Q16P - Introduction to Electrodynamics

Question

A long coaxial cable (Fig. 2.26) carries a uniform volume charge density p on the inner cylinder (radius a ), and a uniform surface charge density on the outer cylindrical shell (radius b ). This surface 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 . 

                                               

Step-by-Step Solution

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 1: Describe the given information

The uniform charge density is p.

The inner radius is a.

The outer radius is b.

Step 2: 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.

Step 3: 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 s and height l is obtained by integrating the charge density from 0 to s , as 

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

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

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

Step 4: 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,

$q_{enclosed}=0$

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