A charge +q is fixed at each of the points $x = x_{0}, x = 3 x_{0}, x = 5 x_{0}$ ,….. upto infinity on the x-axis and a charge –q is fixed at each of points $x = 2 x_{0}, x = 4 x_{0}, x = 6 x_{0}$ , …. upto infinity. Here x₀ is a positive constant. The potential at the origin due to the above system of charges is

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ZERO

$\frac{q}{π ε_{0} x_{0} \log_{e} 2}$

$\infty$

$\frac{q \log_{e} 2}{4 π ε_{0} x_{0}}$

answer is D.

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Detailed Solution

$V = \frac{1}{4 π ε_{0}} \frac{q}{r}$

$V = \frac{q}{4 π ϵ_{0} x_{0}} [1 + \frac{1}{3} + \frac{1}{5} + \dots] - \frac{q}{4 π ϵ_{0} x_{0}} [\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots]$

$\begin{array}{l} = \frac{q}{4 π ϵ_{0} x_{0}} [1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots] \\ = \frac{q}{4 π ϵ_{0} x_{0}} \log_{e} (1 + 1) using \log_{e} (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots] \\ = \frac{q}{4 π ϵ_{0} x_{0}} \log_{e} 2 \end{array}$