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