Question

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) $\mathit{r}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{b}$?

Short answer

a) The magnitude of the electric field at radial distancer = 0, is zero

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

c) The magnitude of the electric field at radial distancer = a is zero

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

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

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

Step-by-step solution

Listing the given quantities

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

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

Understanding the concept of charge density and electric field

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

(a) Explanation

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

(b) Explanation

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

(c) Explanation

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

(d) Calculations for the magnitude of the electric field

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$7.32\mathrm{N}/\mathrm{C}$

(e) Calculations for the magnitude of the electric field at r=b

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$12.1\mathrm{N}/\mathrm{C}$

(f) Calculations for the magnitude of the electric field at r=3.00b

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$1.35\mathrm{N}/\mathrm{C}$