Irodov 3.167: Only the End Values Count – Amit & Deepali Sinha Foundation

Problem Irodov 3.167

The space between plates 1 and 2 of a parallel-plate capacitor is filled with a poorly conducting, inhomogeneous medium. Its permittivity and resistivity change across the gap, from $\varepsilon_1$ and $\rho_1$ at plate 1 to $\varepsilon_2$ and $\rho_2$ at plate 2. A constant voltage across the plates drives a steady current $I$ through the medium from plate 1 to plate 2. Find the total free (extraneous) charge contained in the medium.

The capacitor edge on: the medium has ε_1, ρ_1 at plate 1 and ε_2, ρ_2 at plate 2; the steady current I enters at plate 1 and crosses the gap to plate 2. — The capacitor edge on: the medium has $\varepsilon_1$ , $\rho_1$ at plate 1 and $\varepsilon_2$ , $\rho_2$ at plate 2; the steady current $I$ enters at plate 1 and crosses the gap to plate 2.

Given

$\varepsilon_1$ , $\rho_1$ at plate 1
$\varepsilon_2$ , $\rho_2$ at plate 2
steady current $I$ from plate 1 to plate 2

Find

the total extraneous charge

q

in the medium

Solution

Step 1The same current density at every depth

Let the plates have area $S$ and measure $x$ across the gap from plate 1, up to $x = d$ at plate 2. The properties change only with $x$ , so (edge effects aside) the current flows straight across, and in a steady state the whole current $I$ crosses every plane parallel to the plates:

j = \frac{I}{S}

Ohm’s law in local form, $\vect E = \rho\,\vect j$ , then gives the field at each depth, along $+x$ :

E(x) = \frac{\rho(x)\,I}{S} \tag{1}

Step 2Free charge from the change of D

Gauss’s theorem for $\vect D$ counts free charge only. In differential form, with everything depending on $x$ alone, the free volume density is the slope of $D$ , and $D$ follows from (1):

\begin{gathered} \rho_{\mathrm{free}} = \frac{dD}{dx} \\ D(x) = \varepsilon_0\varepsilon(x)E(x) = \frac{\varepsilon_0 I}{S}\,\varepsilon(x)\rho(x) \end{gathered}

The total charge in the medium is this density integrated over the volume, a layer $S\,dx$ at a time:

q = \int_0^d \rho_{\mathrm{free}}\,S\,dx = S\int_0^d \frac{dD}{dx}\,dx = S\,\bigl[D(d) - D(0)\bigr]

Substituting the two end values of $D$ , the area $S$ cancels:

q = \varepsilon_0 I(\varepsilon_2\rho_2 - \varepsilon_1\rho_1)

(2)

Check

A homogeneous medium has $\varepsilon_1\rho_1 = \varepsilon_2\rho_2$ and must hold no free charge inside: (2) gives $q = 0$ . Dimensions: $\varepsilon_0\rho$ is a time (the medium’s relaxation time), and $I$ times a time is a charge. A two-layer filling is a special case: the charge then sits on the single boundary, and dividing (2) by $S$ with $I/S = j$ gives $\varepsilon_0 j(\varepsilon_2\rho_2 - \varepsilon_1\rho_1)$ , the boundary density of the two-layer capacitor.

Remark

How $\varepsilon$ and $\rho$ vary between the plates does not matter at all: the charge may be spread through the volume or crowded into thin sheets, but its total is fixed by the end values alone, because it is just the difference of the displacement flux leaving through plate 2 and entering through plate 1.