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(D) The electric 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

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

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

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