Question

In Fig. 23-54, a solid sphere of radius $\mathbf{a}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$is concentric with a spherical conducting shell of inner radius $\mathbf{}\mathbf{b}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{a}$ and outer radius $\mathbf{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{a}$. The sphere has a net uniform charge ${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{fC}$ ; the shell has a net charge ${\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{q}}_{\mathbf{1}}$ . What is the magnitude of the electric field at radial distances (a) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$, (b) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{a}\mathbf{/}\mathbf{2}\mathbf{.}\mathbf{00}$, (c) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{a}$, (d) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{a}$, (e) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{a}$, and (f) $\mathbf{r}\mathbf{}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{a}$? What is the net charge on the (g) inner and (h) outer surface of the shell?

Short answer

a) The magnitude of the electric field at $\mathrm{r}=0\mathrm{is}0\mathrm{N}/\mathrm{C}$ .

b) The magnitude of the electric field at $\mathrm{r}=\mathrm{a}/2\mathrm{is}5.62\times {10}^{-2}\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

c) The magnitude of the electric field at $\mathrm{r}=\mathrm{a}\mathrm{is}0.112\mathrm{N}/\mathrm{C}$.

d) The magnitude of the electric field at $\mathrm{r}=1.5\mathrm{a}\mathrm{is}0.0499\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

e) The magnitude of the electric field at $\mathrm{r}=2.3\mathrm{a}\mathrm{is}0\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

f) The magnitude of the electric field at $\mathrm{r}=3.5\mathrm{a}\mathrm{is}0\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

g) The net charge on the inner surface of the shell is $-5.00\mathrm{fC}$.

h) The net charge on the outer surface of the shell is $0\hspace{0.33em}\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

a) The radius of the solid sphere,$\mathrm{a}=0.02\hspace{0.33em}\mathrm{m}$

b) The inner radius of the spherical conducting shell,$\mathrm{b}=0.04\hspace{0.33em}\mathrm{m}$

c) The outer radius of the spherical conducting shell, $\mathrm{c}=0.048\hspace{0.33em}\mathrm{m}$

d) Uniform charge of the sphere,${\mathrm{q}}_1=+5\hspace{0.33em}\mathrm{fC}$

e) The net charge on the shell,${\mathrm{q}}_2=-5\hspace{0.33em}\mathrm{fC}$

Understanding the concept of Gauss’s law

Using the concept of the electric flux of Gauss's law, we can get the magnitude of the electric field for different radial conditions and different charge distributions. Now, using the same concept, we can get the net charge on both the inner and outer surfaces of the shell.

Formula:

The electric flux through a closed surface closes any volume,

$\oint \mathrm{E}\cdot \mathrm{dA}=\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0}$ (1)

a) Calculation of the electric field at r = 0

At all points where there is an electric field, it is radially outward. For each part of the problem, use a Gaussian surface in the form of a sphere that is concentric with the sphere of charge and passes through the point where the electric field is to be found. The field is uniform on the surface, so, the electric flux through a surface is given as:

$\oint \mathrm{E}\cdot \mathrm{dA}=4{\mathrm{\pi r}}^2\mathrm{E}$ (2)

where, r is the radius of the Gaussian surface.

For $\mathrm{r}<\mathrm{a}$, the charge enclosed by the Gaussian surface is given by:

$\mathrm{q}={\mathrm{q}}_1{\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}^3$.

Thus, using this charge value and equation (a) value in equation (2), we can get the electric field in terms of radial distance that is given as:

role="math" localid="1657351141744" $$\begin{gathered} 4{\mathrm{\pi r}}^2\mathrm{E}=\frac{{\mathrm{q}}_1}{{\mathrm{\epsilon }}_0}{\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}^3 \\ \mathrm{E}=\frac{{\mathrm{q}}_1\mathrm{r}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3} \end{gathered}$$……………………..(3)

For $\mathrm{r}=0$, thus from equation (3), we get the electric field as: $\mathrm{E}=0$

Hence, the value of the electric field is$0\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

b) Calculation of the electric field for r = a/2

For$r=a/2$ , we have the magnitude of the electric field using the given data in equation (3) as:

role="math" localid="1657350755209" $$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_1\left(\mathrm{a}/2\right)}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3} \\ =\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(5.00\times {10}^{-15}\mathrm{C}\right)}{2{\left(2.00\times {10}^{-2}\mathrm{m}\right)}^2} \\ =5.62\times {10}^{-2}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is role="math" localid="1657350217590" $5.62\times {10}^{-2}\mathrm{N}/\mathrm{C}$.

c) Calculation of the electric field for r = a

For $\mathrm{r}=\mathrm{a}$, we have the magnitude of the electric field using the given data in equation (3) as follows:

role="math" localid="1657350960073" $$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_1\left(\mathrm{a}/2\right)}{4{\mathrm{\pi \epsilon }}_0{\mathrm{a}}^3} \\ =\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(5.00\times {10}^{-15}\mathrm{C}\right)}{{\left(2.00\times {10}^{-2}\mathrm{m}\right)}^2} \\ =0.112\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $0.112\mathrm{N}/\mathrm{C}$ .

d) Calculation of the electric field for r = 1.5a

In the case where a < r < b, the charge enclosed by the Gaussian surface is $q_1$, so equation (1) gives the electric field as follows:

role="math" localid="1657351957521" $$\begin{gathered} 4{\mathrm{\pi r}}^2\mathrm{E}=\frac{{\mathrm{q}}_1}{{\mathrm{\epsilon }}_0} \\ \mathrm{E}=\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$……………..(4)

For $\mathrm{r}=1.50\mathrm{a}$, we have the magnitude of the electric field using the given data in equation (4) as follows:

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_1}{4{\mathrm{\pi \epsilon }}_0{\left(1.5\mathrm{a}\right)}^2} \\ =\frac{\left(9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(5.00\times {10}^{-15}\mathrm{C}\right)}{{\left(1.50\times 2.00\times {10}^{-2}\mathrm{m}\right)}^2} \\ =0.0499\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $0.0499\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

e) Calculation of the electric field for r = 2.30a

In the region $b<r<c$, since the shell is conducting, the electric field is zero. Thus, for $\mathrm{r}=2.30\mathrm{a}$, we have the value of the electric field is$0\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

f) Calculation of the electric field for r = 3.50a

For $\mathrm{r}>\mathrm{c}$, the charge enclosed by the Gaussian surface is zero. So, the Gauss equation (1) becomes:

$$\begin{gathered} 4{\mathrm{\pi r}}^2\mathrm{E}=0 \\ \mathrm{E}=0 \end{gathered}$$

Thus, the value of the electric field is $0\hspace{0.33em}\mathrm{N}/\mathrm{C}$ .

g) Calculation of the net charge on the inner charge of the shell

Consider a Gaussian surface that lies completely within the conducting shell. Since the electric field is everywhere zero on the surface, thus, equation (1) gives the value:

$\oint \mathrm{E}\cdot \mathrm{dA}=0$……………(5)

If $Q_i$ is the charge on the inner surface of the shell, then the equation (5) gives:

role="math" localid="1657353669805" $$\begin{gathered} {\mathrm{q}}_1+{\mathrm{Q}}_{\mathrm{i}}=0 \\ {\mathrm{Q}}_{\mathrm{i}}=-{\mathrm{q}}_1 \\ =-5.00\mathrm{fc} \end{gathered}$$

Hence, the value of the net charge is $-5.00\mathrm{fc}$.

h) Calculation of the net charge on the outer charge of the shell

Let ${\mathrm{Q}}_0$ be the charge on the outer surface of the shell.

Since the net charge on the shell is$-\mathrm{q}$ .

Thus, equation (5) gives the value:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{i}}+{\mathrm{Q}}_0=-{\mathrm{q}}_1 \\ {\mathrm{Q}}_0=-{\mathrm{q}}_1-{\mathrm{Q}}_{\mathrm{i}} \\ =-{\mathrm{q}}_1-\left(-{\mathrm{q}}_1\right) \\ =0\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the net charge is $0\hspace{0.33em}\mathrm{N}/\mathrm{C}$.