A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :

• A

  $\frac{\text{Q}}{4\pi {\in }_0\text{L}\mathit{\text{ ln }}2}$
• Correct answer
  B

  $\frac{\text{Q}\mathit{\text{ ln }}2}{4\pi {\in }_0\text{L}}$
• C

  $\frac{\text{Q}}{8\pi {\in }_0\text{L}}$
• D

  $\frac{3\text{Q}}{4\pi {\in }_0\text{L}}$

Solution

$\text{V}=\underset{\text{L}}{\overset{2\text{L}}{\int }}\frac{\text{kdq}}{\text{x}}$

$=\underset{\text{L}}{\overset{2\text{L}}{\int }}\frac{1}{4\pi {\epsilon }_0}\frac{\left(\frac{\text{Q}}{\text{L}}\right)\text{dx}}{\text{x}}$

$=\frac{\text{Q}}{4\pi {\epsilon }_0\text{L}}\ell \text{n}\left(2\right)$

Hence, the correct answer is option: 'B'.