Irodov 3.26: A Charge Cloud Flattens the Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.26

A ball of radius $R$ carries a spherically symmetric charge, and the whole space around it holds charge of volume density $\rho = \alpha/r$ , where $\alpha$ is a constant and $r$ is the distance from the ball’s centre. What charge must the ball carry for the magnitude of the electric field outside the ball not to depend on $r$ , and what is that field? Take the permittivity to be unity in the ball and in the space around it.

Section through the centre O: the ball of radius R with charge q, and the surrounding charge, densest near the ball and thinning out as 1/r (drawn for α > 0, and only out to a finite radius). — Section through the centre $O$ : the ball of radius $R$ with charge $q$ , and the surrounding charge, densest near the ball and thinning out as $1/r$ (drawn for $\alpha > 0$ , and only out to a finite radius).

Given

$R$
$\rho = \alpha/r$ for $r > R$
$\varepsilon = 1$

Find

the ball’s charge

q

for which

E

outside is constant, and that

E

Solution

Step 1Gauss’s law outside the ball

Everything is spherically symmetric, so the field is radial and has the same magnitude all over a concentric sphere. On a sphere of radius $r > R$ , Gauss’s law reads

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

where $q$ is the ball’s charge and $q_{\mathrm{out}}(r)$ the surrounding charge between $R$ and $r$ .

A concentric Gaussian sphere S of radius r > R. It encloses the ball's charge q and the part of the surrounding charge between R and r. — A concentric Gaussian sphere $S$ of radius $r > R$ . It encloses the ball’s charge $q$ and the part of the surrounding charge between $R$ and $r$ .

Step 2The surrounding charge enclosed

Adding shells of radius $r'$ and volume $4\pi r'^2\,dr'$ from the ball’s surface out to $r$ :

q_{\mathrm{out}}(r) = \int_R^r \frac{\alpha}{r'}\,4\pi r'^2\,dr' = 4\pi\alpha\Bigl[\frac{r'^2}{2}\Bigr]_R^r = 2\pi\alpha\,(r^2 - R^2)

Step 3Make the field independent of

r

Putting $q_{\mathrm{out}}$ into (1) and dividing by $4\pi\varepsilon_0 r^2$ :

E = \frac{q + 2\pi\alpha r^2 - 2\pi\alpha R^2}{4\pi\varepsilon_0 r^2}

Splitting off the part that grows with $r^2$ :

E = \frac{q - 2\pi\alpha R^2}{4\pi\varepsilon_0 r^2} + \frac{\alpha}{2\varepsilon_0} \tag{2}

The second term is already constant. The field is independent of $r$ for every $r > R$ only if the first term vanishes, that is if its numerator is zero:

q = 2\pi\alpha R^2

(3)

The ball then carries charge of the same sign as $\alpha$ . With the first term of (2) gone, the field outside is

E = \frac{\alpha}{2\varepsilon_0}

(4)

Check

The required charge (3) is the charge the surrounding cloud would hold inside radius $R$ if it continued in to the centre:

\int_0^R\frac{\alpha}{r}\,4\pi r^2\,dr = 2\pi\alpha R^2

So, seen from outside the ball, ball and cloud together act like one cloud with $\rho = \alpha/r$ everywhere, whose enclosed charge $2\pi\alpha r^2$ grows exactly as the sphere’s area; Gauss’s law then gives the same constant field:

E = \frac{2\pi\alpha r^2}{4\pi\varepsilon_0 r^2} = \frac{\alpha}{2\varepsilon_0}

Units: $\alpha$ is a charge per area, so $\alpha/\varepsilon_0$ is a field.

Remark

A field of constant magnitude pointing straight outward is not a uniform field: its direction turns from point to point. Its divergence is nonetheless exactly what Gauss’s law in differential form requires for the given density:

\frac{1}{r^2}\frac{d(r^2E)}{dr} = \frac{2E}{r} = \frac{\alpha}{\varepsilon_0 r} = \frac{\rho}{\varepsilon_0}