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 distance $L$ from the end $A$ is
A$\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}$
B$\;\frac{Q}{{8\pi {\varepsilon _0}L}}$
C$\;\frac{{3Q}}{{4\pi {\varepsilon _0}L}}$
D$\;\frac{{3Q}}{{4\pi {\varepsilon _0}Lln2}}$
JEE MAIN 2013, Medium
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A$\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}$
a Electric potential is given by,
$V=\int_{L}^{2 L} \frac{k d q}{x}=\int_{L}^{2 L} \frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\frac{q}{L}\right) d x}{x}=\frac{q}{4 \pi \varepsilon_{0} L} \ln (2)$
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