Irodov 3.81: A Charged Dielectric Ball Inside Out – Amit & Deepali Sinha Foundation

Problem Irodov 3.81

A ball of radius $R$ made of uniform isotropic dielectric of permittivity $\varepsilon$ carries free (extraneous) charge spread uniformly through its volume with density $\rho > 0$ . Find:

the magnitude of the electric field as a function of the distance $r$ from the center, and sketch the graphs of $E(r)$ and of the potential $\varphi(r)$ ;
the volume density and the surface density of the bound charge.

The dielectric ball of radius R and permittivity ε with the free charge density ρ spread through it. — The dielectric ball of radius $R$ and permittivity $\varepsilon$ with the free charge density $\rho$ spread through it.

Given

ball radius $R$
permittivity $\varepsilon$
free charge density $\rho > 0$ , uniform

Find

(a)

E(r)

, graphs of

E(r)

and

\varphi(r)

; (b)

\rho'

and

\sigma'

Solution

Step 1The displacement from Gauss’s theorem

The arrangement is spherically symmetric, so $\vect D$ is radial and the same all over a concentric sphere of radius $r$ . Gauss’s theorem for $\vect D$ counts the free charge inside it, $\rho\cdot 4\pi r^3/3$ for $r < R$ and $\rho\cdot 4\pi R^3/3$ for $r > R$ :

D\cdot 4\pi r^2 = \begin{cases} \dfrac{4}{3}\pi r^3\rho, & r < R\\[8pt] \dfrac{4}{3}\pi R^3\rho, & r > R \end{cases}

D = \begin{cases} \dfrac{\rho r}{3}, & r < R\\[8pt] \dfrac{\rho R^3}{3r^2}, & r > R \end{cases} \tag{1}

Step 2(a) The field

The field is $D/(\varepsilon_0\varepsilon)$ in the ball and $D/\varepsilon_0$ outside. From (1):

E = \begin{cases} \dfrac{\rho r}{3\varepsilon_0\varepsilon}, & r < R\\[10pt] \dfrac{\rho R^3}{3\varepsilon_0 r^2}, & r > R \end{cases}

(2)

At the surface $E$ jumps up by the factor $\varepsilon$ , from $\rho R/(3\varepsilon_0\varepsilon)$ to $\rho R/(3\varepsilon_0)$ .

Step 3(a) The potential

With the potential zero at infinity, outside the ball it is that of a point charge, $\varphi = \rho R^3/(3\varepsilon_0 r)$ . Inside, add to its surface value the integral of the inner field from $r$ out to $R$ :

\begin{aligned} \varphi &= \frac{\rho R^2}{3\varepsilon_0} + \int_r^R\frac{\rho r'}{3\varepsilon_0\varepsilon}\,dr'\\ &= \frac{\rho R^2}{3\varepsilon_0} + \frac{\rho\,(R^2 - r^2)}{6\varepsilon_0\varepsilon} \end{aligned}

so that

\varphi = \begin{cases} \dfrac{\rho R^2}{3\varepsilon_0}\left(1 + \dfrac{1 - r^2/R^2}{2\varepsilon}\right), & r \le R\\[10pt] \dfrac{\rho R^3}{3\varepsilon_0 r}, & r \ge R \end{cases} \tag{3}

E

jumps up at

R

;

\varphi

is continuous there, with a kink.

Step 4(b) The bound charges

In the uniform dielectric $\vect P = (\varepsilon - 1)\vect D/\varepsilon$ , so by (1) the polarization inside is radial:

\vect P = \frac{\varepsilon - 1}{\varepsilon}\,\frac{\rho}{3}\,\vect r \tag{4}

The volume density of bound charge is minus its divergence, and the surface density is its radial component at the surface, the outward normal there being radial:

\begin{gathered} \rho' = -\nabla\cdot\vect P\\ \sigma' = P_r(R) \end{gathered}

With $\nabla\cdot\vect r = 3$ and $|\vect r| = R$ at the surface, (4) gives

\begin{gathered} \rho' = -\frac{\varepsilon - 1}{\varepsilon}\,\rho\\ \sigma' = \frac{\varepsilon - 1}{3\varepsilon}\,\rho R \end{gathered}

(5)

The polarization points outward and grows in proportion to r; it leaves a uniform positive bound charge σ' on the surface, balancing the negative bound charge spread through the volume. — The polarization points outward and grows in proportion to $r$ ; it leaves a uniform positive bound charge $\sigma'$ on the surface, balancing the negative bound charge spread through the volume.

Check

The ball’s bound charge totals zero, as it must:

\rho'\cdot\frac{4}{3}\pi R^3 + \sigma'\cdot 4\pi R^2 = -\frac{\varepsilon - 1}{\varepsilon}\,\frac{4\pi\rho R^3}{3} + \frac{\varepsilon - 1}{3\varepsilon}\,4\pi\rho R^3 = 0

The jump of $E$ at the surface is $\rho R(\varepsilon - 1)/(3\varepsilon_0\varepsilon)$ , which is $\sigma'/\varepsilon_0$ , as Gauss’s theorem for $\vect E$ over a pillbox requires. For $\varepsilon = 1$ both bound densities vanish and $E$ is continuous, the familiar uniformly charged ball.

Remark

Inside, the free and bound charge add to $\rho/\varepsilon$ , so the ball’s interior behaves like a vacuum ball of density $\rho/\varepsilon$ , and the screened charge $(\varepsilon - 1)/\varepsilon$ of the total sits on the surface. Seen from outside the dielectric changes nothing; inside it lowers the field by $\varepsilon$ and the potential at the center from $\rho R^2/(2\varepsilon_0)$ toward the surface value $\rho R^2/(3\varepsilon_0)$ .