91影视

The uniformcharge density is p.

The inner radius is a.

The outer radius is b.

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

$\mathbf{鈭�}\mathit{E}\mathbf{.}\mathit{d}\mathit{a}\frac{\mathbf{q}}{{\mathbf{蔚}}_{\mathbf{0}}}$

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

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

$$\begin{gathered} q_{enclosed}={鈭�}_0^{r}pd味\frac{{饾湑}^2惟}{饾湑u^2} \\ ={鈭�}_0^l{鈭�}_0^{2\mathrm{蟺}}{鈭�}_0^{s}p\left(sds\right)d蠁dz \\ =\left(2\mathrm{蟺辫濒}\right)\left[\frac{s^2}{2}\right] \\ =2{\mathrm{蟺辫濒s}}^2 \\ \mathrm{Apply}\mathrm{Gauss}\mathrm{law}\mathrm{on}\mathrm{the}\mathrm{Gaussian}\mathrm{surface},\mathrm{by}\mathrm{substituting}2蟺pl{s}^2\mathrm{for}{q}_{enclosed} \\ \mathrm{and}2蟺sl\mathrm{for}da\mathrm{into}鈭�E.da=\frac{q_{enclosed}}{{蔚}_0}. \end{gathered}$$

$$\begin{gathered} 鈭�E.da=\frac{q_{enclosed}}{{蔚}_0} \\ E\left(2\mathrm{蟺蝉濒}\right)=\frac{{\mathrm{蟺辫濒s}}^2}{{蔚}_0} \\ E=\frac{{\mathrm{蟺辫濒s}}^2}{{蔚}_0\left(2\mathrm{蟺蝉濒}\right)} \\ E=\frac{ps}{2{蔚}_0}\hat{s} \end{gathered}$$

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

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

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

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

$q_{enclosed}=0$

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