a larger share of the voltage and weighs more in the resistance. Because $E$ changes across the gap while $j$ does not, the medium also holds a steady volume charge, the price of a graded conductor carrying a uniform current.