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. $\frac{\lambda}{2 \pi \varepsilon_{0}}\left\{\ln \frac{(d-a)(d-b)}{a b}\right\}$
• B. $\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \left(\frac{d^{2}}{a b}\right)$
• C. $\frac{\lambda}{4 \pi \varepsilon_{0}} \ln \left(\frac{(d-a)(d-b)}{a b}\right)$
• D. $\frac{\lambda}{2 \pi \varepsilon_{0}} \ln \left(\frac{d-b}{b} / \frac{d-a}{a}\right)$

Answer 1

Answer: (A)

The potential difference between $A$ and $B$ is
$V_{B}-V_{A}=-\int \limits_{A}^{B}\vec{E}.d\vec{r}$
$V_{B}-V_{A}=-\frac{\lambda }{2 \pi \left(\epsilon \right)_{0}}\int\limits _{A}^{B}\left(\frac{1}{r} + \frac{1}{d - r}\right)dr$
$V_{B}-V_{A}=-\frac{\lambda }{2 \pi \epsilon _{0}}\left\{\int \limits_{a}^{d - b} \frac{d r}{r} + \int \limits_{a}^{d - b} \frac{d r}{d - r}\right\}$
$V_{A}-V_{B}=\frac{\lambda }{2 \pi \left(\epsilon \right)_{0}}\left\{ln \left(\frac{d - b}{a}\right) - ln \left(\frac{b}{d - a}\right)\right\}$
$V_{A}-V_{B}=\frac{\lambda }{2 \pi \left(\epsilon \right)_{0}}\left\{ln \frac{\left(d - a\right) \left(d - b\right)}{a b}\right\}$