Irodov 3.85: Side by Side, One Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.85

A parallel-plate capacitor with air between its plates has the field strength $E_0$ in the gap. A block of uniform isotropic dielectric of permittivity $\varepsilon$ is then inserted so that it fills the whole height of the gap over one half of the plate area; the boundary between the dielectric and the air is perpendicular to the plates. Call the air-filled half part 1 and the dielectric half 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 gap is split at the middle of the plates: air on the left (part 1), the dielectric ε filling the full gap height on the right (part 2). — The capacitor edge on. The gap is split at the middle of the plates: air on the left (part 1), the dielectric $\varepsilon$ filling the full gap height on the right (part 2).

Given

field $E_0$ in the air-filled gap
dielectric $\varepsilon$ filling the gap over half the plate area

Find

E_1

E_2

D_1

D_2

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

Solution

Step 1One field in both halves

Neglecting edge effects, the field in each half is uniform and normal to the plates. Each plate is a single conductor, so both halves span the same potential difference $V$ across the same gap, which we call $d$ . A uniform field across a gap is the voltage over the gap, so the two fields are equal.

E_1 = E_2 = E = \frac{V}{d} \tag{1}

The displacements then differ by the factor $\varepsilon$ :

D_1 = \varepsilon_0E, \qquad D_2 = \varepsilon_0\varepsilon E \tag{2}

The field with the dielectric in place, drawn for ε = 3: the same line density in both halves, while the plates over the dielectric carry the denser free charge. — The field with the dielectric in place, drawn for $\varepsilon = 3$ : the same line density in both halves, while the plates over the dielectric carry the denser free charge.

Since $D$ equals the free surface charge on the plate above it, the plates over the dielectric carry $\varepsilon$ times the charge density of the plates over the air.

Step 2(a) The voltage is held fixed

The voltage was $E_0d$ before the dielectric went in and is unchanged, so (1) gives $E = E_0d/d$ , and (2) the displacements:

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

(3)

Step 3(b) The plate charges are held fixed

Let each plate have area $S$ . Before, the whole plate carried the charge $\varepsilon_0E_0S$ . Afterwards the half over the air carries $D_1S/2$ and the half over the dielectric $D_2S/2$ , and their sum is the same charge. Substituting (2):

\begin{aligned} \frac{S}{2}\left(\varepsilon_0E + \varepsilon_0\varepsilon E\right) &= \varepsilon_0E_0S\\ E\,(1 + \varepsilon) &= 2E_0 \end{aligned}

Solving for $E$ and using (2) again:

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

(4)

Check

With $\varepsilon = 1$ both cases return the empty gap, $E = E_0$ and $D_1 = D_2 = \varepsilon_0E_0$ . In (b) the charge really is conserved: $(S/2)(D_1 + D_2) = (S/2)\cdot 2\varepsilon_0E_0(1 + \varepsilon)/(\varepsilon + 1) = \varepsilon_0E_0S$ .

Remark

The two halves are capacitors in parallel, where the layered arrangement of Irodov 3.84 put two layers in series. In series the displacement is shared and the fields differ; in parallel the field is shared and the displacements, and so the free charges, differ. At constant charge the free charge migrates along each plate toward the dielectric half until it holds the fraction $\varepsilon/(\varepsilon + 1)$ of the total. The capacitance has risen by the factor $(\varepsilon + 1)/2$ , so the stored energy $q^2/2C$ has fallen by it, and that drop in energy is why a dielectric is drawn into a charged capacitor.