Question

An infinite cylinder of radius $r_0$, carrying linear charge density $\lambda$ The equation of the equipotential surface for this cylinder is

• A. $r=r_0{e}^{\pi {\epsilon }_0\left[V\left(r\right)+V\left(r_0\right)\right]\lambda }$
• B. $r=r_0{e}^{2\pi {\epsilon }_0\left[V\left(r\right)-V\left(r_0\right)\right]{\lambda }^2}$
• C. $r=r_0{e}^{-2\pi {\epsilon }_0\left[V\left(r\right)-V\left(r_0\right)\right]/\lambda }$
• D. $r=r_0{e}^{-2\pi {\epsilon }_0\left[V\left(r\right)-V\left(r_0\right)\right]\lambda }$

Answer 1

Answer: (C)

Gaussian surface of radius $r$ and length $l$ According to Gauss’s theorem

$\oint \vec{E}.\overrightarrow{ds}=\frac{q}{{\epsilon }_0}=\frac{\lambda l}{{\epsilon }_0}$
$E\left(2\pi rl\right)=\frac{\lambda l}{{\epsilon }_0}$
or $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}\dots \left(i\right)$
$\therefore V\left(r\right)-V\left(r_0\right)=-{\int }_{r_0}^r\vec{E}.\overrightarrow{dl}=\frac{\lambda }{2\pi {\epsilon }_0}lo{g}_e\frac{r_0}{r}$
For an equipotential surface of given $V\left(r\right)$,
$lo{g}_e\frac{r}{r_0}=\frac{2\pi {\epsilon }_0}{\lambda }\left[V\left(r\right)-V\left(r_0\right)\right]$
$\therefore r=r_0{e}^{-2\pi {\epsilon }_0\left[V\left(r\right)-V\left(r_0\right)\right]/\lambda }$