91Ó°ÊÓ

Inner radius a = 2.00 cm;

Outer radius b = 2.40 cm

The small ball of charge q = 45.0 fC is located at that center

To find an expression for the electric field inside the shell in terms of A and the distance from the center of the shell, choose A so the field does not depend on the distance. We use a Gaussian surface in the form of a sphere with a radius ${\mathbf{r}}_{\mathbf{g}}$, concentric with the spherical shell and within it (a <${\mathit{r}}_{\mathbf{g}}$ < b). Gauss’ law will be used to find the magnitude of the electric field at a distance rg from the shell center. The charge that is both in the shell and within the Gaussian sphere is given by the integral ${\mathbf{q}}_{\mathbf{s}}\mathbf{=}\mathbf{∫}\mathbf{}\mathbf{ÒÏ»å³Õ}$over the portion of the shell within the Gaussian surface. Since the charge distribution has spherical symmetry, we may take dV to be the volume of a spherical shell with radius r and infinitesimal thickness dr, $\mathbf{dV}\mathbf{=}\mathbf{4}{\mathbf{Ï€°ù}}^{\mathbf{2}}\mathbf{dr}\mathbf{.}$Thus,

$$\begin{gathered} {\mathrm{q}}_{\mathrm{s}}=4\mathrm{Ï€}{∫}_{\mathrm{a}}^{{\mathrm{r}}_{\mathrm{g}}}{\mathrm{pr}}^2\mathrm{dr} \\ =4\mathrm{Ï€}{∫}_{\mathrm{a}}^{{\mathrm{r}}_{\mathrm{g}}}\frac{\mathrm{A}}{\mathrm{r}}{\mathrm{r}}^2\mathrm{dr} \\ =4\mathrm{Ï€´¡}{∫}_{\mathrm{a}}^{{\mathrm{r}}_{\mathrm{g}}}\mathrm{rdr} \\ =2\mathrm{Ï€´¡}\left({\mathrm{r}}_{\mathrm{g}}^2-{\mathrm{a}}^2\right) \end{gathered}$$

The total charge inside the Gaussian surface is

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}=\mathrm{q}+{\mathrm{q}}_{\mathrm{s}} \\ =\mathrm{q}+2\mathrm{Ï€´¡}\left({\mathrm{r}}_{\mathrm{g}}^2-{\mathrm{a}}^2\right) \end{gathered}$$

The electric field is radial, so the flux through the Gaussian surface is $\mathrm{Ï•}=4{\mathrm{Ï€°ù}}_{\mathrm{g}}^2\mathrm{E}$ where E is the magnitude of the field. Gauss’ law yields

role="math" localid="1657364518829" $$\begin{gathered} \mathrm{Ï•}=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{Î\mu }}_0} \\ =4{\mathrm{Ï€Î\mu }}_0{{\mathrm{Er}}_{\mathrm{g}}}^2 \\ =\mathrm{q}+2\mathrm{Ï€´¡}\left({\mathrm{r}}_{\mathrm{g}}^2-{\mathrm{a}}^2\right) \end{gathered}$$

We solve for E:

role="math" localid="1657363815610" $\mathrm{E}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\left|\frac{\mathrm{q}}{{\mathrm{r}}_{\mathrm{g}}^2}+2\mathrm{Ï€´¡}-\frac{2{\mathrm{Ï€´¡a}}^2}{{\mathrm{r}}_{\mathrm{g}}^2}\right|$

For the field to be uniform, the first and last terms in the brackets must cancel.

They do if $\mathrm{q}-2{\mathrm{Ï€´¡a}}^2=0$or $\mathrm{A}=\frac{\mathrm{q}}{2{\mathrm{Ï€²¹}}^2}$ .

With a = 0.02 m and$\mathrm{q}=45.0×{10}^{-15}\mathrm{C}$, we have$\mathrm{A}=1.79×{10}^{-11}\mathrm{C}/{\mathrm{m}}^2.$ .

The value we have found for A ensures the uniformity of the field strength inside the shell. Using the result found above, we can readily show that the electric field in the region$a≤r≤b$is

$$\begin{gathered} \mathrm{E}=\frac{2\mathrm{Ï€´¡}}{4{\mathrm{Ï€Î\mu }}_0} \\ =\frac{\mathrm{A}}{2{\mathrm{Î\mu }}_0} \\ =\frac{1.79×{10}^{-11}\mathrm{C}/{\mathrm{m}}^2}{2\left(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)} \\ =1.01\mathrm{N}/\mathrm{C} \end{gathered}$$

The electric field in the region $\mathrm{a}≤\mathrm{r}≤\mathrm{b}$is uniform for the area $\mathrm{A}=1.79×{10}^{-11}\mathrm{C}/{\mathrm{m}}^2$