Irodov 3.104: Lines That End Inside the Filling – Amit & Deepali Sinha Foundation

Problem Irodov 3.104

A parallel-plate capacitor has plates of area $S$ a distance $d$ apart. The gap is filled with an isotropic dielectric whose permittivity changes linearly across the gap, perpendicular to the plates, from $\varepsilon_1$ at one plate to $\varepsilon_2$ at the other, with $\varepsilon_2 > \varepsilon_1$ . Find:

the capacitance;
the volume density of bound charge as a function of $\varepsilon$ , when the capacitor carries the charge $q$ and its field $\vect E$ points toward increasing $\varepsilon$ .

The capacitor edge on. The permittivity of the filling grows linearly with the distance x from the lower plate, from ε_1 there to ε_2 at the upper plate, a distance d away. — The capacitor edge on. The permittivity of the filling grows linearly with the distance $x$ from the lower plate, from $\varepsilon_1$ there to $\varepsilon_2$ at the upper plate, a distance $d$ away.

Given

plate area $S$
gap $d$
$\varepsilon$ linear from $\varepsilon_1$ to $\varepsilon_2 > \varepsilon_1$
charge $q$
$\vect E$ toward growing $\varepsilon$

Find

(a)

C

· (b)

\rho'(\varepsilon)

Solution

Step 1The permittivity profile and the displacement

Measure $x$ across the gap from the plate where the permittivity is $\varepsilon_1$ . A linear profile with the given end values is

\varepsilon(x) = \varepsilon_1 + \frac{\varepsilon_2 - \varepsilon_1}{d}\,x \tag{1}

Let that plate carry $q$ and the other $-q$ . Edge effects aside, the fields are uniform along the plates and normal to them, and Gauss’s theorem for $\vect D$ sees only the free charge, so $D$ is the same at every $x$ :

D = \frac{q}{S}

Step 2The field and the voltage

Dividing by $\varepsilon_0\varepsilon$ gives the field at each depth:

E(x) = \frac{q}{\varepsilon_0 S\,\varepsilon(x)} \tag{2}

The voltage is its integral across the gap. With (1), change the variable to $\varepsilon$ , using $dx = d\,d\varepsilon/(\varepsilon_2 - \varepsilon_1)$ :

\begin{aligned} V &= \int_0^d E\,dx = \frac{q}{\varepsilon_0 S}\int_0^d \frac{dx}{\varepsilon(x)} = \frac{q}{\varepsilon_0 S}\cdot\frac{d}{\varepsilon_2 - \varepsilon_1}\int_{\varepsilon_1}^{\varepsilon_2}\frac{d\varepsilon}{\varepsilon}\\ &= \frac{q\,d}{\varepsilon_0 S(\varepsilon_2 - \varepsilon_1)}\,\ln\frac{\varepsilon_2}{\varepsilon_1} \end{aligned}

Step 3The capacitance

The capacitance is $C = q/V$ , and $q$ cancels:

C = \dfrac{\varepsilon_0(\varepsilon_2 - \varepsilon_1)S}{d\,\ln(\varepsilon_2/\varepsilon_1)}

(3)

Step 4The bound volume charge

The field points toward growing $\varepsilon$ , so it runs along $+x$ and the plate at $x = 0$ is the positive one, as assumed. The polarization is $\vect P = \vect D - \varepsilon_0\vect E$ , along $+x$ ; with (2):

P = \frac{q}{S} - \frac{q}{S\,\varepsilon} = \frac{q}{S}\left(1 - \frac{1}{\varepsilon}\right)

The bound volume density is $\rho' = -\nabla\cdot\vect P$ , which here is $-dP/dx$ . Differentiating through $\varepsilon(x)$ , whose slope is $(\varepsilon_2 - \varepsilon_1)/d$ by (1):

\frac{dP}{dx} = \frac{q}{S}\cdot\frac{1}{\varepsilon^2}\,\frac{d\varepsilon}{dx} = \frac{q(\varepsilon_2 - \varepsilon_1)}{S\,d\,\varepsilon^2}

so that

\rho' = -\dfrac{q(\varepsilon_2 - \varepsilon_1)}{S\,d\,\varepsilon^2}

(4)

The field for a positive lower plate, drawn for ε growing from 1 to 4. The number of lines falls as 1/ε, and each line that stops ends on a negative bound charge inside the filling; the positive bound charge on the upper face balances them. — The field for a positive lower plate, drawn for $\varepsilon$ growing from 1 to 4. The number of lines falls as $1/\varepsilon$ , and each line that stops ends on a negative bound charge inside the filling; the positive bound charge on the upper face balances them.

The bound charge is negative throughout the filling, densest where $\varepsilon$ is smallest, next to the positive plate.

Check

The bound charge must total zero. The volume charge is $S\int_0^d \rho'\,dx$ ; changing the variable to $\varepsilon$ as before,

S\int_0^d \rho'\,dx = -q\int_{\varepsilon_1}^{\varepsilon_2}\frac{d\varepsilon}{\varepsilon^2} = -q\left(\frac{1}{\varepsilon_1} - \frac{1}{\varepsilon_2}\right).

The surface bound charges are $-P(0)S = -q(1 - 1/\varepsilon_1)$ at the positive plate and $+P(d)S = q(1 - 1/\varepsilon_2)$ at the negative one, and the three sum to zero. Uniform limit: as $\varepsilon_2 \to \varepsilon_1$ , $\ln(\varepsilon_2/\varepsilon_1) \approx (\varepsilon_2 - \varepsilon_1)/\varepsilon_1$ , so (3) tends to $\varepsilon_0\varepsilon_1 S/d$ , and (4) vanishes.

Remark

The capacitance is the plain parallel-plate value $\varepsilon_0\bar\varepsilon S/d$ with an effective permittivity $\bar\varepsilon = (\varepsilon_2 - \varepsilon_1)/\ln(\varepsilon_2/\varepsilon_1)$ , the logarithmic mean of the two end values. A series stack averages $1/\varepsilon$ , not $\varepsilon$ , so this mean always falls below the arithmetic mean $(\varepsilon_1 + \varepsilon_2)/2$ : the weak end of the filling costs more than the strong end gains.