Irodov 3.103: Two Layers Share One Displacement – Amit & Deepali Sinha Foundation

Problem Irodov 3.103

A parallel-plate capacitor with plates of area $S$ has its gap filled by two dielectric layers lying one on top of the other, parallel to the plates: layer 1 of thickness $d_1$ and permittivity $\varepsilon_1$ , and layer 2 of thickness $d_2$ and permittivity $\varepsilon_2$ . Find:

the capacitance;
the surface density $\sigma'$ of bound charge on the plane where the two layers meet, when the voltage across the capacitor is $V$ and the field points from layer 1 into layer 2.

The capacitor edge on: layer 1 (thickness d_1, permittivity ε_1) against one plate, layer 2 (thickness d_2, permittivity ε_2) against the other. — The capacitor edge on: layer 1 (thickness $d_1$ , permittivity $\varepsilon_1$ ) against one plate, layer 2 (thickness $d_2$ , permittivity $\varepsilon_2$ ) against the other.

Given

plate area $S$
layer 1: $d_1$ , $\varepsilon_1$
layer 2: $d_2$ , $\varepsilon_2$
voltage $V$
$\vect E$ from layer 1 to layer 2

Find

(a)

C

· (b)

\sigma'

on the boundary

Solution

Step 1One displacement through both layers

Let the plate next to layer 1 carry the free charge $q$ , with surface density $\sigma = q/S$ , and the other plate $-q$ . Neglecting edge effects, every field is uniform and normal to the plates. Gauss’s theorem for $\vect D$ on a pillbox with one face inside the positive plate counts only the free charge, so $D$ is the same in both layers:

D = \sigma \tag{1}

Step 2The field in each layer

In each isotropic layer $\vect D = \varepsilon_0\varepsilon\vect E$ , so dividing (1) by $\varepsilon_0\varepsilon$ layer by layer:

\begin{gathered} E_1 = \frac{\sigma}{\varepsilon_0\varepsilon_1} \\ E_2 = \frac{\sigma}{\varepsilon_0\varepsilon_2} \end{gathered} \tag{2}

Step 3The voltage and the capacitance

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

V = E_1 d_1 + E_2 d_2 = \frac{\sigma}{\varepsilon_0}\left(\frac{d_1}{\varepsilon_1} + \frac{d_2}{\varepsilon_2}\right) \tag{3}

The capacitance is the charge per unit voltage, $C = \sigma S/V$ , and $\sigma$ cancels:

C = \dfrac{\varepsilon_0 S}{d_1/\varepsilon_1 + d_2/\varepsilon_2}

(4)

Step 4The bound charge on the boundary

The field with the plate under layer 1 positive, drawn for ε_1 < ε_2: E_1 is the longer vector, and the boundary carries a negative bound charge σ'. — The field with the plate under layer 1 positive, drawn for $\varepsilon_1 < \varepsilon_2$ : $\vect E_1$ is the longer vector, and the boundary carries a negative bound charge $\sigma'$ .

Take the normal from layer 1 into layer 2, along the field. The bound charge on the boundary is the jump of the normal component of $\varepsilon_0\vect E$ across it, because the free charge there is zero. Using (2):

\sigma' = \varepsilon_0(E_2 - E_1) = \sigma\left(\frac{1}{\varepsilon_2} - \frac{1}{\varepsilon_1}\right) = \sigma\,\frac{\varepsilon_1 - \varepsilon_2}{\varepsilon_1\varepsilon_2}

The free density follows from the given voltage by solving (3) for $\sigma$ :

\sigma = \frac{\varepsilon_0 V}{d_1/\varepsilon_1 + d_2/\varepsilon_2} = \frac{\varepsilon_0 V\varepsilon_1\varepsilon_2}{\varepsilon_1 d_2 + \varepsilon_2 d_1}

Substituting it, the product $\varepsilon_1\varepsilon_2$ cancels:

\sigma' = \varepsilon_0 V\,\dfrac{\varepsilon_1 - \varepsilon_2}{\varepsilon_1 d_2 + \varepsilon_2 d_1}

(5)

The bound charge is negative when the field passes into the layer of larger permittivity ( $\varepsilon_2 > \varepsilon_1$ ) and positive in the opposite case.

Check

Count the bound charge face by face instead. Each layer’s polarization points along the field, $P = \sigma(1 - 1/\varepsilon)$ , so layer 1 shows $+P_1$ on its face at the boundary and layer 2 shows $-P_2$ on its face there:

P_1 - P_2 = \sigma\left(1 - \frac{1}{\varepsilon_1}\right) - \sigma\left(1 - \frac{1}{\varepsilon_2}\right) = \sigma\left(\frac{1}{\varepsilon_2} - \frac{1}{\varepsilon_1}\right),

the same as before. Equal permittivities give $\sigma' = 0$ , as they must for one uniform slab, and (4) then reduces to $\varepsilon_0\varepsilon S/(d_1 + d_2)$ .

Remark

The field is larger in the layer of smaller permittivity, in the ratio $E_1/E_2 = \varepsilon_2/\varepsilon_1$ . That is why a thin air film left between a plate and a solid dielectric is dangerous in high-voltage work: the air takes a field $\varepsilon$ times stronger than the solid, and it is the air that breaks down first.