A charge q is fixed at each of the points
$\large x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$
..... infinite, on the x-axis and a charge -q is fixed at each of the points
$\large x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$
..... infinite. Here $x_{0}$ is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be
$\large Q/(4\pi {\varepsilon _0}r)$
. Then, the potential at the origin due to the above system of charges is

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$\large \frac{q}{{8\pi {\varepsilon _0}{x_0}\ln 2}}$

$\large \infty \$

$\large \frac{{q\ln 2}}{{4\pi {\varepsilon _0}{x_0}}}$

answer is D.

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Detailed Solution

$\large V = \frac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {1 + \frac{1}{3} + \frac{1}{5} + ...} \right] - \frac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ...} \right]$
$\large = \frac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ....} \right] = \frac{q}{{4\pi {\varepsilon _0}{x_0}}}{\log _e}2$

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