Question

Two points are at distances $a$ and $b$ $(a<b)$ from a long string having linear charge density $\lambda$. The potential difference between the points is proportional to

• A. $\frac{b^2}{a^2}$
• B. $\sqrt{\frac{b}{a}}$
• C. $\frac{b}{a}$
• D. $\ln \left(\frac{b}{a}\right)$

Answer 1

The correct option is D $\ln \left(\frac{b}{a}\right)$

Electric field due to a long string of charge per unit length $\lambda$ is

$E=\frac{\lambda }{2\pi {ϵ}_{o}r}$ (From Gauss Law)

Now, using relation between potential and electric field

$E=\frac{-dV}{dr}$

Integrating both side, we get

$\Rightarrow {\int }_{V_a}^{V_b}dV=-{\int }_a^{b}Edr$

$\Rightarrow V_b-V_a=-{\int }_a^b\frac{\lambda }{2\pi {ϵ}_o}\frac{dr}{r}$

$\Rightarrow V_a-V_b=\frac{\lambda }{2\pi {ϵ}_o}\ln \left(\frac{b}{a}\right)$

$\Rightarrow V_a-V_b\propto \ln \left(\frac{b}{a}\right)$

So, option $\left(d\right)$ is correct.

Tip: $\ln b-\ln a=\ln \left(\frac{b}{a}\right)$