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Question

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:

31851.PNG


A
3Q4πε0L
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B
Q4πε0Lln2
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C
Qln24πε0L
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D
Q8πε0L
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Solution

The correct option is B $$\dfrac{Q\ln 2}{4\pi\varepsilon _{0}L}$$
$$
V=\int_{x=L}^{x=2L}\dfrac{k}{x}(\frac{Q}{L})dx=\dfrac{Q\ln 2}{4\pi\varepsilon _{0}L}
$$
114252_31851_ans.png

Physics
NCERT
Standard XII

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