---
title: "An isolated positive point charge \\(+Q\\) is fixed in space. Point \\(P\\) is located a distance \\(r\\) from the charge, with the electric potential defined to be zero at infinitely far distances (\\(V(\\infty) = 0\\)). What is the electric potential \\(V\\) at point \\(P\\), and what is the sign of the work done by the electric field as a small positive test charge \\(+q_0\\) is moved from infinity to point \\(P\\)?"
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/118036/"
date_modified: "2026-08-04T08:04:36+00:00"
---

# An isolated positive point charge \(+Q\) is fixed in space. Point \(P\) is located a distance \(r\) from the charge, with the electric potential defined to be zero at infinitely far distances (\(V(\infty) = 0\)). What is the electric potential \(V\) at point \(P\), and what is the sign of the work done by the electric field as a small positive test charge \(+q_0\) is moved from infinity to point \(P\)?

An isolated positive point charge \(+Q\) is fixed in space. Point \(P\) is located a distance \(r\) from the charge, with the electric potential defined to be zero at infinitely far distances (\(V(\infty) = 0\)). What is the electric potential \(V\) at point \(P\), and what is the sign of the work done by the electric field as a small positive test charge \(+q_0\) is moved from infinity to point \(P\)?

- **A.** Electric Potential: \(+\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\) | Work Done by Field: Positive
- **B.** Electric Potential: \(+\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\) | Work Done by Field: Negative
- **C.** Electric Potential: \(+\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\) | Work Done by Field: Positive
- **D.** Electric Potential: \(+\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}\) | Work Done by Field: Negative

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118036/*
