Irodov 3.27: A Cloud That Ends Without an Edge – Amit & Deepali Sinha Foundation

Problem Irodov 3.27

All of space is filled with charge of volume density $\rho = \rho_0 e^{-\alpha r^3}$ , where $\rho_0$ and $\alpha$ are positive constants and $r$ is the distance from the centre of the distribution. Find the magnitude of the electric field as a function of $r$ , and examine its form for small and for large $r$ , that is for $\alpha r^3 \ll 1$ and for $\alpha r^3 \gg 1$ .

Given

$\rho = \rho_0 e^{-\alpha r^3}$
$\rho_0 > 0$ , $\alpha > 0$

Find

E(r)

, and its forms for

\alpha r^3 \ll 1

and

\alpha r^3 \gg 1

Solution

Step 1Gauss’s law on a concentric sphere

The density depends on $r$ alone, so by symmetry the field is radial and has the same magnitude all over any sphere about the centre. On a sphere of radius $r$ enclosing the charge $q(r)$ :

E\cdot 4\pi r^2 = \frac{q(r)}{\varepsilon_0} \tag{1}

Step 2The charge inside radius

r

Add up shells of radius $r'$ and volume $4\pi r'^2\,dr'$ :

q(r) = \int_0^r \rho_0 e^{-\alpha r'^3}\,4\pi r'^2\,dr'

The factor $r'^2\,dr'$ is, up to a constant, the differential of the exponent, so substitute $u = \alpha r'^3$ :

du = 3\alpha r'^2\,dr'

and the integral becomes elementary:

q(r) = \frac{4\pi\rho_0}{3\alpha}\int_0^{\alpha r^3} e^{-u}\,du = \frac{4\pi\rho_0}{3\alpha}\Bigl[-e^{-u}\Bigr]_0^{\alpha r^3} = \frac{4\pi\rho_0}{3\alpha}\bigl(1 - e^{-\alpha r^3}\bigr)

Step 3The field

Dividing by $4\pi\varepsilon_0 r^2$ as (1) requires:

E = \frac{\rho_0}{3\varepsilon_0\alpha r^2}\bigl(1 - e^{-\alpha r^3}\bigr)

(2)

The field is radial, pointing outward since $\rho_0 > 0$ .

The field magnitude against r: a straight rise near the centre, a single peak, then a 1/r^2 fall once nearly all the charge is enclosed. — The field magnitude against $r$ : a straight rise near the centre, a single peak, then a $1/r^2$ fall once nearly all the charge is enclosed.

Step 4Small and large distances

Near the centre, $\alpha r^3 \ll 1$ , expand the exponential to first order:

1 - e^{-\alpha r^3} \approx \alpha r^3 \quad\Longrightarrow\quad E \approx \frac{\rho_0}{3\varepsilon_0\alpha r^2}\,\alpha r^3

Far out, $\alpha r^3 \gg 1$ , the exponential is negligible beside $1$ , and (2) keeps only its first factor:

\begin{gathered} E \approx \frac{\rho_0 r}{3\varepsilon_0} \quad (\alpha r^3 \ll 1)\\ E \approx \frac{\rho_0}{3\varepsilon_0\alpha r^2} \quad (\alpha r^3 \gg 1) \end{gathered}

(3)

Near the centre the field grows in proportion to $r$ , as inside a uniform ball of density $\rho_0$ ; far away it falls as $1/r^2$ , as the field of a point charge.

Check

The total charge is finite: letting $r \to \infty$ in $q(r)$ gives it, and its Coulomb field is exactly the far form:

\begin{gathered} Q = \frac{4\pi\rho_0}{3\alpha}\\ \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{\rho_0}{3\varepsilon_0\alpha r^2} \end{gathered}

Units: $\alpha$ is an inverse volume, so $\rho_0/\alpha$ is a charge and $\rho_0/(\varepsilon_0\alpha r^2)$ a field.

Remark

The cloud has no edge, yet it behaves like a ball of density $\rho_0$ and radius about $\alpha^{-1/3}$ : the field peaks near that radius, where the two limiting forms cross. Indeed a uniform ball of density $\rho_0$ with the same total charge $4\pi\rho_0/(3\alpha)$ has radius exactly $\alpha^{-1/3}$ , and its two field branches are precisely the two limits found here.