---
title: "An infinitely long, thin cylindrical shell of radius \\(R\\) carries a uniform positive linear charge density \\(\\lambda\\). Point 1 is located at a radial distance \\(r_1 = \\dfrac{R}{2}\\) from the central axis of the cylinder, and Point 2 is located at a radial distance \\(r_2 = 3R\\) from the central axis. What is the electric potential difference \\(V_1 – V_2\\) between Point 1 and Point 2?"
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url: "https://nerd-notes.com/ubq/118080/"
date_modified: "2026-08-04T08:05:01+00:00"
---

# An infinitely long, thin cylindrical shell of radius \(R\) carries a uniform positive linear charge density \(\lambda\). Point 1 is located at a radial distance \(r_1 = \dfrac{R}{2}\) from the central axis of the cylinder, and Point 2 is located at a radial distance \(r_2 = 3R\) from the central axis. What is the electric potential difference \(V_1 – V_2\) between Point 1 and Point 2?

An infinitely long, thin cylindrical shell of radius \(R\) carries a uniform positive linear charge density \(\lambda\). Point 1 is located at a radial distance \(r_1 = \dfrac{R}{2}\) from the central axis of the cylinder, and Point 2 is located at a radial distance \(r_2 = 3R\) from the central axis. What is the electric potential difference \(V_1 - V_2\) between Point 1 and Point 2?

![A cross-sectional view of an infinitely long thin cylindrical shell centered on a central axis dot labeled O. The shell is depicted as a dashed circular ring of radius R. A point labeled 1 sits on the horizontal axis to the right of O at radial distance R/2, inside the ring. A point labeled 2 sits on the horizontal axis to the right of O at radial distance 3R, outside the ring. A thin line with arrowheads extends from O to the ring labeled R. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830701-eEeZWT.jpg)

- **A.** \(\dfrac{\lambda \ln(3/2)}{2\pi \varepsilon_0}\)
- **B.** \(\dfrac{\lambda \ln 3}{2\pi \varepsilon_0}\)
- **C.** \(\dfrac{\lambda \ln 6}{2\pi \varepsilon_0}\)
- **D.** \(\dfrac{\lambda \ln 3}{\pi \varepsilon_0}\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118080/*