Question

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

• A. $\frac{\lambda }{2\pi {\epsilon }_0}\ln 2$
• B. $-\frac{\lambda }{2\pi {\epsilon }_0}\ln 2$
• C. $\frac{\lambda }{4\pi {\epsilon }_0}\ln 2$
• D. $-\frac{\lambda }{4\pi {\epsilon }_0}\ln 2$

Answer 1

The correct option is A $\frac{\lambda }{2\pi {\epsilon }_0}\ln 2$

As we know that the potential difference between two points in electric field is given by,

$V_A-V_B=-{\int }_B^A\overrightarrow{E}.\overrightarrow{dr}$

$\because$ The Electric field due to infinitely long linear charge density is,

$E=\frac{2K\lambda }{r}=\frac{2\lambda }{4\pi {\epsilon }_{0}r}$

$\Rightarrow V_A-V_B=-{\int }_{2r}^r\overrightarrow{E}.\overrightarrow{dr}$

$=-{\int }_{2r}^r\frac{2\lambda }{4\pi {\epsilon }_{0}r}dr$

$=\frac{2\lambda }{4\pi {\epsilon }_0}{\int }_r^{2r}\frac{1}{r}dr$

$=\frac{2\lambda }{4\pi {\epsilon }_0}\ln \frac{2r}{r}$

$=\frac{2\lambda }{4\pi {\epsilon }_0}\ln 2=\frac{\lambda }{2\pi {\epsilon }_0}\ln 2$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.