---
title: "An isolated metallic pear-shaped conductor carries a net positive charge in electrostatic equilibrium. Point \\(P\\) is located at the narrow, sharply curved end of the conductor, and Point \\(R\\) is located at the wide, gently rounded end. Which of the following claims correctly compares the surface charge densities \\(\\sigma_P\\) and \\(\\sigma_R\\) at these points and provides the correct physical justification?"
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url: "https://nerd-notes.com/ubq/118199/"
date_modified: "2026-08-04T08:08:07+00:00"
---

# An isolated metallic pear-shaped conductor carries a net positive charge in electrostatic equilibrium. Point \(P\) is located at the narrow, sharply curved end of the conductor, and Point \(R\) is located at the wide, gently rounded end. Which of the following claims correctly compares the surface charge densities \(\sigma_P\) and \(\sigma_R\) at these points and provides the correct physical justification?

An isolated metallic pear-shaped conductor carries a net positive charge in electrostatic equilibrium. Point \(P\) is located at the narrow, sharply curved end of the conductor, and Point \(R\) is located at the wide, gently rounded end. Which of the following claims correctly compares the surface charge densities \(\sigma_P\) and \(\sigma_R\) at these points and provides the correct physical justification?

![A single continuous line forming an asymmetric, smooth pear-shaped outer boundary on a plain white background. The left portion of the boundary is wide with a gentle curvature, and the right portion tapers smoothly into a narrower region with a sharp curvature. A solid black dot on the outermost edge of the sharp right tip is labeled P. A second solid black dot on the outermost edge of the broad left end is labeled R. The interior of the shape is unshaded white. No arrows, axes, charge symbols, or field lines are included anywhere in the diagram. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1785830886-EVNtcE.jpg)

- **A.** \(\sigma_P < \sigma_R\), because positive charge carriers repel each other more strongly at the narrow end and are pushed toward the wider region.
- **B.** \(\sigma_P > \sigma_R\), because the entire conductor is an equipotential surface, which requires a higher surface charge density where the radius of curvature is smaller.
- **C.** \(\sigma_P > \sigma_R\), because the electric potential at point \(P\) is greater than the electric potential at point \(R\).
- **D.** \(\sigma_P = \sigma_R\), because net charge distributes uniformly over the outer surface of any isolated conductor in electrostatic equilibrium.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/118199/*