Question

A long coaxial cable (Fig. 2.26) carries a uniform volume charge density pon the inner cylinder (radius a ), and a uniform surface charge density on the outer cylindrical shell (radius b ). Thissurface charge is negative and is of just the right magnitude that the cable as a whole is electrically neutral. Find the electric field in each of the three regions: (i) inside the inner cylinder$(s<a)$,(ii) between the cylinders$(a<s<b)$(iii) outside the cable$(s>b)$Plot lEI as a function of s.

Short answer

(i)The electric field inside the inner cylinder is $E=\frac{k{r}^2}{4{\epsilon }_0}\hat{r}$.

(ii)The electric field between the cylinder is $E=\frac{p{r}^2}{2S{\epsilon }_0}\hat{s}.$

(iii)The electric field outside the cable ( s > b ) is E = 0. The electric field is plotted as follows:

Step-by-step solution

 Step 1: Describe the given information

The uniformcharge density is p.

The inner radius is a.

The outer radius is b.

Define the Gauss law

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

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

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

Obtain the electric field inside the inner cable.

The Gaussian cylinder is shown as

It is known that the charge density inside the inner cylinder is p. So, the charge enclosed by the inner cylinder of radius sand height l is obtained by integrating the charge density from 0 to s , as

$$\begin{gathered} q_{enclosed}={\int }_0^{r}pd\zeta \frac{{𝜕}^2\Omega }{𝜕u^2} \\ ={\int }_0^l{\int }_0^{2\mathrm{\pi }}{\int }_0^{s}p\left(sds\right)d\phi dz \\ =\left(2\mathrm{\pi pl}\right)\left[\frac{s^2}{2}\right] \\ =2{\mathrm{\pi pls}}^2 \\ \mathrm{Apply}\mathrm{Gauss}\mathrm{law}\mathrm{on}\mathrm{the}\mathrm{Gaussian}\mathrm{surface},\mathrm{by}\mathrm{substituting}2\pi pl{s}^2\mathrm{for}{q}_{enclosed} \\ \mathrm{and}2\pi sl\mathrm{for}da\mathrm{into}\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0}. \end{gathered}$$

$$\begin{gathered} \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ E\left(2\mathrm{\pi sl}\right)=\frac{{\mathrm{\pi pls}}^2}{{\epsilon }_0} \\ E=\frac{{\mathrm{\pi pls}}^2}{{\epsilon }_0\left(2\mathrm{\pi sl}\right)} \\ E=\frac{ps}{2{\epsilon }_0}\hat{s} \end{gathered}$$

$\mathrm{Thus},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{inside}\mathrm{the}\mathrm{inner}\mathrm{cylinder}\mathrm{is}\mathrm{E}=\frac{{\mathrm{kr}}^2}{4{\mathrm{s}}_0}\hat{\mathrm{r}},$

Obtain the electric field between the cylinder.

It is known that the charge density inside the inner cylinder is p . So, the charge enclosed between the cylinder a < s < b is obtained by integrating the charge density as

$$\begin{gathered} q_{enclosed}={\int }_0^l{\int }_0^{2\mathrm{\pi }}{\int }_0^{a}p\left(sds\right)d\varphi dz \\ =p\left({\mathrm{\pi a}}^2\right)l \\ \mathrm{Apply}\mathrm{Gauss}\mathrm{law}\mathrm{on}\mathrm{the}\mathrm{Gaussian}\mathrm{surface},\mathrm{by}\mathrm{substituting}p\left({\mathrm{\pi a}}^2\right)l\mathrm{for} \\ {\mathrm{q}}_{\mathrm{enclosed}}\mathrm{and}2\mathrm{\pi sl}\mathrm{for}da\mathrm{into}\oint \mathrm{E}.\mathrm{da}=\frac{{\mathrm{q}}_{\mathrm{enclosed}}}{{\mathrm{\epsilon }}_0} \\ \oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ E\left(2\pi sl\right)=\frac{p\left(\pi a^2\right)l}{{\epsilon }_0\left(2\pi sl\right)} \\ E=\frac{p\left(\pi a^2\right)l}{{\epsilon }_0\left(2\pi sl\right)} \\ E=\frac{p{a}^2}{2s{\epsilon }_0}\hat{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{between}\mathrm{the}\mathrm{cylinder}\mathrm{is}E=\frac{p{a}^2}{2s{\epsilon }_0}\hat{s} \end{gathered}$$

 Step 5: Obtain the electric field outside the cable.

Outside the cable, s > b, the cable is electrically neutral. So, the charge enclosed outside the cable is 0. Thus,

$q_{enclosed}=0$

$$\begin{gathered} \mathrm{Apply}\mathrm{Gauss}\mathrm{law}\mathrm{on}\mathrm{the}\mathrm{Gaussian}\mathrm{surface},\mathrm{by}\mathrm{substituting}p\left(\pi a^2\right)l\mathrm{for} \\ {\mathrm{q}}_{\mathrm{enclosed}}\mathrm{and}2\pi sl\mathrm{for}da\mathrm{into}\oint E.da=\frac{q_{enclosed}}{{\epsilon }_0} \\ \oint E.da=\frac{0}{{\epsilon }_0} \\ E\left(2\pi sl\right)=\frac{0}{{\epsilon }_0} \\ E=0 \\ \mathrm{Thus},\mathrm{the}\mathrm{electric}\mathrm{field}\mathrm{outside}\mathrm{the}\mathrm{cable}(s>b)\mathrm{is}E=0.\mathrm{thus} \\ \mathrm{electric}\mathrm{field}\left|\mathrm{E}\right|\mathrm{is}\mathrm{plotted}\mathrm{against}\mathrm{the}\mathrm{distance}s\mathrm{as}, \end{gathered}$$