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Question

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

Vedclass · Accepted Answer

$\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}$

Vedclass · Answer

Electric potential is given by,

$V=\int_{L}^{2 L} \frac{k d q}{x}=\int_{L}^{2 L} \frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\frac{q}{L}\right) d x}{x}=\frac{q}{4 \pi \varepsilon_{0} L} \ln (2)$

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

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