Irodov 3.25: Thinning Charge Moves the Peak Inside – Amit & Deepali Sinha Foundation

Problem Irodov 3.25

A ball of radius $R$ carries positive charge whose density depends only on the distance $r$ from its centre, as $\rho = \rho_0(1 - r/R)$ with $\rho_0$ a constant. The permittivity is unity both in the ball and around it. Find:

the magnitude of the electric field as a function of $r$ , inside and outside the ball;
the largest value $E_{\max}$ of the field and the distance $r_m$ at which it occurs.

Section of the ball, radius R: the charge density is largest at the centre and falls linearly to zero at the surface. — Section of the ball, radius $R$ : the charge density is largest at the centre and falls linearly to zero at the surface.

Given

$R$
$\rho = \rho_0(1 - r/R)$
$\varepsilon = 1$

Find

(a)

E(r)

inside and outside; (b)

E_{\max}

and

r_m

Solution

Step 1Gauss’s law on a concentric sphere

The charge depends on $r$ alone, so by symmetry the field is radial and has the same magnitude all over any concentric sphere of radius $r$ . Gauss’s law on that sphere relates $E$ to the charge $q(r)$ it encloses:

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

A concentric Gaussian sphere S of radius r < R inside the ball. The field is radial and the same all over it. — A concentric Gaussian sphere $S$ of radius $r < R$ inside the ball. The field is radial and the same all over it.

Step 2The charge enclosed, inside the ball

For $r \le R$ , add up spherical shells of radius $r'$ and volume $4\pi r'^2\,dr'$ :

\begin{aligned} q(r) &= \int_0^r \rho_0\Bigl(1 - \frac{r'}{R}\Bigr)4\pi r'^2\,dr'\\ &= 4\pi\rho_0\Bigl[\frac{r'^3}{3} - \frac{r'^4}{4R}\Bigr]_0^r = 4\pi\rho_0\Bigl(\frac{r^3}{3} - \frac{r^4}{4R}\Bigr) \end{aligned}

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

E = \frac{\rho_0}{\varepsilon_0}\Bigl(\frac{r}{3} - \frac{r^2}{4R}\Bigr)

Step 3Outside the ball

For $r \ge R$ the sphere encloses the whole charge, the value of $q(r)$ at $r = R$ :

q(R) = 4\pi\rho_0\Bigl(\frac{R^3}{3} - \frac{R^3}{4}\Bigr) = \frac{\pi\rho_0 R^3}{3}

and (1) gives the field of a point charge:

E = \frac{\pi\rho_0 R^3/3}{4\pi\varepsilon_0 r^2}

Collecting both regions:

E = \begin{cases} \dfrac{\rho_0 r}{3\varepsilon_0}\Bigl(1 - \dfrac{3r}{4R}\Bigr), & r \le R\\[10pt] \dfrac{\rho_0 R^3}{12\varepsilon_0 r^2}, & r \ge R \end{cases}

(2)

Step 4The maximum

Outside the ball $E$ only falls, so the maximum lies inside. Setting the derivative of the inner branch of (2) to zero:

\frac{dE}{dr} = \frac{\rho_0}{\varepsilon_0}\Bigl(\frac{1}{3} - \frac{r}{2R}\Bigr) = 0 \quad\Longrightarrow\quad r_m = \frac{2R}{3}

The derivative is positive below $r_m$ and negative above it, and $r_m < R$ , so this is the maximum. Substituting $r_m$ :

E_{\max} = \frac{\rho_0}{\varepsilon_0}\Bigl(\frac{2R}{9} - \frac{4R^2}{9\cdot 4R}\Bigr) = \frac{\rho_0}{\varepsilon_0}\Bigl(\frac{2R}{9} - \frac{R}{9}\Bigr)

\begin{gathered} E_{\max} = \frac{\rho_0 R}{9\varepsilon_0}\\ r_m = \frac{2}{3}R \end{gathered}

(3)

The field against r: it rises from zero at the centre, peaks at r_m = 2R/3 inside the ball, joins the outside branch continuously at R, and falls as 1/r^2 beyond. — The field against $r$ : it rises from zero at the centre, peaks at $r_m = 2R/3$ inside the ball, joins the outside branch continuously at $R$ , and falls as $1/r^2$ beyond.

Check

At $r = R$ both branches of (2) give $\rho_0 R/(12\varepsilon_0)$ : the inner one through the factor $1 - 3/4 = 1/4$ , the outer one through $R^3/R^2 = R$ , so the field is continuous at the surface, as it must be with no surface charge. Near the centre the density is about $\rho_0$ , and the field $\rho_0 r/(3\varepsilon_0)$ is that of a uniform ball of density $\rho_0$ .

Remark

For a uniformly charged ball the field peaks right at the surface. Here the charge thins toward the surface, so beyond $r_m$ the added charge grows more slowly than $r^2$ and the field already starts to fall inside the ball. The surface value is three quarters of the peak:

\frac{E(R)}{E_{\max}} = \frac{\rho_0 R/(12\varepsilon_0)}{\rho_0 R/(9\varepsilon_0)} = \frac{3}{4}