---
title: "A solid insulating sphere of radius \\(R\\) has a non-uniform volume charge density \\(\\rho(r)\\) and a total net charge of \\(-Q\\). A small particle of mass \\(m\\) and positive charge \\(+q\\) is released from rest at a distance very far from the sphere (\\(r \\to \\infty\\)). Assuming gravitational interactions are negligible, what is the speed of the particle when it reaches a distance \\(r = 2R\\) from the center of the sphere?"
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url: "https://nerd-notes.com/ubq/118126/"
date_modified: "2026-08-04T08:05:18+00:00"
---

# A solid insulating sphere of radius \(R\) has a non-uniform volume charge density \(\rho(r)\) and a total net charge of \(-Q\). A small particle of mass \(m\) and positive charge \(+q\) is released from rest at a distance very far from the sphere (\(r \to \infty\)). Assuming gravitational interactions are negligible, what is the speed of the particle when it reaches a distance \(r = 2R\) from the center of the sphere?

A solid insulating sphere of radius \(R\) has a non-uniform volume charge density \(\rho(r)\) and a total net charge of \(-Q\). A small particle of mass \(m\) and positive charge \(+q\) is released from rest at a distance very far from the sphere (\(r \to \infty\)). Assuming gravitational interactions are negligible, what is the speed of the particle when it reaches a distance \(r = 2R\) from the center of the sphere?

![A solid circle representing a sphere of radius R is centered at the origin, shaded light gray and labeled with total charge -Q. A dashed circular arc of radius 2R is drawn concentric with the sphere, labeled 2R at the top. On the horizontal axis to the right, a small black circle represents a particle labeled +q with mass m, with a horizontal velocity arrow pointing to the left toward the sphere. A horizontal axis line connects the center of the sphere to the particle. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830717-HreaYC.jpg)

- **A.** \(\sqrt{\dfrac{qQ}{8\pi\varepsilon_0 m R}}\)
- **B.** \(\sqrt{\dfrac{qQ}{4\pi\varepsilon_0 m R}}\)
- **C.** \(\sqrt{\dfrac{qQ}{2\pi\varepsilon_0 m R}}\)
- **D.** \(\sqrt{\dfrac{qQ}{\pi\varepsilon_0 m R}}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118126/*