Irodov 3.82: A Thin Disc Barely Depolarizes – Amit & Deepali Sinha Foundation

Problem Irodov 3.82

A thin round disc of dielectric, of radius $R$ and thickness $d$ with $d \ll R$ , carries a frozen-in uniform polarization $\vect P$ that lies in the plane of the disc. Find the electric field $\vect E$ at the center of the disc.

The disc face on, uniformly polarized along P in its own plane; O is its center. — The disc face on, uniformly polarized along $\vect P$ in its own plane; $O$ is its center.

The disc edge on: thickness d, diameter 2R, with P in the plane. — The disc edge on: thickness $d$ , diameter $2R$ , with $\vect P$ in the plane.

Given

radius $R$
thickness $d \ll R$
uniform $\vect P$ in the plane of the disc

Find

\vect E

at the center

Solution

Step 1Where the bound charge sits

A uniform polarization has no divergence, so the volume density of bound charge $\rho' = -\nabla\cdot\vect P$ is zero, and all the bound charge is on the surface, with density $\sigma' = \vect P\cdot\vect n$ for the outward normal $\vect n$ . On the two flat faces $\vect n$ is perpendicular to $\vect P$ , so they carry nothing. On the rim, with $\varphi$ the angle of the outward normal measured from $\vect P$ :

\sigma' = P\cos\varphi \tag{1}

The rim is positive where $\vect P$ leaves the disc and negative where it enters.

Step 2The field of one rim element

The bound charge on the rim, positive where P leaves and thinning to zero at right angles to it. The element dq' at angle φ makes the field d E at O, directed from the element through O. — The bound charge on the rim, positive where $\vect P$ leaves and thinning to zero at right angles to it. The element $dq'$ at angle $\varphi$ makes the field $d\vect E$ at $O$ , directed from the element through $O$ .

Cut the rim into strips of height $d$ and arc length $R\,d\varphi$ . Because $d \ll R$ , every point of a strip is at distance $R$ from $O$ , so each strip acts as a point charge there. By (1) the strip at angle $\varphi$ carries

dq' = \sigma'\,R\,d\varphi\,d = Pd\,R\cos\varphi\,d\varphi

Its field at $O$ has magnitude $dq'/(4\pi\varepsilon_0R^2)$ and points from the strip toward $O$ , so its component along $\vect P$ carries the factor $-\cos\varphi$ . The strips at $\varphi$ and $-\varphi$ carry equal charge, so the components across $\vect P$ cancel in pairs and only the component along $\vect P$ survives.

dE_P = -\frac{dq'\cos\varphi}{4\pi\varepsilon_0R^2} = -\frac{Pd}{4\pi\varepsilon_0R}\cos^2\varphi\,d\varphi

Step 3Adding up the rim

Integrating once around the rim, $\varphi$ from $0$ to $2\pi$ , with $\int_0^{2\pi}\cos^2\varphi\,d\varphi = \pi$ :

E_P = -\frac{Pd}{4\pi\varepsilon_0R}\int_0^{2\pi}\cos^2\varphi\,d\varphi = -\frac{Pd}{4\pi\varepsilon_0R}\cdot\pi

The field is along $\vect P$ with this component, so as a vector:

\vect E = -\frac{d}{4\varepsilon_0R}\,\vect P

(2)

The field at the center has magnitude $Pd/(4\varepsilon_0R)$ and points opposite to $\vect P$ .

The field at the center points against the polarization: away from the positive rim charge on the side P points to. — The field at the center points against the polarization: away from the positive rim charge on the side $\vect P$ points to.

Check

Dimensions: $P/\varepsilon_0$ is a field (C/m $^2$ over F/m gives V/m), and $d/R$ is a pure number. As $d/R \to 0$ the field vanishes, as it must for an infinite film polarized in its plane, which has no surface across $\vect P$ to carry bound charge. Against a disc polarized normal to its faces, where the faces carry $\pm P$ and the inside field is $-\vect P/\varepsilon_0$ , the in-plane field is smaller by the factor $d/(4R)$ .

Remark

The field a polarized body produces inside itself opposes the polarization, and its strength depends only on the body’s shape. Written as $\vect E = -N\vect P/\varepsilon_0$ , the factor $N$ is $1/3$ for a ball, $1$ for a plate polarized across its thickness, and here $d/(4R)$ at the center of a thin disc polarized in its plane. This is why the polarization of a thin ferroelectric film tends to lie in its plane: there the self-field fighting it almost disappears.