Irodov 3.84: Series Layers Share One Displacement – Amit & Deepali Sinha Foundation

Problem Irodov 3.84

A parallel-plate capacitor with air between its plates has the field strength $E_0$ in the gap. A slab of uniform isotropic dielectric of permittivity $\varepsilon$ is then laid on one plate, parallel to the plates and covering their whole area, so that it fills half the thickness of the gap. Call the remaining air layer part 1 and the dielectric layer part 2. Find the magnitudes of $\vect E$ and $\vect D$ in parts 1 and 2 if inserting the dielectric

leaves the voltage across the plates unchanged;
leaves the charges on the plates unchanged.

The capacitor edge on. The dielectric layer ε (part 2) lies on the lower plate and fills the lower half of the gap over the whole plate area; part 1 is the air above it. — The capacitor edge on. The dielectric layer $\varepsilon$ (part 2) lies on the lower plate and fills the lower half of the gap over the whole plate area; part 1 is the air above it.

Given

field $E_0$ in the air-filled gap
a layer of permittivity $\varepsilon$ filling half the gap thickness

Find

E_1

E_2

D_1

D_2

at (a) constant voltage, (b) constant charge

Solution

Step 1One displacement through both layers

Neglecting edge effects, every field in the gap is uniform and normal to the plates. The interface between the layers carries no free charge, so Gauss’s theorem for $\vect D$ over a flat pillbox straddling it makes the normal component of $\vect D$ continuous: the two layers carry the same $D$ .

D_1 = D_2 = D \tag{1}

In each layer $D = \varepsilon_0\varepsilon E$ , with $\varepsilon = 1$ in the air, so the two fields follow from the one displacement.

E_1 = \frac{D}{\varepsilon_0}, \qquad E_2 = \frac{D}{\varepsilon_0\varepsilon} \tag{2}

The field after the layer goes in, drawn for ε = 3: fewer lines cross the layer than the air, and the lines that stop end on negative bound charge on its upper face. — The field after the layer goes in, drawn for $\varepsilon = 3$ : fewer lines cross the layer than the air, and the lines that stop end on negative bound charge on its upper face.

Step 2(a) The voltage is held fixed

Let the gap be $d$ . Before the dielectric goes in, the voltage is $E_0d$ . Afterwards each layer has thickness $d/2$ , and the voltage is the sum of the two drops, which must still equal $E_0d$ . Substituting (2):

\begin{aligned} E_1\frac{d}{2} + E_2\frac{d}{2} &= E_0d\\ \frac{D}{\varepsilon_0}\left(1 + \frac{1}{\varepsilon}\right) &= 2E_0 \end{aligned}

Solving for $D$ , then dividing it by $\varepsilon_0$ for the air and by $\varepsilon_0\varepsilon$ for the dielectric, as in (2):

\begin{gathered} E_1 = \frac{2\varepsilon E_0}{\varepsilon + 1}, \qquad E_2 = \frac{2E_0}{\varepsilon + 1}\\ D_1 = D_2 = \frac{2\varepsilon\varepsilon_0E_0}{\varepsilon + 1} \end{gathered}

(3)

The field in the air layer has grown: $E_1 > E_0$ whenever $\varepsilon > 1$ .

Step 3(b) The plate charges are held fixed

Now the free surface charge $\sigma$ on the plates does not change. Gauss’s theorem for $\vect D$ over a pillbox enclosing a piece of the upper plate gives $D = \sigma$ in the layer next to it, and the air-filled gap had $\sigma = \varepsilon_0E_0$ :

D = \sigma = \varepsilon_0E_0

By (1) this is the displacement in both layers, and (2) gives the fields:

\begin{gathered} E_1 = E_0, \qquad E_2 = \frac{E_0}{\varepsilon}\\ D_1 = D_2 = \varepsilon_0E_0 \end{gathered}

(4)

Check

With $\varepsilon = 1$ the “dielectric” is air, and both cases must return the original gap. In (a), $E_1 = E_2 = 2E_0/2 = E_0$ and $D = \varepsilon_0E_0$ ; in (b), $E_1 = E_2 = E_0$ and $D = \varepsilon_0E_0$ . The opposite limit, $\varepsilon \to \infty$ , makes the layer act as a slab of metal, which a separate argument settles. At fixed voltage the whole voltage then falls across the air half, so its field must be $E_0d/(d/2) = 2E_0$ ; the boxed (a) gives $E_1 \to 2E_0$ and $E_2 \to 0$ . At fixed charge the plates’ charge still faces the air, so the air keeps $E_0$ ; the boxed (b) gives $E_1 = E_0$ and $E_2 \to 0$ .

Remark

At constant voltage the dielectric does not protect the air: it pushes the air layer’s field up toward $2E_0$ as $\varepsilon$ grows, because the dielectric takes almost none of the voltage and the thin air layer must carry nearly all of it. This is why a thin air film trapped against a high-permittivity insulator is where breakdown tends to start.