Irodov 3.100: A Dielectric Cylinder Across a Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.100

A long round cylinder of uniform dielectric with permittivity $\varepsilon$ is placed in a uniform electric field $\vect E_0$ perpendicular to its axis, and it becomes uniformly polarized. Using the field of a uniformly polarized cylinder (Problem 3.99), find the field $\vect E$ inside the cylinder and the polarization $\vect P$ of the dielectric.

A section of the long dielectric cylinder of permittivity ε, its axis normal to the page, in the uniform field E_0. — A section of the long dielectric cylinder of permittivity $\varepsilon$ , its axis normal to the page, in the uniform field $\vect E_0$ .

Given

uniform field $\vect E_0$ perpendicular to the axis
long dielectric cylinder, permittivity $\varepsilon$
the cylinder polarizes uniformly

Find

\vect E

and

\vect P

inside the cylinder

Solution

Step 1The field inside as a superposition

The field inside is the applied $\vect E_0$ plus the field of the cylinder’s own bound charge. The polarization is uniform and across the axis, so by Problem 3.99 the bound charge adds the uniform field $-\vect P/2\varepsilon_0$ inside.

\vect E = \vect E_0 - \frac{\vect P}{2\varepsilon_0} \tag{1}

Step 2The dielectric closes the loop

The polarization is set by the field inside, through the linear relation of an isotropic dielectric.

\vect P = \varepsilon_0(\varepsilon - 1)\vect E \tag{2}

Substituting (2) into (1) and collecting the terms in $\vect E$ :

\begin{aligned} \vect E &= \vect E_0 - \frac{\varepsilon - 1}{2}\,\vect E\\ \frac{\varepsilon + 1}{2}\,\vect E &= \vect E_0 \end{aligned}

\vect E = \frac{2\vect E_0}{\varepsilon + 1}

(3)

The field lines for ε = 3. Inside, the field is uniform and half of E_0; of the lines that reach the cylinder, one in three goes through, and the others end on negative bound charge on the near side and start again from positive bound charge on the far side. — The field lines for $\varepsilon = 3$ . Inside, the field is uniform and half of $E_0$ ; of the lines that reach the cylinder, one in three goes through, and the others end on negative bound charge on the near side and start again from positive bound charge on the far side.

Step 3The polarization

Substituting (3) into (2):

\vect P = \frac{2\varepsilon_0(\varepsilon - 1)}{\varepsilon + 1}\,\vect E_0

(4)

Check

With $\varepsilon = 1$ , $\vect E = \vect E_0$ and $\vect P = 0$ . As $\varepsilon \to \infty$ the cylinder behaves as a conductor: $\vect E \to 0$ and $\vect P \to 2\varepsilon_0\vect E_0$ , so the bound surface charge $P\cos\phi$ tends to $2\varepsilon_0 E_0\cos\phi$ , the charge induced on a conducting cylinder across a uniform field. The figure’s count also checks. The normal component of $\vect D$ is continuous at the surface, so all the flux of $\vect D$ that arrives goes through; inside, $\vect E$ is $\vect D/\varepsilon_0\varepsilon$ , while outside it is $\vect D/\varepsilon_0$ . The flux of $\vect E$ through the cylinder is therefore $1/\varepsilon$ of the flux that arrives, one line in three for $\varepsilon = 3$ .

Remark

The cylinder keeps less of the field than a ball does: $2/(\varepsilon + 1)$ against $3/(\varepsilon + 2)$ , for $\varepsilon = 3$ a half against three fifths. Its depolarizing factor is $1/2$ against the ball’s $1/3$ (Problems 3.96 and 3.98), so its own bound charge cancels more of the applied field.