---
title: "An infinite array of point charges is fixed along the positive \\(x\\)-axis. Positive point charges \\(+q\\) are located at \\(x = a, 3a, 5a, \\dots\\) and negative point charges \\(-q\\) are located at \\(x = 2a, 4a, 6a, \\dots\\), where \\(a > 0\\). Given that \\(1 – \\dfrac{1}{2} + \\dfrac{1}{3} – \\dfrac{1}{4} + \\dots = \\ln 2\\), what is the electric potential at the origin \\(x = 0\\) due to this infinite collection of charges?"
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url: "https://nerd-notes.com/ubq/118152/"
date_modified: "2026-08-04T08:05:32+00:00"
---

# An infinite array of point charges is fixed along the positive \(x\)-axis. Positive point charges \(+q\) are located at \(x = a, 3a, 5a, \dots\) and negative point charges \(-q\) are located at \(x = 2a, 4a, 6a, \dots\), where \(a > 0\). Given that \(1 – \dfrac{1}{2} + \dfrac{1}{3} – \dfrac{1}{4} + \dots = \ln 2\), what is the electric potential at the origin \(x = 0\) due to this infinite collection of charges?

An infinite array of point charges is fixed along the positive \(x\)-axis. Positive point charges \(+q\) are located at \(x = a, 3a, 5a, \dots\) and negative point charges \(-q\) are located at \(x = 2a, 4a, 6a, \dots\), where \(a > 0\). Given that \(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dots = \ln 2\), what is the electric potential at the origin \(x = 0\) due to this infinite collection of charges?

![A horizontal line represents the positive x-axis, starting from a tick mark at x = 0 labeled as the origin. Tick marks are placed at x = a, 2a, 3a, 4a, and 5a. A positive point charge labeled +q is located at x = a, x = 3a, and x = 5a. A negative point charge labeled -q is located at x = 2a and x = 4a. Three dots extend to the right of x = 5a to indicate that the pattern continues infinitely. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830731-PLt8r9.jpg)

- **A.** \(\dfrac{q \ln 2}{4\pi\varepsilon_0 a}\)
- **B.** \(\dfrac{q}{4\pi\varepsilon_0 a}\)
- **C.** \(\dfrac{q \ln 2}{2\pi\varepsilon_0 a}\)
- **D.** \(\dfrac{q}{8\pi\varepsilon_0 a}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118152/*