JEE MAIN - Physics (2013 (Offline) - No. 17)
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 distance $$L$$ from the end $$A$$ is
_en_17_1.png)
$${Q \over {8\pi {\varepsilon _0}L}}$$
$${{3Q} \over {4\pi {\varepsilon _0}L}}$$
$${Q \over {4\pi {\varepsilon _0}L\,\ln \,2}}$$
$${{Q\ln \,2} \over {4\pi {\varepsilon _0}L\,{}^s}}$$
Explanation
_en_17_2.png)
Electric potential is given by,
$$V = \int\limits_L^{2L} {{{kdq} \over x}} $$
$$ = {{2L} \over L}{1 \over {4\pi {\varepsilon _0}}}{{\left( {{q \over L}} \right)dx} \over x}$$
$$ = {q \over {4\pi {\varepsilon _0}L}}\ln \left( 2 \right)$$
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