Question

Two points are at distances a and b (a< b) from a long string of charge per unit length $\lambda$. The potential difference between the points is proportional to

• A. $\frac{b}{a}$
• B. $\frac{b^2}{a^2}$
• C. $\sqrt{\frac{b}{a}}$
• D. $\log \frac{b}{a}$

Answer 1

Answer: (D)

$\log \frac{b}{a}$

Consider a Gaussian cylinder of radius $r$ and length $l$. Using Gauss's law the electric field at a distance $r$ from the string is $E.\left(2\pi rl\right)=\frac{\lambda l}{{ϵ}_0}$ where $\lambda$, line charge density.
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
$E=-\frac{dV}{dr}=\frac{\lambda }{2\pi {ϵ}_{0}r}$
$dV=-\frac{\lambda }{2\pi {ϵ}_{0}r}dr$
integrating, ${\int }_{V_a}^{V_b}=-\frac{\lambda }{2\pi {ϵ}_0}{\int }_a^b\frac{dr}{r}$
$V_b-V_a=-\frac{\lambda }{2\pi {ϵ}_0}ln\left(\frac{b}{a}\right)$
$V_a-V_b=\frac{\lambda }{2\pi {ϵ}_0}ln\left(\frac{b}{a}\right)$
The potential difference between the points a and b is $(V_a-V_b)\propto ln\left(\frac{b}{a}\right)$