question_answer

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:     AIEEE Solevd Paper-2013

A) \[\frac{Q}{8\pi {{\varepsilon }_{0}}L}\]  

B)        \[\frac{3Q}{4\pi {{\varepsilon }_{0}}L}\]

C)        \[\frac{Q}{4\pi {{\varepsilon }_{0}}L\ell n2}\]     

D)        \[\frac{Q\ell n2}{4\pi {{\varepsilon }_{0}}L}\]

Correct Answer: D

Solution :

\[V=\int{\frac{KdQ}{x}}\] \[=K\int{\left( \frac{Q}{L} \right)}\frac{1}{x}dx=\frac{KQ}{L}\int\limits_{L}^{2L}{\frac{1}{x}dx}\] \[=\frac{KQ}{L}(\ell nx)_{L}^{2L}=\frac{KQ}{L}\ell n2\] \[=\frac{Q\ell n2}{4\pi {{\varepsilon }_{0}}L}\]