Question

An infinite cylinder of radius $r_{0}$, carrying linear charge density $\lambda$ The equation of the equipotential surface for this cylinder is

• A. $r=r_{0} e^{\pi\varepsilon_{0} \left[V\left(r\right)+V\left(r_{0}\right)\right]\lambda}$
• B. $r=r_{0} e^{2\pi\varepsilon_{0} \left[V\left(r\right)-V\left(r_{0}\right)\right]\lambda^2}$
• C. $r=r_{0} e^{-2\pi\varepsilon_{0} \left[V\left(r\right)-V\left(r_{0}\right)\right]/\lambda}$
• D. $r=r_{0} e^{-2\pi\varepsilon_{0} \left[V\left(r\right)-V\left(r_{0}\right)\right]\lambda}$

Answer 1

Answer: (C)

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

$\oint \vec{E} . \overrightarrow{ds}=\frac{q}{\varepsilon_{0}}=\frac{\lambda l}{\varepsilon_{0}}$
$E\left(2\pi rl\right)=\frac{\lambda l}{\varepsilon_{0}} $
or $E=\frac{\lambda}{2\pi\varepsilon_{0}r} \ldots\left(i\right)$
$\therefore V \left(r\right)-V\left(r_{0}\right)=-\int_{r_0}^{r} \vec{E}. \vec{dl} =\frac{\lambda}{2\pi\varepsilon_{0}} log_{e} \frac{r_{0}}{r}$
For an equipotential surface of given $V\left(r\right)$,
$log_{e} \frac{r}{r_{0}}=\frac{2\pi\varepsilon_{0}}{\lambda} \left[V\left(r\right)-V\left(r_{0}\right)\right]$
$\therefore r=r_{0}e^{-2\pi\varepsilon_{0} \left[V\left(r\right)-V\left(r_{0}\right)\right]/\lambda}$