Irodov 3.94: A Shorted Electret Opposes Itself – Amit & Deepali Sinha Foundation

Problem Irodov 3.94

A flat plate of thickness $h$ , made of a uniformly and permanently polarized dielectric, lies inside a parallel-plate capacitor, parallel to its plates. The capacitor’s plates, a distance $d$ apart, are joined by a conductor. The polarization $\vect P$ is perpendicular to the plates. Find the vectors $\vect E$ and $\vect D$ inside the dielectric plate and in the gap outside it.

The capacitor with its plates joined by a wire, and the permanently polarized plate of thickness h inside it; P points toward the upper plate and d is the plate separation. — The capacitor with its plates joined by a wire, and the permanently polarized plate of thickness $h$ inside it; $\vect P$ points toward the upper plate and $d$ is the plate separation.

Given

polarization $\vect P$ perpendicular to the plates
plate thickness $h$
separation $d$
capacitor plates connected

Find

\vect E

and

\vect D

inside the dielectric plate and outside it

Solution

Step 1One displacement throughout

The plates are large, so every field is uniform in each layer and perpendicular to the plates. There is no free charge between the plates, so the normal component of $\vect D$ , which is all of it, is continuous at both faces of the dielectric plate.

\vect D_1 = \vect D_2 = \vect D \tag{1}

Here and below the subscript 1 marks the gap and 2 the dielectric plate.

Step 2The field in each layer

In vacuum $\vect D = \varepsilon_0\vect E$ . In the plate the definition $\vect D = \varepsilon_0\vect E + \vect P$ holds with the given, permanent $\vect P$ , which does not depend on $\vect E$ . Solving each for the field and using (1):

\begin{align*} \vect E_1 &= \frac{\vect D}{\varepsilon_0} \\ \vect E_2 &= \frac{\vect D - \vect P}{\varepsilon_0} \tag{2}\end{align*}

Step 3The connecting wire sets the voltage to zero

The plates are joined by a conductor, so they are at one potential and the integral of $\vect E$ across the gap vanishes. Measured along $\vect P$ , the vacuum layers have total thickness $d - h$ and the plate $h$ .

E_1(d - h) + E_2 h = 0

Substituting (2) and multiplying by $\varepsilon_0$ :

D(d - h) + (D - P)h = 0

D\,d = P\,h

(3)

Step 4The two vectors in both regions

Solving (3) for $D$ , and with every vector along $\vect P$ :

\vect D_1 = \vect D_2 = \frac{h}{d}\,\vect P

(4)

Putting this into (2), the gap field follows directly; inside the plate $\vect D - \vect P = -(1 - h/d)\vect P$ .

\begin{gathered} \vect E_1 = \frac{h}{\varepsilon_0 d}\,\vect P\\ \vect E_2 = -\left(1 - \frac{h}{d}\right)\frac{\vect P}{\varepsilon_0} \end{gathered}

(5)

The field in the gaps points along $\vect P$ , the field in the plate against it.

The solved fields: D (open heads) the same in the gap and in the plate; E along P in the gap and reversed, longer, in the plate; free charge on the capacitor plates, bound charge on the faces. The thin lower gap carries the same fields as the upper one. — The solved fields: $\vect D$ (open heads) the same in the gap and in the plate; $\vect E$ along $\vect P$ in the gap and reversed, longer, in the plate; free charge on the capacitor plates, bound charge on the faces. The thin lower gap carries the same fields as the upper one.

Check

The voltage really is zero: with the boxed fields,

E_1(d - h) + E_2 h = \frac{P h}{\varepsilon_0}\left(1 - \frac{h}{d}\right) - \frac{P h}{\varepsilon_0}\left(1 - \frac{h}{d}\right) = 0

If the plate fills the capacitor, $h = d$ , then $\vect D = \vect P$ and $\vect E_2 = 0$ : the free charge drawn onto the plates exactly cancels the bound charge of the faces. As $h \to 0$ the gap field vanishes and $\vect E_2 \to -\vect P/\varepsilon_0$ , the field of an isolated thin polarized sheet.

Remark

The connecting wire lets the plates take the free charge $\sigma = \pm D = \pm P h/d$ , negative on the plate that $\vect P$ points toward. That charge weakens the depolarizing field, the field in the dielectric that points against $\vect P$ , from $P/\varepsilon_0$ for an isolated sheet to $(1 - h/d)P/\varepsilon_0$ . Where the plate sits between the capacitor plates does not matter: only the total gap $d - h$ enters.