Irodov 3.99: A Cylinder Depolarizes by One Half – Amit & Deepali Sinha Foundation

Problem Irodov 3.99

An infinitely long round dielectric cylinder carries a uniform, permanent polarization $\vect P$ perpendicular to its axis. Find the electric field $\vect E$ inside the dielectric.

A section of the long cylinder, its axis normal to the page, with the uniform permanent polarization P across the axis. — A section of the long cylinder, its axis normal to the page, with the uniform permanent polarization $\vect P$ across the axis.

Given

uniform permanent polarization $\vect P$
$\vect P$ perpendicular to the axis
infinitely long round cylinder

Find

\vect E

inside the dielectric

Solution

Step 1Two charged cylinders, slightly apart

Picture the polarization as all the positive charge, density $+\rho$ , filling one cylinder with axis $O_+$ , and all the negative charge, $-\rho$ , an equal cylinder with axis $O_-$ , shifted by the small $\vect a$ from $O_-$ to $O_+$ , across the axis. The moment per volume is then

\vect P = \rho\,\vect a \tag{1}

The polarized cylinder, seen end on, as two cylinders of charge density +ρ and -ρ whose axes are shifted by a (drawn far larger than it is). Only the thin crescents carry charge. — The polarized cylinder, seen end on, as two cylinders of charge density $+\rho$ and $-\rho$ whose axes are shifted by $\vect a$ (drawn far larger than it is). Only the thin crescents carry charge.

Step 2The field inside one uniformly charged cylinder

By symmetry the field of a long cylinder of density $\rho$ points straight away from its axis and is the same all round a coaxial cylinder. Gauss’s theorem over a coaxial cylinder of radius $s$ and length $L$ , with $\vect s$ the perpendicular vector from the axis to the point:

E\cdot 2\pi s L = \frac{1}{\varepsilon_0}\,\rho\,\pi s^2 L

\vect E = \frac{\rho\,\vect s}{2\varepsilon_0} \tag{2}

Step 3The two fields add to a uniform field

A point inside both cylinders lies at $\vect s_+$ from $O_+$ and $\vect s_-$ from $O_-$ , with $\vect s_- - \vect s_+ = \vect a$ . Applying (2) to each, the second with density $-\rho$ :

\vect E = \frac{\rho\,\vect s_+}{2\varepsilon_0} - \frac{\rho\,\vect s_-}{2\varepsilon_0} = -\frac{\rho\,(\vect s_- - \vect s_+)}{2\varepsilon_0} = -\frac{\rho\,\vect a}{2\varepsilon_0}

Substituting (1):

\vect E = -\frac{\vect P}{2\varepsilon_0}

(3)

The field is uniform and points against $\vect P$ .

The uniform field inside the cylinder, against P with magnitude P/2ε_0, and the bound surface charge Pφ that produces it. — The uniform field inside the cylinder, against $\vect P$ with magnitude $P/2\varepsilon_0$ , and the bound surface charge $P\cos\phi$ that produces it.

Check

The field on the axis can be summed directly from the surface charge $\sigma' = P\cos\phi$ , $\phi$ measured from $\vect P$ . A strip of the surface of width $R\,d\phi$ , with $R$ the cylinder’s radius, is a long thread of linear density $\sigma' R\,d\phi$ , whose field at the distance $R$ is $\sigma' R\,d\phi/2\pi\varepsilon_0 R$ . Its component along $-\vect P$ is that times $\cos\phi$ :

E = \int_0^{2\pi}\frac{P\cos^2\phi\,d\phi}{2\pi\varepsilon_0} = \frac{P}{2\pi\varepsilon_0}\cdot\pi = \frac{P}{2\varepsilon_0}

in agreement with (3).

Remark

The coefficient is a depolarizing factor $N$ in $\vect E = -N\vect P/\varepsilon_0$ , set by the body’s shape: $1/3$ for a ball (Problem 3.96), $1/2$ for a cylinder polarized across its axis, $0$ along the axis of a long needle and $1$ across a flat slab. For an ellipsoid the three factors along its axes always add to $1$ : a ball has three thirds, a long cylinder $1/2 + 1/2 + 0$ .