Question

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

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

Answer 1

The correct option is C $r=r_0{e}^{-2\pi {\epsilon }_0\lfloor V\left(r\right)-V\left(r_0\right)\rfloor /\lambda }$

Gaussian surface of radius $r$ and length $l$.

According to Gauss's theorem

$\oint \overrightarrow{E}.\overrightarrow{ds}=\frac{q}{{\epsilon }_0}=\frac{\lambda l}{{\epsilon }_0}$

$E\left(2\pi rl\right)=\frac{\lambda l}{{\epsilon }_0}....\left(i\right)$

$orE=\frac{\lambda }{2\pi {\epsilon }_0}{\log }_e\frac{r_0}{r}$

$\therefore V\left(r\right)-V\left(r_0\right)=-{\oint }_{r_0}^r\overrightarrow{E}.\overrightarrow{dl}=\frac{\lambda }{2\pi {\epsilon }_0}{\log }_e\frac{r_0}{r}$

For an equipotential surface of given $V\left(r\right)$

${\log }_e\frac{r_0}{r}=\frac{2\pi {\epsilon }_0}{\lambda }\left[V\right(r)-V(r_0\left)\right]$

$\therefore r=r_0{e}^{-2\pi {\epsilon }_0\left[V\right(r)-V(r_0\left)\right]\lambda }$