Figure 23-57 shows a spherical shell with uniform volume charge density $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{84}\mathbf{}\mathit{n}\mathbf{C}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$, inner radius localid="1657346086449" $\mathit{a}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$, and outer radius $\mathit{b}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{a}$. What is the magnitude of the electric field at radial distances (a)localid="1657346159507" $\mathit{r}\mathbf{=}\mathbf{0}$; (b) $\mathit{r}\mathbf{=}\mathit{a}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{00}$, (c) $\mathit{r}\mathbf{=}\mathit{a}$, (d) $\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathit{a}$, (e) $\mathit{r}\mathbf{=}\mathit{b}$, and (f) $\mathit{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{b}$?

$\mathrm{\rho }=1.84\mathrm{nC}/{\mathrm{m}}^3$

Inner radius a = 10.0 cm, and outer radius b = 2.00a

Using the concept of charge density for the given region the magnitude of the electric field is determined.

The field is zero for $0\le \mathrm{r}\le \mathrm{a}$as a result of Eq. 23-16. Thus

E = 0 at r = 0,

The magnitude of the electric field at radial distance r = 0, is zero

The field is zero for $0\le \mathrm{r}\le \mathrm{a}$as a result of Eq. 23-16. Thus

E = 0 at r = a/2.

The magnitude of the electric field at radial distance r = a/2.00 is zero

The field is zero for $0\le \mathrm{r}\le \mathrm{a}$as a result of Eq. 23-16.

Thus E = 0 at r = a

The magnitude of the electric field at radial distance r = a is zero

For $\mathrm{a}\le \mathrm{r}\le \mathrm{b}$the enclosed charge ${\mathrm{q}}_{\mathrm{enc}}$(for $\mathrm{a}\le \mathrm{r}\le \mathrm{b}$) is related to the volume by

$q_{enc}=\mathrm{\rho }\left|\frac{4\mathrm{\pi }{r}^3}{3}-\frac{4\mathrm{\pi }{a}^3}{3}\right|$

The electric field will be,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \\ =\frac{\mathrm{\rho }}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}\left|\frac{4{\mathrm{\pi r}}^2}{3}-\frac{4{\mathrm{\pi a}}^3}{3}\right| \\ =\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\frac{{\mathrm{r}}^3-{\mathrm{a}}^3}{{\mathrm{r}}^2} \end{gathered}$$

For$\mathrm{a}\le \mathrm{r}\le \mathrm{b}$

For r = 1.50a, we have

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\left(\frac{{\left(1.50\mathrm{a}\right)}^3-{\mathrm{a}}^3}{{\left(1.50\mathrm{a}\right)}^2}\right) \\ =\frac{\mathrm{\rho a}}{3{\mathrm{\epsilon }}_0}\left(\frac{2.375}{2.25}\right) \\ =\frac{\left(1.84\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^3\right)\left(0.10\mathrm{m}\right)}{3\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)}\left(\frac{2.375}{2.25}\right) \\ =7.32\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at radial distance r=1.50a is$7.32\mathrm{N}/\mathrm{C}$

For r = b = 2.00a, the electric field is

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\left(\frac{{\left(2.00\mathrm{a}\right)}^3-{\mathrm{a}}^3}{{\left(2.00\mathrm{a}\right)}^2}\right) \\ =\frac{\mathrm{\rho a}}{3{\mathrm{\epsilon }}_0}\left(\frac{7}{4}\right) \\ =\frac{\left(1.84\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^3\right)\left(0.10\mathrm{m}\right)}{3\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)}\left(\frac{7}{4}\right) \\ =12.1\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at radial distance r=b is$12.1\mathrm{N}/\mathrm{C}$

For $r\ge b$we have $E=\frac{{\mathrm{q}}_{\mathrm{total}}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$or

$\mathrm{E}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\left(\frac{{\mathrm{b}}^3-{\mathrm{a}}^3}{{\mathrm{r}}^2}\right)$

Thus, for r = 3.00b = 6.00a, the electric field is

$$\begin{gathered} \mathrm{E}=\frac{\mathrm{\rho }}{3{\mathrm{\epsilon }}_0}\left(\frac{{\left(2.00\mathrm{a}\right)}^3-{\mathrm{a}}^3}{{\left(6.00\mathrm{a}\right)}^2}\right) \\ =\frac{\mathrm{\rho a}}{3{\mathrm{\epsilon }}_0}\left(\frac{7}{36}\right) \\ =\frac{\left(1.84\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^3\right)\left(0.10\mathrm{m}\right)}{3\left(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)}\left(\frac{7}{36}\right) \\ =1.35\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at radial distance r = 3.00b is$1.35\mathrm{N}/\mathrm{C}$