Irodov 3.95: Uniform Charge from Growing Polarization

Problem Irodov 3.95

A long round dielectric cylinder is polarized so that $\vect P = \alpha\vect r$ , where $\alpha$ is a positive constant and $\vect r$ is the radius vector drawn perpendicular to the axis, from the axis to the point. Find the volume density $\rho'$ of bound charge as a function of the distance $r$ from the axis.

A section of the cylinder across its axis (the dot). The polarization points radially outward and grows in proportion to the distance from the axis.

Given

$\vect P = \alpha\vect r$ , radial from the axis
$\alpha > 0$

Find

the bound volume charge density

\rho'(r)

Solution

Step 1Bound charge is the divergence of the polarization

Where the polarization varies, the dipoles fail to cancel and leave a net bound charge, minus the divergence of $\vect P$ .

\rho' = -\nabla\cdot\vect P \tag{1}

Step 2The divergence of a radial field about an axis

$\vect P$ points away from the axis and its magnitude $P_r = \alpha r$ depends on $r$ only, so in cylindrical coordinates the divergence has only its radial term.

\nabla\cdot\vect P = \frac{1}{r}\frac{d}{dr}\left(r P_r\right) = \frac{1}{r}\frac{d}{dr}\left(\alpha r^2\right) = 2\alpha

Substituting into (1):

\rho' = -2\alpha

(2)

The density is the same at every distance from the axis.

The bound charge: a uniform negative density through the volume and a uniform positive density on the surface, equal in total for each length of cylinder.

Check

The same result comes from the integral form. The bound charge inside a coaxial cylinder of radius $r$ and length $L$ is minus the outward flux of $\vect P$ through it; only the curved side contributes, where $P = \alpha r$ is normal to it.

q' = -\alpha r\cdot 2\pi r L = -2\pi\alpha r^2 L

Dividing by the volume $\pi r^2 L$ gives $-2\alpha$ at every $r$ , the boxed density.

Remark

With $R$ the cylinder’s radius, the surface carries $\sigma' = P_r(R) = \alpha R$ . Per length $L$ the surface holds $\alpha R\cdot 2\pi R L$ and the volume $-2\alpha\cdot\pi R^2 L$ , which cancel, as they must for a body that was neutral before it was polarized. The same $\vect P = \alpha\vect r$ about a point instead of an axis would give $-3\alpha$ : the divergence counts the dimensions the field spreads into.