---
title: "Two point charges, \\(+q\\) and \\(-q\\), are fixed on the \\(x\\)-axis at positions \\(x = +\\dfrac{d}{2}\\) and \\(x = -\\dfrac{d}{2}\\), respectively. A point \\(P\\) is located on the positive \\(x\\)-axis at position \\(x = r\\), where \\(r > \\dfrac{d}{2}\\). What is the exact electric potential at point \\(P\\) due to the two charges, relative to zero potential at infinity?"
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url: "https://nerd-notes.com/ubq/118064/"
date_modified: "2026-08-04T08:04:57+00:00"
---

# Two point charges, \(+q\) and \(-q\), are fixed on the \(x\)-axis at positions \(x = +\dfrac{d}{2}\) and \(x = -\dfrac{d}{2}\), respectively. A point \(P\) is located on the positive \(x\)-axis at position \(x = r\), where \(r > \dfrac{d}{2}\). What is the exact electric potential at point \(P\) due to the two charges, relative to zero potential at infinity?

Two point charges, \(+q\) and \(-q\), are fixed on the \(x\)-axis at positions \(x = +\dfrac{d}{2}\) and \(x = -\dfrac{d}{2}\), respectively. A point \(P\) is located on the positive \(x\)-axis at position \(x = r\), where \(r > \dfrac{d}{2}\). What is the exact electric potential at point \(P\) due to the two charges, relative to zero potential at infinity?

![A horizontal axis labeled x contains an origin marked by a tick line and label O. Exactly two point charges are rendered as small filled circles on the axis. To the left of the origin at x = -d/2 is a circle labeled -q. To the right of the origin at x = +d/2 is a circle labeled +q. A small dot labeled P is positioned on the axis to the right of +q at position x = r. A horizontal dashed dimension line below the axis extends from x = -d/2 to x = +d/2 and is labeled d. Another horizontal dashed dimension line above the axis extends from the origin O to point P and is labeled r. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830697-iEPUyp.jpg)

- **A.** \(\dfrac{qd}{2\pi\varepsilon_0 r^2}\)
- **B.** \(\dfrac{qd}{4\pi\varepsilon_0 \left(r^2 - \dfrac{d^2}{4}\right)}\)
- **C.** \(\dfrac{2qr}{4\pi\varepsilon_0 \left(r^2 - \dfrac{d^2}{4}\right)}\)
- **D.** \(\dfrac{qd}{4\pi\varepsilon_0 \left(r - \dfrac{d}{2}\right)^2}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118064/*