A long conducting cylinder of radius R has a uniform linear charge density $λ$ . The electric field at a point at a perpendicular distance r from the cylinder is $E = \frac{λ}{2 π \in_{0} r}$ . The electric potential at that point is

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$\frac{λ}{2 π ϵ_{0}}$

$\frac{λ}{2 π \in_{0}} \ln (\frac{r}{R})$

$\frac{λ}{2 π \in_{0}} \ln (\frac{R}{r})$

$- \frac{λ R}{2 π ϵ_{0} r^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{rcl} E & = & - \frac{d V}{d r} \\ d V & = & - E d r \\ V & = & - \int_{R}^{r} E d r \\ = & - \frac{λ}{2 π \in_{0}} \int_{R}^{r} \frac{d r}{r} \\ = & - \frac{λ}{2 π \in_{0}} \ln (\frac{r}{R}) \\ = & \frac{λ}{2 π ϵ_{0}} \ln (\frac{R}{r}) \end{array}$