91Ó°ÊÓ

The potential of wall,$V_{wall}=0$.

The charge density, $\mathrm{ÒÏ}=-1.1×{10}^{-3}C/m^3$

Consider the radial field produced at points within a uniform cylindrical distribution of charge.

The volume enclosed by a Gaussian surface in this case is$L\mathrm{Ï€}{r}^2$.

Thus, Gauss’ law leads to,

$$\begin{gathered} E=\frac{\left|q_{enc}\right|}{A_{cylinder}{\mathrm{Î\mu }}_0} \\ =\frac{\mathrm{ÒÏ}\left(L\mathrm{Ï€}{r}^2\right)}{{\mathrm{Î\mu }}_0\left(2\mathrm{Ï€}rL\right)} \\ =\frac{\left|\mathrm{ÒÏ}\right|r}{2{\mathrm{Î\mu }}_0} \end{gathered}$$

After reading the question, Electric potential can be defined as the work needed to transfer any positive charge from a reference point to a particular point inside the field without the production of any acceleration.

It is asked to find an expression for the electric potential,$V$as a function of radial distance,$r$from the center of the pipe. Assume electric potential to be zero on the grounded pipe wall.

The required potential can be calculated as,

role="math" localid="1662554546234" $V_{wall}-V=-{∫}_r^{R}Edr$

Here,$E$is the electric field.

$0-V=-{∫}_r^R\frac{\mathrm{ÒÏ}r}{2{\mathrm{Î\mu }}_0}$

Integrate the above equation, and you have

$$\begin{gathered} -V=\frac{\mathrm{ÒÏ}}{2{\mathrm{Î\mu }}_0}{\left[\frac{r^2}{2}\right]}_r^R \\ =-\frac{\mathrm{ÒÏ}\left(R^2-r^2\right)}{4{\mathrm{Î\mu }}_0} \end{gathered}$$

$V=\frac{\mathrm{ÒÏ}\left(R^2-r^2\right)}{4{\mathrm{Î\mu }}_0}$

Thus, an expression for the electric potential comes out to be$V=\frac{\mathrm{ÒÏ}\left(R^2-r^2\right)}{4{\mathrm{Î\mu }}_0}$ .

The difference in the electric potential,$V_{center}$between the pipe’s center and its inside wall can be calculated as ,

$V_{center}=\frac{\mathrm{ÒÏ}\left(R^2-r^2\right)}{4{\mathrm{Î\mu }}_0}$.

Substitute, $-1.1×{10}^{-3}C/m^3$for $p$, $0.05\text{m}$ for $R$, $0$for $r$ and $8.85×{10}^{-12}C/V·m$for ${\mathrm{Î\mu }}_0$ and solve for $V_{center}$

$V_{center}=\frac{-1.1×{10}^{-3}C/m^3\left({(0.05m)}^2-0\right)}{4\left(8.85×{10}^{-12}C/V·m\right)}$

$V_{center}=\frac{2.75×{10}^{-6}C/m}{35.4×{10}^{-12}C/V·m}$

$V_{center}=7.8×{10}^{4}V$

Thus, the difference in the electric potential between the pipe's center and its inside wall comes out to be $7.8×{10}^{4}V$