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A charge $ + q$ is fixed at each of the points $x = {x_0},\,x = 3{x_0},\,x = 5{x_0}$..... $\infty$, on the $x - $axis and a charge $ - q$ is fixed at each of the points $x = 2{x_0},\,x = 4{x_0},x = 6{x_0}$,..... $\infty$. 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 $Q/(4\pi {\varepsilon _0}r)$. Then, the potential at the origin due to the above system of charges is
$0$
$\frac{q}{{8\pi {\varepsilon _0}{x_0}\ln 2}}$
$\infty $
$\frac{{q\ln 2}}{{4\pi {\varepsilon _0}{x_0}}}$
Solution
(d)$V = \frac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {1 + \frac{1}{3} + \frac{1}{5} + …} \right] – \frac{q}{{4\pi {\varepsilon _0}{x_0}}}\left[ {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + …} \right]$
$ = \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$
