Question

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

• A. $r = r_{o}e^{\pi\varepsilon_{o}}\lbrack V(r) + V(r_{o})\rbrack\lambda$
• B. $r = r_{o}e^{2\pi\varepsilon_{o}}\lbrack V(r) - V(r_{o})\rbrack\lambda^{2}$
• C. $r = r_{o}e^{- 2\pi\varepsilon_{o}}\lbrack V(r) - V(r_{o})\rbrack\lambda$
• D. $r = r_{o}e^{- 2\pi\varepsilon_{o}}\lbrack V(r) - V(r_{o})\rbrack\lambda$

Answer 1

Answer: (C)

$r = r_{o}e^{- 2\pi\varepsilon_{o}}\lbrack V(r) - V(r_{o})\rbrack\lambda$

Answer 2

:

Gaussial surface of radius r and length l. according to Gauss’s theorem

$\oint_{}^{}{\overset{\rightarrow}{E}\overset{\rightarrow}{ds} = \frac{q}{\varepsilon_{0}} = \frac{\lambda l}{\varepsilon_{0}}}$

$E(2\pi rl) = \frac{\lambda l}{\varepsilon_{0}}$ or $E = \frac{\lambda}{2\pi\varepsilon_{0}r}$ … (i)

$\therefore V(r) - V(r_{0}) = - \int_{r_{0}}^{r}{\overset{\rightarrow}{E}.\overset{\rightarrow}{dl} = \frac{\lambda}{2\pi\varepsilon_{0}}\log_{e}\frac{r_{0}}{r}}$

For an equipotential surface of given V (r),

$\log_{e}\frac{r}{r_{0}} = \frac{2\pi\varepsilon_{0}}{\lambda}\lbrack V(r) - V(r_{0})\rbrack$

$\therefore r = r_{0}e^{- 2\pi\varepsilon_{0}\lbrack V(r) - V(r_{0})\rbrack}$