Irodov 3.136: Energy Stored in a Dielectric Shell – Amit & Deepali Sinha Foundation

Problem Irodov 3.136

A point charge $q = 3.0\ \mu$ C sits at the center of a spherical layer of uniform isotropic dielectric of permittivity $\varepsilon = 3.0$ . The layer’s inner radius is $a = 250$ mm and its outer radius is $b = 500$ mm. Find the electrostatic field energy contained in the dielectric layer.

The point charge q at the center of the dielectric layer of permittivity ε, inner radius a, outer radius b, in section (drawn for q > 0). — The point charge $q$ at the center of the dielectric layer of permittivity $\varepsilon$ , inner radius $a$ , outer radius $b$ , in section (drawn for $q > 0$ ).

Given

$q = 3.0\ \mu$ C at the center
$\varepsilon = 3.0$
$a = 250$ mm
$b = 500$ mm

Find

the field energy

W

in the layer

Solution

Step 1The field inside the layer

The only free charge is $q$ and the arrangement is spherically symmetric, so $\vect D$ is radial with the same magnitude all over a concentric sphere. Gauss’s theorem for $\vect D$ over a sphere of radius $r$ inside the layer, and $D = \varepsilon_0\varepsilon E$ there:

\begin{aligned} D\cdot 4\pi r^2 &= q\\ E &= \frac{D}{\varepsilon_0\varepsilon} = \frac{q}{4\pi\varepsilon_0\varepsilon r^2} \end{aligned}

Step 2The energy density

In a linear isotropic dielectric the field energy per unit volume is $w = \varepsilon_0\varepsilon E^2/2$ . With the field above:

w = \frac{\varepsilon_0\varepsilon}{2}\left(\frac{q}{4\pi\varepsilon_0\varepsilon r^2}\right)^2 = \frac{q^2}{32\pi^2\varepsilon_0\varepsilon\,r^4} \tag{1}

Step 3The energy in the layer

Integrating (1) over thin spherical shells of volume $4\pi r^2\,dr$ from $a$ to $b$ :

\begin{aligned} W &= \int_a^b\frac{q^2}{32\pi^2\varepsilon_0\varepsilon\,r^4}\,4\pi r^2\,dr = \frac{q^2}{8\pi\varepsilon_0\varepsilon}\int_a^b\frac{dr}{r^2} \end{aligned}

Numerically, with $1/(8\pi\varepsilon_0) = 4.5\times10^9$ m/F, $q^2 = 9.0\times10^{-12}$ C $^2$ , $1/a = 4.0$ m $^{-1}$ and $1/b = 2.0$ m $^{-1}$ :

W = \frac{4.5\times10^{9}\times 9.0\times10^{-12}}{3.0}\,(4.0 - 2.0)\ \text{J}

W = \frac{q^2}{8\pi\varepsilon_0\varepsilon}\left(\frac1a - \frac1b\right) \approx 27\text{ mJ}

(2)

Check

Dimensions: $q^2/(\varepsilon_0 a)$ is an energy, and $\varepsilon$ is a pure number. Limits: as $b \to a$ the layer vanishes and so does $W$ ; for $\varepsilon = 1$ the formula is the vacuum field energy between radii $a$ and $b$ , and the dielectric cuts it by exactly the factor $\varepsilon$ , because $E$ falls by $\varepsilon$ while the density $\varepsilon_0\varepsilon E^2/2$ gains one factor $\varepsilon$ back.

Remark

The dielectric lowers the energy of the region it fills, which is why a dielectric is drawn into the field of a charge: moving it inward, toward stronger field, reduces the total field energy further. Outside the layer the field is exactly that of $q$ in vacuum, since the layer’s two bound surface charges are equal and opposite.