91Ó°ÊÓ

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

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

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

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

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

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,

$âˆ\circledR \mathrm{E}â‹\dots \mathrm{dA}=\frac{\mathrm{Q}}{{\mathrm{Î\mu }}_0}$ (1)

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:

$âˆ\circledR \mathrm{E}â‹\dots \mathrm{dA}=4{\mathrm{Ï€°ù}}^2\mathrm{E}$ (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" $$\begin{gathered} 4{\mathrm{Ï€°ù}}^2\mathrm{E}=\frac{{\mathrm{q}}_1}{{\mathrm{Î\mu }}_0}{\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}^3 \\ \mathrm{E}=\frac{{\mathrm{q}}_1\mathrm{r}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{a}}^3} \end{gathered}$$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(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$0 \mathrm{N}/\mathrm{C}$ .

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

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

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

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" $$\begin{gathered} \mathrm{E}=\frac{{\mathrm{q}}_1\left(\mathrm{a}/2\right)}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{a}}^3} \\ =\frac{\left(9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(5.00×{10}^{-15}\mathrm{C}\right)}{{\left(2.00×{10}^{-2}\mathrm{m}\right)}^2} \\ =0.112\mathrm{N}/\mathrm{C} \end{gathered}$$

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

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" $$\begin{gathered} 4{\mathrm{Ï€°ù}}^2\mathrm{E}=\frac{{\mathrm{q}}_1}{{\mathrm{Î\mu }}_0} \\ \mathrm{E}=\frac{{\mathrm{q}}_1}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \end{gathered}$$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(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:

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

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

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$0 \mathrm{N}/\mathrm{C}$ .

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

$$\begin{gathered} 4{\mathrm{Ï€°ù}}^2\mathrm{E}=0 \\ \mathrm{E}=0 \end{gathered}$$

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

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:

$âˆ\circledR \mathrm{E}â‹\dots \mathrm{dA}=0$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(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 $-5.00\mathrm{fc}$.

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 $0 \mathrm{N}/\mathrm{C}$.