Question

An infinitely long line of linear charge density $\text{\lambda }$ is shown in the figure. The potential difference $V_{A}-V_{B}$ between the two points $A$ and $B$ is

• A. $\frac{\text{\lambda }}{ \, \text{2\pi ε}_{\text{0}}} \, l\text{n 2}$
• B. $-\frac{\text{\lambda }}{ \, \text{2\pi ε}_{\text{0}}}l\text{ n 2}$
• C. $\frac{\text{\lambda }}{ \, \text{4\pi ε}_{\text{0}}}l\text{ n 2}$
• D. $-\frac{\text{\lambda }}{ \, \text{4\pi ε}_{\text{0}}}l\text{ n 2}$

Answer 1

Answer: (A)

$\text{V}_{\text{A}}-\text{ V}_{\text{B}}\text{ =}-\displaystyle \int _{B}^{A} \overset{ \rightarrow }{E} . \overset{ \rightarrow }{d r}$
$\text{V}_{\text{A}}-\text{ V}_{\text{B}}\text{ =}-\displaystyle \int _{\text{2r}}^{\text{r}} \frac{\text{2\lambda }}{\text{4\pi } \text{ε}_{\text{0}} \text{r}} \text{dr}$
$\text{V}_{\text{A}}-\text{ V}_{\text{B}}\text{ =}\frac{\text{2\lambda }}{\text{4\pi } \text{ε}_{\text{0}} \text{r}} \, \displaystyle \int _{\text{r}}^{\text{2r}} \frac{\text{1}}{\text{r}} \text{dr}$
$=\frac{\text{2\lambda }}{\text{4\pi } \text{ε}_{\text{0}}}\text{ ln }\frac{\text{2r}}{\text{r}}$
$=\frac{\text{2\lambda }}{\text{4\pi } \text{ε}_{\text{0}}}\text{ln 2}$