Irodov 3.168: A Linear Resistivity Gives Uniform Charge

Problem Irodov 3.168

A parallel-plate capacitor with gap $d$ is filled with a poorly conducting, inhomogeneous medium whose resistivity changes linearly across the gap, perpendicular to the plates; the largest resistivity is $\eta$ times the smallest. Take the permittivity to be $1$ everywhere. Find the volume density of charge in the gap when a voltage $V$ is applied across the plates.

The capacitor edge on, gap d, with the leads across which the voltage is applied. The resistivity ρ of the filling changes linearly with the distance x across the gap. — The capacitor edge on, gap $d$ , with the leads across which the voltage is applied. The resistivity $\rho$ of the filling changes linearly with the distance $x$ across the gap.

Given

gap $d$
resistivity linear across the gap
$\rho_{\mathrm{max}}/\rho_{\mathrm{min}} = \eta$
voltage $V$
$\varepsilon = 1$
steady state (assumed)

Find

the volume charge density in the gap

Solution

Step 1The resistivity profile and the current density

Measure $x$ from the plate where the resistivity has its least value $\rho_0$ ; at the other plate it is $\eta\rho_0$ , so the linear profile is

\rho(x) = \rho_0\left[1 + (\eta - 1)\frac{x}{d}\right] \tag{1}

In the steady state (assumed) no charge accumulates any longer, so, edge effects aside, the same current density $j$ crosses every plane parallel to the plates. Take the plate at $x = 0$ as the positive one, so $j$ points along $+x$ . Ohm’s law in local form, $\vect E = \rho\,\vect j$ , gives the field at each depth:

E(x) = \rho(x)\,j \tag{2}

Step 2The current density from the voltage

The voltage is the field integrated across the gap; with (1) the integral of a linear profile is its mean value times $d$ :

V = \int_0^d E\,dx = j\int_0^d \rho(x)\,dx = j\rho_0 d\left(1 + \frac{\eta - 1}{2}\right) = \frac{j\rho_0 d\,(\eta + 1)}{2}

so that

j = \frac{2V}{\rho_0 d\,(\eta + 1)} \tag{3}

Step 3The charge from the slope of the field

With $\varepsilon = 1$ , Gauss’s theorem in differential form gives the volume charge density $\rho_q$ as $\varepsilon_0$ times the slope of the field. Differentiating (2), where only the resistivity (1) changes with $x$ :

\rho_q = \varepsilon_0\frac{dE}{dx} = \varepsilon_0 j\,\frac{d\rho}{dx} = \varepsilon_0 j\,\frac{\rho_0(\eta - 1)}{d}

Substituting the current density (3), $\rho_0$ cancels:

\rho_q = \frac{2\varepsilon_0 V(\eta - 1)}{(\eta + 1)\,d^2}

(4)

This is positive when the current flows toward increasing resistivity, as taken here; with the opposite polarity the charge has the same magnitude and is negative.

Check

A uniform medium, $\eta = 1$ , holds no charge, and (4) vanishes. Dimensions: $\varepsilon_0 V/d^2$ is (F/m)(V)/m $^2$ , which is C/m $^3$ , a volume density. A weak gradient: for $\eta = 1 + \delta$ with $\delta$ small, (4) tends to $\varepsilon_0 V\delta/d^2$ . That is the first-order result for any slowly varying resistivity, $\varepsilon_0 E\,d(\ln\rho)/dx$ , with the field $V/d$ and the logarithmic slope $\delta/d$ .

Remark

The space charge exists only because current flows: in a perfect insulator with $\varepsilon = 1$ the field between the plates would be uniform and the gap empty of charge. Here the medium has to raise the field where it conducts worse, and the only way the field can change between the plates is by charge sitting in the gap.