Irodov 3.166: Equal Relaxation Times Leave No Charge

Problem Irodov 3.166

The gap of a parallel-plate capacitor holds two slightly conducting dielectric layers, one on top of the other: layer 1 of thickness $d_1$ , permittivity $\varepsilon_1$ and resistivity $\rho_1$ , and layer 2 of thickness $d_2$ , permittivity $\varepsilon_2$ and resistivity $\rho_2$ . A constant voltage $V$ is kept across the plates, and the field points from layer 1 into layer 2. Find the surface density $\sigma$ of free (extraneous) charge on the boundary between the layers, and the condition for $\sigma$ to vanish.

The capacitor edge on: layer 1 (ε_1, ρ_1, thickness d_1) on the positive plate, layer 2 (ε_2, ρ_2, thickness d_2) on the negative one, so the field runs from layer 1 into layer 2. — The capacitor edge on: layer 1 ( $\varepsilon_1$ , $\rho_1$ , thickness $d_1$ ) on the positive plate, layer 2 ( $\varepsilon_2$ , $\rho_2$ , thickness $d_2$ ) on the negative one, so the field runs from layer 1 into layer 2.

Given

layer 1: $d_1$ , $\varepsilon_1$ , $\rho_1$
layer 2: $d_2$ , $\varepsilon_2$ , $\rho_2$
voltage $V$
$\vect E$ from layer 1 to layer 2
steady state (assumed)

Find

\sigma

on the boundary · the condition for

\sigma = 0

Solution

Step 1One current density through both layers

With a dc voltage the system settles into a steady state (assumed), in which no charge piles up anywhere any longer. Edge effects aside, the current flows straight across the gap, and the same current density $j$ must cross every plane parallel to the plates, or charge would keep accumulating between them. Ohm’s law in local form, $\vect E = \rho\,\vect j$ , then gives the field in each layer:

\begin{gathered} E_1 = \rho_1 j \\ E_2 = \rho_2 j \end{gathered} \tag{1}

Step 2The current density from the voltage

The voltage is the sum of the drops across the two layers:

V = E_1 d_1 + E_2 d_2 = j\,(\rho_1 d_1 + \rho_2 d_2)

so that

j = \frac{V}{\rho_1 d_1 + \rho_2 d_2} \tag{2}

Step 3The free charge on the boundary

Take the normal along the field, from layer 1 into layer 2. Gauss’s theorem for $\vect D$ on a flat pillbox straddling the boundary counts only the free charge on it, so $\sigma$ is the jump of the normal component of $\vect D = \varepsilon_0\varepsilon\vect E$ . With (1):

\sigma = D_2 - D_1 = \varepsilon_0(\varepsilon_2 E_2 - \varepsilon_1 E_1) = \varepsilon_0 j\,(\varepsilon_2\rho_2 - \varepsilon_1\rho_1)

Substituting the current density (2):

\sigma = \frac{\varepsilon_0 V(\varepsilon_2\rho_2 - \varepsilon_1\rho_1)}{\rho_1 d_1 + \rho_2 d_2}

(3)

The charge is positive when $\varepsilon_2\rho_2 > \varepsilon_1\rho_1$ and negative in the opposite case.

Step 4When the boundary stays uncharged

The factor in the numerator is the only one that can vanish, so $\sigma = 0$ exactly when

\varepsilon_1\rho_1 = \varepsilon_2\rho_2

(4)

Check

Dimensions: $\varepsilon_0 V\rho/(\rho d)$ is $\varepsilon_0 V/d$ , a surface charge density. A limit: let layer 2 become a perfect insulator, $\rho_2 \to \infty$ . Then (3) tends to $\varepsilon_0\varepsilon_2 V/d_2$ ; the current stops, layer 1 carries no field and acts as an extension of the positive plate, and the boundary must hold exactly the charge of a capacitor of gap $d_2$ filled with $\varepsilon_2$ at voltage $V$ , which is $\varepsilon_0\varepsilon_2 V/d_2$ .

Remark

The product $\varepsilon_0\varepsilon\rho$ is the charge relaxation time of a medium, the time in which a free charge placed inside it leaks away. When the voltage is first switched on, the field divides between the layers as in a pure dielectric ( $D$ continuous); later it divides as in a resistor chain ( $j$ continuous). Free charge must flow onto the boundary to carry one division into the other, unless the two agree from the start, which is precisely the condition of equal relaxation times.