Question

Two infinitely long wires carry linear charge densities $+\lambda$ and $-\lambda$ respectively, as shown.The potential difference between points $A$ (at a distance $a$ from the first wire) and $B$ (at a distance $b$ from the second wire) is

• A. $\frac{\lambda }{2\pi {\epsilon }_0}\left\{\ln \frac{(d-a)(d-b)}{ab}\right\}$
• B. $\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{d^2}{ab}\right)$
• C. $\frac{\lambda }{4\pi {\epsilon }_0}\ln \left(\frac{(d-a)(d-b)}{ab}\right)$
• D. $\frac{\lambda }{2\pi {\epsilon }_0}\ln \left(\frac{d-b}{b}/\frac{d-a}{a}\right)$

Answer 1

Answer: (A)

The potential difference between $A$ and $B$ is
$V_B-V_A=-\underset{A}{\overset{B}{\int }}\vec{E}.d\vec{r}$
$V_B-V_A=-\frac{\lambda }{2\pi {\left(ϵ\right)}_0}\underset{A}{\overset{B}{\int }}\left(\frac{1}{r}+\frac{1}{d-r}\right)dr$
$V_B-V_A=-\frac{\lambda }{2\pi {ϵ}_0}\left\{\underset{a}{\overset{d-b}{\int }}\frac{dr}{r}+\underset{a}{\overset{d-b}{\int }}\frac{dr}{d-r}\right\}$
$V_A-V_B=\frac{\lambda }{2\pi {\left(ϵ\right)}_0}\left\{ln\left(\frac{d-b}{a}\right)-ln\left(\frac{b}{d-a}\right)\right\}$
$V_A-V_B=\frac{\lambda }{2\pi {\left(ϵ\right)}_0}\left\{ln\frac{\left(d-a\right)\left(d-b\right)}{ab}\right\}$