Question

A thick spherical shell carries charge density

$\mathit{p}\mathbf{=}\frac{\mathbf{k}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathit{a}\mathbf{<}\mathit{r}\mathbf{<}\mathit{b}\mathbf{)}$

(Fig. 2.25). Find the electric field in the three regions: (i) r< a,(ii) a< r< b,(iii) r> b.Plot lEI as a function of r,for the case b=2a.

Short answer

(i) The electric field for the regionr < a is E = 0 .

(ii) The electric field between the region a < r < b is $E=\frac{k}{{\epsilon }_0}\left[\frac{1}{r}-\frac{a}{r^2}\right]$

(iii)The electric field for the region ( r > b ) is $E=\frac{k(b-a)}{{\epsilon }_0\left(r^2\right)}.$. The electric field $\left|E\right|$is plotted against the distance ras,

Step-by-step solution

Describe the given information

It is given that hallow spherical shell has inner radius a and outer radius b.

It carries the volume charge density of ${}^{P=\frac{K}{r^2}}$

Define the Gauss law

If there is a surface area enclosing a volume, possessing a charge inside the volume then the electric field due to the surface or volume charge is given as

$\mathbf{\oint }\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$

Here q is the elemental surface area, ${\mathit{\epsilon }}_{\mathbf{0}}$is the permittivity of free surface.

Obtain the electric field for the region (i)

(i)

Consider a Gaussian sphere of radius a such that r < a inside the solid sphere as shown below:

It is known that the charge density is present only on the surface of the sphere whereas, inside the solid sphere, the charge enclosed is 0. Thus, the total charge inside the Gaussian sphere is 0.

Apply Gauss law, on the Gaussian surface, as,

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ =\frac{0}{{\epsilon }_0} \\ =0 \end{gathered}$$

Thus the electric field for the region r < a is E = 0 .

Obtain the electric field for the region (ii)

(ii)

It is known that the charge density inside the radius a < r < bis p . So, the charge enclosed between a < r < bis obtained by integrating the charge density from ato b as

$$\begin{gathered} q_{enclosed}={\int }_a^{r}p\left(4{\mathrm{\pi r}}^2\right)\mathrm{dr} \\ ={\int }_{\mathrm{a}}^{\mathrm{r}}\frac{\mathrm{k}}{{\mathrm{r}}^2}\left(4{\mathrm{\pi r}}^2\right)\mathrm{dr} \\ ={\int }_{\mathrm{a}}^{\mathrm{r}}\left(4{\mathrm{\pi r}}^2\right)\mathrm{dr} \\ =4\mathrm{\pi k}(\mathrm{r}-\mathrm{a}) \end{gathered}$$

Apply Gauss law on the Gaussian surface, by substitutinglocalid="1654489491971" $=4\mathrm{\pi k}(\mathrm{r}-\mathrm{a})$forlocalid="1654489502377" $q_{enclosed}$andlocalid="1654489517890" $4{\mathrm{\pi r}}^2$for da into $\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0}$

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ Solveas, \\ E\left(4{\mathrm{\pi r}}^2\right)=\frac{4\mathrm{\pi k}(\mathrm{r}-\mathrm{a})}{{\epsilon }_0} \\ E=\frac{4\mathrm{\pi k}(\mathrm{r}-\mathrm{a})}{{\epsilon }_0\left(4{\mathrm{\pi r}}^2\right)} \\ E=\frac{k(\mathrm{r}-\mathrm{a})}{{\epsilon }_0\left(r^2\right)} \\ E=\frac{k}{{\epsilon }_0}\left(\frac{1}{r}-\frac{a}{r^2}\right) \\ \mathrm{Thus},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{between}\mathrm{the}\mathrm{region}a<r<bisE=\frac{k}{{\epsilon }_0}\left(\frac{1}{r}-\frac{a}{r^2}\right). \end{gathered}$$

Obtain the electric field for the region (iii)

(iii)

Outside the shell,$r>b$, consider a Gaussian sphere of radius outside the spherical shell. The differential volume of the spherical shell at radius r and thickness dr is$dv=4{\mathrm{\pi r}}^2\mathrm{dr}$. The differential charge enclosed by this shell is obtained by multiplying the charge density with differential volume as

$$\begin{gathered} dp=p\times 4{\mathrm{\pi r}}^2\mathrm{dr} \\ =\frac{\mathrm{k}}{{\mathrm{r}}^2}\times 4{\mathrm{\pi r}}^2\mathrm{dr} \\ =4\mathrm{\pi kdr} \\ \mathrm{Integrate}\mathrm{above}\mathrm{equation}\mathrm{from}\mathrm{a}\mathrm{to}\mathrm{b}\mathrm{as}, \\ \int \mathrm{dp}={\int }_{\mathrm{a}}^{\mathrm{b}}4\mathrm{\pi kdr} \\ \mathrm{q}=4\mathrm{\pi k}{\left(\mathrm{r}\right)}_{\mathrm{a}}^{\mathrm{b}} \\ =4\mathrm{\pi k}(\mathrm{b}-\mathrm{a}) \\ \mathrm{Apply}\mathrm{Gauss}\mathrm{law}\mathrm{on}\mathrm{the}\mathrm{Gaussian}\mathrm{surface},\mathrm{by}\mathrm{substituting}4\mathrm{\pi k}(\mathrm{b}-\mathrm{a})\mathrm{for} \\ {\mathrm{q}}_{\mathrm{enclosed}},\mathrm{and}4{\mathrm{\pi r}}^2\mathrm{for}da\mathrm{into}\oint \mathrm{E}.\mathrm{da}=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{\epsilon }}_0} \\ \oint \mathrm{E}.\mathrm{da}=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{\epsilon }}_0} \\ \mathrm{E}\left(4{\mathrm{\pi r}}^2\right)=\frac{4\mathrm{\pi k}(\mathrm{b}-\mathrm{a})}{{\mathrm{\epsilon }}_0} \\ \mathrm{E}=\frac{4\mathrm{\pi k}(\mathrm{b}-\mathrm{a})}{{\mathrm{\epsilon }}_0\left(4{\mathrm{\pi r}}^2\right)} \\ \mathrm{E}=\frac{\mathrm{k}(\mathrm{b}-\mathrm{a})}{{\mathrm{\epsilon }}_0\left({\mathrm{r}}^2\right)} \\ \mathrm{Thus},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{for}\mathrm{the}\mathrm{region}(\mathrm{r}>\mathrm{b})\mathrm{is}\mathrm{E}=\frac{\mathrm{k}(\mathrm{b}-\mathrm{a})}{{\mathrm{\epsilon }}_0\left({\mathrm{r}}^2\right)}.\mathrm{The}\mathrm{electric}\mathrm{field}\left|\mathrm{E}\right|\mathrm{is}\mathrm{plotted} \\ \mathrm{against}\mathrm{the}\mathrm{distance}r\mathrm{as}, \end{gathered}$$