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(D) The potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.For th

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$\frac{q \log_e 2}{4\pi\varepsilon_0 x_0}$

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(D) The potential $V$ at the origin due to a system of point charges is given by the sum $V = \sum \frac{1}{4\pi\varepsilon_0} \frac{q_i}{r_i}$.For the given system,the charges $+q$ are at $x = x_0, 3x_0, 5x_0, \dots$ and charges $-q$ are at $x = 2x_0, 4x_0, 6x_0, \dots$.The potential at the origin is:$V = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{x_0} - \frac{1}{2x_0} + \frac{1}{3x_0} - \frac{1}{4x_0} + \dots \right)$$V = \frac{q}{4\pi\varepsilon_0 x_0} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \right)$Using the Taylor series expansion for $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$,for $x=1$,we get $\log_e(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$.Substituting this into the expression for $V$:$V = \frac{q}{4\pi\varepsilon_0 x_0} \log_e 2$.

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