Question

The electric potential due to an infinite uniformly charge of linear charge density $\lambda$ at a distance $r$ from it is given by
$r_0$ is the reference point

• A. $\frac{2\lambda }{4\pi {ϵ}_{0}r}$
• B. $\frac{2\lambda }{4\pi {ϵ}_0{r}_0}$
• C. $\frac{2\lambda (\ln r-\ln r_0)}{4\pi {ϵ}_0}$
• D. $\frac{2\lambda \left(\ln r\right)}{4\pi {ϵ}_{0}r}$

Answer 1

The correct option is C $\frac{2\lambda (\ln r-\ln r_0)}{4\pi {ϵ}_0}$

The electric field due to an infinite uniformly charge.

density charge $=\lambda$

distance $=r$

reference point $=r_0$

Potential of any point a with respect to any other $b$,

$V_a-V_b=\frac{\lambda }{2\pi {\in }_0}[ln\left(\frac{2a}{r_a}\right)-ln\left(\frac{2a}{r_b}\right)]$

$=\frac{\lambda }{2\pi {\in }_0}ln\frac{r_b}{r_a}⟶\left(1\right)$

Suppose $V_b=0$ at $r_b=r_0$ and $V_a=V$ at $r_a=r$,

then, $V=\frac{\lambda }{2\pi {\in }_0}ln\frac{r_0}{r}$

$=\frac{2\lambda }{4\pi {\in }_0}$