Irodov 3.77: Two Ways to Charge a Shell – Amit & Deepali Sinha Foundation

Problem Irodov 3.77

A spherical layer of uniform isotropic dielectric has inner radius $a$ and outer radius $b$ . It carries some positive free (extraneous) charge, spread uniformly

over the layer’s inner surface;
through the layer’s volume.

In each case sketch the electric field strength $E$ and the potential $\varphi$ as functions of the distance $r$ from the center.

(a) The positive charge spread over the inner surface of the layer.

(b) The positive charge spread through the layer.

Given

dielectric layer, radii $a < b$
positive free charge, uniform (a) on the inner surface, (b) through the volume

Find

the graphs of

E(r)

and

\varphi(r)

in each case

Solution

Step 1Gauss’s theorem for D, and E from it

Call the total free charge $q$ . Everything is spherically symmetric, so $\vect D$ is radial and the same all over a concentric sphere of radius $r$ , and Gauss’s theorem for $\vect D$ counts only the free charge $q(r)$ inside that sphere:

D\cdot 4\pi r^2 = q(r) \tag{1}

The field is $D/(\varepsilon_0\varepsilon)$ inside the dielectric, $\varepsilon$ being its permittivity, and $D/\varepsilon_0$ in the cavity and outside. The potential, zero at infinity, is the integral of $E$ inward from infinity:

\varphi(r) = \int_r^\infty E\,dr' \tag{2}

Step 2(a) Charge on the inner surface: the field

Every sphere with $r > a$ encloses all of $q$ , and a sphere with $r < a$ encloses none of it. Dividing (1) by $4\pi r^2$ and by $\varepsilon_0\varepsilon$ in the layer, $\varepsilon_0$ elsewhere:

E = \begin{cases} 0, & r < a\\[4pt] \dfrac{q}{4\pi\varepsilon_0\varepsilon r^2}, & a < r < b\\[10pt] \dfrac{q}{4\pi\varepsilon_0 r^2}, & r > b \end{cases}

(3)

$E$ jumps up from zero at $r = a$ , where the free charge sits, and jumps up again by the factor $\varepsilon$ at $r = b$ , where the dielectric ends.

Step 3(a) Charge on the inner surface: the potential

Outside the layer (2) gives the point-charge potential $q/(4\pi\varepsilon_0 r)$ . Inside the layer, between $a$ and $b$ , add the integral of the layer’s field from $r$ out to $b$ :

\begin{aligned} \varphi &= \frac{q}{4\pi\varepsilon_0 b} + \int_r^b \frac{q\,dr'}{4\pi\varepsilon_0\varepsilon r'^2}\\ &= \frac{q}{4\pi\varepsilon_0 b} + \frac{q}{4\pi\varepsilon_0\varepsilon}\left(\frac{1}{r} - \frac{1}{b}\right) \end{aligned}

In the cavity $E = 0$ , so the potential keeps its value at $r = a$ . It is continuous everywhere, with kinks at $a$ and $b$ where $E$ jumps.

Case (a), drawn for ε = 2 and b = 2a: E is zero in the cavity, jumps up at a and again at b; φ is flat in the cavity, then falls, with kinks at a and b. — Case (a), drawn for $\varepsilon = 2$ and $b = 2a$ : $E$ is zero in the cavity, jumps up at $a$ and again at $b$ ; $\varphi$ is flat in the cavity, then falls, with kinks at $a$ and $b$ .

(a)

E

jumps at

r = a

and at

r = b

;

\varphi

is continuous.

Step 4(b) Charge through the volume: the field

Now the charge inside a sphere of radius $r$ in the layer is the fraction of the layer’s volume it takes in:

q(r) = q\,\frac{r^3 - a^3}{b^3 - a^3}, \qquad a \le r \le b

with none for $r < a$ and all of $q$ for $r > b$ . By (1):

E = \begin{cases} 0, & r < a\\[4pt] \dfrac{q}{4\pi\varepsilon_0\varepsilon}\,\dfrac{r^3 - a^3}{(b^3 - a^3)\,r^2}, & a < r < b\\[10pt] \dfrac{q}{4\pi\varepsilon_0 r^2}, & r > b \end{cases}

(4)

Now $E$ grows continuously from zero at $r = a$ and jumps only at $r = b$ , again by the factor $\varepsilon$ .

Step 5(b) Charge through the volume: the potential

Outside, the potential is again $q/(4\pi\varepsilon_0 r)$ . In the layer, between $a$ and $b$ , integrating the middle line of (4) from $r$ out to $b$ :

\begin{aligned} \varphi &= \frac{q}{4\pi\varepsilon_0 b} + \frac{q}{4\pi\varepsilon_0\varepsilon(b^3 - a^3)}\int_r^b\left(r' - \frac{a^3}{r'^2}\right)dr'\\ &= \frac{q}{4\pi\varepsilon_0 b} + \frac{q}{4\pi\varepsilon_0\varepsilon(b^3 - a^3)}\left[\frac{b^2 - r^2}{2} + a^3\left(\frac{1}{b} - \frac{1}{r}\right)\right] \end{aligned}

In the cavity it keeps its value at $r = a$ . This time $\varphi$ leaves the cavity with zero slope, because $E$ starts from zero at $a$ .

Case (b), drawn for ε = 2 and b = 2a: E is zero in the cavity, rises through the layer and jumps up at b; φ is flat in the cavity and falls smoothly, with a kink only at b. — Case (b), drawn for $\varepsilon = 2$ and $b = 2a$ : $E$ is zero in the cavity, rises through the layer and jumps up at $b$ ; $\varphi$ is flat in the cavity and falls smoothly, with a kink only at $b$ .

(b)

E

rises from zero at

a

, jumps only at

b

Check

At $r = b$ there is no free surface charge in either case, so $D$ is continuous there and $E$ must jump by exactly the factor $\varepsilon$ , as both fields do. For $\varepsilon = 1$ the jump at $b$ disappears and case (a) becomes the vacuum shell: zero field inside, $q/(4\pi\varepsilon_0 r^2)$ outside, the potential constant inside at $q/(4\pi\varepsilon_0 a)$ .

Remark

The jumps of $E$ mark bound charge. At $r = b$ the dielectric’s outer surface carries the positive bound charge $q(\varepsilon - 1)/\varepsilon$ , which lifts the field to its vacuum value; in case (a) the inner surface carries $-q(\varepsilon - 1)/\varepsilon$ beside the free charge, so the net charge there is only $q/\varepsilon$ and the jump at $a$ is $\varepsilon$ times smaller than in vacuum. The potential never jumps, because a finite field over a vanishing distance does no work.