- Home
- Standard 12
- Physics
Find the equation of the equipotential for an infinite cylinder of radius ${{r_0}}$, carrying charge of linear density $\lambda $.
Solution
$\int_{0}^{2 \pi r l} \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{S}} =\frac{Q}{\epsilon_{0}}$$=\frac{\lambda l}{\epsilon_{0}}$
From Gauss's law,
$\left[ E _{r} S \cos \theta\right]_{0}^{2 \pi r l} =\frac{\lambda l}{\epsilon_{0}}$
$E _{r} \times 2 \pi r l =\frac{\lambda l}{\epsilon_{0}} \quad\left[\theta=0 \therefore \cos 0^{\circ}=1\right]$
$\therefore E _{r}=\frac{\lambda}{2 \pi \epsilon_{0} r}$
The radius of infinite cylinder is $r_{0}$,
$\mathrm{V}(r)-\mathrm{V}\left(r_{0}\right)=-\int_{r_{0}}^{r} \mathrm{E} d l$
$=-\frac{\lambda}{2 \pi \in_{0}} \log _{e} \frac{r}{r_{0}}=\frac{\lambda}{2 \pi \in_{0}} \log _{e} \frac{r_{0}}{r}$
$\text { because, } \int_{r_{0}}^{r} \frac{\lambda}{2 \pi \in_{0} r} d r=\frac{\lambda}{2 \pi \in_{0}} \int_{r_{0}}^{r} \frac{1}{r} d r$
$\quad \mathrm{~V}=\frac{\lambda}{2 \pi \in_{0}} \log _{e} \frac{r}{r_{0}}$
For given $V$,
$\log _{e} \frac{r}{r_{0}}=-\frac{2 \pi \epsilon_{0}}{\lambda} \times\left[\mathrm{V}(r)-\mathrm{V}\left(r_{0}\right)\right] r=r_{0} e^{-\frac{2 \pi \epsilon_{0}}{\lambda}\left[\mathrm{V}(r)-\mathrm{V}\left(r_{0}\right)\right]}$
$\therefore r=r_{0} e^{-\frac{2 \pi \epsilon_{0}}{\lambda}\left[\mathrm{V}(r)-\mathrm{V}\left(r_{0}\right)\right]}$
